Éléments de Géométrie Algébrique — English

§4. Base field change in algebraic preschemes

4.1. Dimension of algebraic preschemes

We shall develop in §5 the general theory of the dimension of preschemes; but the theory of the dimension of algebraic preschemes can be developed in a more elementary way, and since it presents in addition many special features, we give here a rapid exposition of it, independent of the general theory.

If $K$ is a field and $L$ an extension of $K$, we shall denote by $deg.t{r}_{K}L$ the transcendence degree of $L$ over $K$.

Definition (4.1.1).

Let $k$ be a field, $X$ a prescheme locally of finite type over $k$. We call dimension of $X$ the number

  dim(X) = sup_{x} deg.tr_k k(x)                                    (4.1.1.1)

where $x$ runs through the set of maximal points of $X$.

We shall see in §5 (5.2.2) that the definition (4.1.1) depends only apparently on the base field $k$ over which $X$ is locally of finite type, and that the number $\dim (X)$ so defined coincides with the dimension of the underlying space $X$ (defined in (0, 14.1.2)). One evidently has dim(X) = dim(X_red).

We note that each $k(x)$ is an extension of finite type of $k$ (I, 6.3.3) hence has finite transcendence degree over $k$. If $X$ is of finite type over $k$ and non-empty, it is Noetherian (I, 6.3.7), hence has only a finite number of irreducible components, and consequently $\dim (X)$ is finite and $\ge 0$; definition (4.1.1) gives

$$\dim (\varnothing )=-\infty .$$

It is clear that, if one denotes by $(X_{\alpha })$ the family of reduced closed sub-preschemes of $X$ having as underlying spaces the irreducible components of $X$ (I, 5.2.1), one has

  dim(X) = sup_α dim(X_α).                                          (4.1.1.2)

This therefore reduces the calculation of the dimension to the case of integral preschemes (locally of finite type over $k$). Finally, one evidently has

$$\dim (X)=\dim (U)(\mathrm{4.1.1.3})$$

for every sub-prescheme induced on an everywhere dense open set $U$ of $X$; this finally reduces the notion of dimension to the case of an affine scheme of finite type over $k$.

Theorem (4.1.2).

Let $X$, $Y$ be two preschemes locally of finite type over a field $k$, $f:X\to Y$ a $k$-morphism.

(i) If $f$ is quasi-compact and dominant, one has $\dim (Y)\le \dim (X)$.

(ii) If $f$ is quasi-finite, one has $\dim (X)\le \dim (Y)$.

(iii) Suppose $X$ is of finite type over $k$. For $\dim (X)\le n$ (resp. $\dim (X)\le n$, $\dim (X)=n$), it is necessary and sufficient that there exist a dense open set $U$ in $X$, and a $k$-morphism $g:U\to V_n^{(k)}$ ($=\mathrm{Spec}(k[T_1,\cdots ,T_n])$, which we shall also denote by $V_n^{(k)}$) which is surjective (resp. finite, resp. finite and surjective). If $X$ is an affine scheme, one can take $U=X$.

(i) Let $y$ be a maximal point of $Y$; one knows that $f^{-1}(y)$ contains a maximal point $x$ of $X$ (1.1.5), hence $k(x)$ is an extension of $k(y)$, whence the inequality deg.tr_k k(y) ≤ deg.tr_k k(x) ≤ dim(X), which proves (i).

(ii) If $x$ is a maximal point of $X$, $k(x)$ is a finite extension of $k(f(x))$ (II, 6.2.2), hence has the same transcendence degree over $k$. Considering the reduced closed sub-prescheme of $Y$ having $\overline{f(x)}$ as underlying space, one sees that one is reduced to proving the

Corollary (4.1.2.1).

Let $Y$ be a $k$-prescheme locally of finite type; for every sub-prescheme $Z$ of $Y$, one has $\dim (Z)\le \dim (Y)$. Suppose in addition that the irreducible components of $Y$ all have the same dimension; then, for $Z$ to be rare in $Y$, it is necessary and sufficient that $\dim (Z)<\dim (Y)$.

Indeed, let $z$ be a maximal point of $Z$; by considering one of the reduced closed sub-preschemes of $Y$ having as underlying space one of the irreducible components of $Y$ containing $z$, one reduces to the case where $Y$ is integral with generic point $y$, then, by considering in $Y$ an affine open set containing $z$, to the case where $Y$ is affine and of finite type over $k$, say $Y=\mathrm{Spec}(A)$, and $Z$ closed in $Y$, say $Z=\mathrm{Spec}(A/\mathfrak{a})$ where $\mathfrak{a}$ is an ideal of the integral ring $A$, distinct from $A$. By virtue of the normalization lemma (Bourbaki, Alg. comm., chap. V, §3, n° 1, th. 1) there exists a finite sequence $(x_i{)}_{1\le i\le n}$ of elements of $A$, algebraically independent over $k$, such that $A$ is integral over the ring $B=k[x_1,\cdots ,x_n]$ and that $\mathfrak{a}\cap B$ is generated by a subfamily $(x_i{)}_{1\le i\le p}$ (possibly empty) of $(x_i{)}_{1\le i\le n}$. Let $g:Y\to \mathrm{Spec}(B)=V_n^{(k)}$ be the finite dominant morphism corresponding to the canonical injection $B\to A$; since $C=B/(\mathfrak{a}\cap B)$ is isomorphic to $k[x_{p+1},\cdots ,x_n]$, $g$ induces on $Z$ a finite dominant morphism $Z\to V_{n-p}^{(k)}$; one has therefore $deg.t{r}_{k}k(y)=n$ and $deg.t{r}_{k}k(z)=n-p$, since $k(y)$ (resp. $k(z)$) is a finite extension of $k(g(y))$ (resp. $k(g(z))$). This proves the first assertion of (4.1.2.1). Furthermore, if $p=0$, one necessarily has $z=y$, since $y$ is the only point of $Y$ whose image by $g$ is the generic point of $V_n^{(k)}$; in this case, $Z$ therefore contains a non-empty open set of $Y$. If on the contrary $p>0$, one necessarily has $z\ne y$ and $\overline{z}$ is therefore rare in $\overline{y}$, which completes the proof of (4.1.2.1).

(iii) The fact that the conditions stated are sufficient follows at once from (i) and (ii) and from (1.5.4, (v)). To prove that they are necessary, one may consider a dense open set $U$ of $X$, union of pairwise disjoint and irreducible affine open sets $U_i$ $(1\le i\le m)$, each of which contains one of the maximal points of $X$ $(0_I,\mathrm{2.1.6})$; one may in addition suppose $X$ reduced and if $X$ is affine, one may of course take $U=X$. The same

reasoning as in (4.1.2.1) shows that if $\dim (U_i)=n_i$, there exists a $k$-morphism $g_i:U_i\to V_{n_i}^{(k)}$ which is finite and dominant, hence surjective (II, 6.1.10). Let $n=\dim (X)={\sup }_i{n}_i$; for each $n_i$, there is then a morphism $h_i:V_{n_i}^{(k)}\to V_n^{(k)}$ which is a closed immersion corresponding to the canonical homomorphism $k[T_1,\cdots ,T_n]\to k[T_1,\cdots ,T_{n_i}]$, and which is the identity for $n_i=n$. Since $U$ is the sum of the $U_i$, one takes for $g$ the morphism which on each $U_i$ coincides with $h_i\circ g_i$; it is evidently finite and surjective. When $n^{\prime }\ge n$ one will have a finite morphism $U\to V_{n^{\prime }}^{(k)}$ by composing the canonical closed immersion $V_n^{(k)}\to V_{n^{\prime }}^{(k)}$ with $g$; when $n^{\prime }\le n$, one composes likewise with $g$ the canonical morphism $p:V_n^{(k)}\to V_{n^{\prime }}^{(k)}$ corresponding to the canonical injection $k[T_1,\cdots ,T_{n^{\prime }}]\to k[T_1,\cdots ,T_n]$, noting that $p$ is faithfully flat, hence surjective.

Remark (4.1.3).

The corollary (4.1.2.1) shows that in formula (4.1.1.1), one may suppose that $x$ runs through the set of all points of $X$.

Corollary (4.1.4).

Let $X$ be a prescheme locally of finite type over a field $k$, $K$ an extension of $k$; then $\dim (X{\otimes }_{k}K)=\dim (X)$.

One may evidently restrict to the case where $X$ is of finite type over $k$; then the morphism $u:X{\otimes }_{k}K\to X$ is faithfully flat (2.2.13, (i)), hence if $U$ is an everywhere dense open set in $X$, $u^{-1}(U)=U{\otimes }_{k}K$ is dense in $X{\otimes }_{k}K$ (2.3.10); if $g:U\to V_n^{(k)}$ is finite and surjective, so is $g_{(K)}:U_{(K)}\to V_n^{(K)}$ (I, 3.5.2 and II, 6.1.5), whence the corollary.

Corollary (4.1.5).

Let $X$ and $Y$ be two preschemes locally of finite type over a field $k$; then dim(X ×_k Y) = dim(X) + dim(Y).

It suffices to prove that if $U$ (resp. $V$) is an affine neighbourhood of a point $x$ (resp. $y$) of $X$ (resp. $Y$) such that $deg.t{r}_{k}k(x)=m=\dim (U)$ and $deg.t{r}_{k}k(y)=n=\dim (V)$, then $\dim (U{\times }_{k}V)=m+n$, in other words, one may, by virtue of (4.1.2), suppose that there exist finite surjective $k$-morphisms $f:X\to V_m^{(k)}$, $g:Y\to V_n^{(k)}$; then $f\times g:X{\times }_{k}Y\to V_{m+n}^{(k)}$ is finite and surjective (I, 3.5.2 and II, 6.1.5), whence the corollary.

4.2. Associated prime cycles on algebraic preschemes

Proposition (4.2.1).

Let $K$ and $L$ be two extensions of a field $k$, such that $K{\otimes }_{k}L$ is Noetherian. Then the prime ideals associated to $K{\otimes }_{k}L$ are minimal, and if $E$ is the residue field of the local ring of such an ideal, one has

  deg.tr_K E = deg.tr_k L,    deg.tr_L E = deg.tr_k K               (4.2.1.1)

whence

  deg.tr_k E = deg.tr_k K + deg.tr_k L.                             (4.2.1.2)

One knows that $K$ is an algebraic extension of a purely transcendental extension $K^{\prime }=k(\mathbf{t})$, where $\mathbf{t}=(t_{\iota }{)}_{\iota \in I}$ is a family of indeterminates; the ring $k[\mathbf{t}]{\otimes }_{k}L=L[\mathbf{t}]$ is integral, hence so is $K^{\prime }{\otimes }_{k}L$, which is a ring of fractions of it, and the field of fractions of $K^{\prime }{\otimes }_{k}L$ is $L(\mathbf{t})$; one has the commutative diagram of canonical homomorphisms

  K  ──→  K ⊗_k L  ──→  K ⊗_{K'} L(𝐭)
  ↑        ↑              ↑
  K'  ──→  K' ⊗_k L  ──→  L(𝐭)        (4.2.1.3)
  ↑        ↑
  k  ───→  L

Since $K$ is faithfully flat over $K^{\prime }$, $K{\otimes }_{k}L=K{\otimes }_{K^{\prime }}(K^{\prime }{\otimes }_{k}L)$ is faithfully flat over $K^{\prime }{\otimes }_{k}L$, hence $K^{\prime }{\otimes }_{k}L$ is Noetherian $(0_I,\mathrm{6.5.2})$; in addition $K^{\prime }{\otimes }_{k}L$ is identified with a sub-ring of $K{\otimes }_{k}L$; the trace on $K^{\prime }{\otimes }_{k}L$ of a prime ideal $\mathfrak{p}$ associated to $K{\otimes }_{k}L$ is the ideal 0, an element $\ne 0$ of $K^{\prime }{\otimes }_{k}L$ not being a zero-divisor in $K{\otimes }_{k}L$ (0_I, 6.3.4 and Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 3 of prop. 2). As in addition $K$ is algebraic over $K^{\prime }$, $K{\otimes }_{k}L$ is an integral algebra over $K^{\prime }{\otimes }_{k}L$, and the prime ideals of $K{\otimes }_{k}L$ inducing 0 on $K^{\prime }{\otimes }_{k}L$ are necessarily without mutual inclusion relation (Bourbaki, Alg. comm., chap. V, §2, n° 1, cor. 1 of prop. 1); this proves the first assertion (Bourbaki, Alg. comm., chap. IV, §1, n° 3, cor. 1 of prop. 7). Moreover, the residue field $E$ of $\mathfrak{p}$ is algebraic over the residue field of the ideal (0) of $K^{\prime }{\otimes }_{k}L$, that is $L(\mathbf{t})$; hence deg.tr_K E = deg.tr_K L(𝐭) = Card(I) = deg.tr_k K, in other words one has the first relation (4.2.1.1); exchanging the roles of $K$ and $L$, one has the second relation (4.2.1.1), whence (4.2.1.2).

Corollary (4.2.2).

Under the hypotheses of (3.3.6), if the preschemes $T_{x,y}$ are locally Noetherian, they have no embedded associated prime cycle.

Corollary (4.2.3).

Under the hypotheses of (3.3.6) (resp. (3.3.7)), if the $T_{x,y}$ are locally Noetherian and if $\mathcal{F}$ and the ${\mathcal{G}}_s$ (for $s\in S$) (resp. $\mathcal{F}$ and $\mathcal{G}$) are without embedded associated prime cycle, the same holds for $\mathcal{F}{\otimes }_{\mathcal{S}}\mathcal{G}$.

This results from (4.2.2), (3.3.2) and from the proof of (3.3.6). In particular, since every prescheme over a field $k$ is flat over $k$, we have thus proved assertion (i) of the

Proposition (4.2.4).

Let $k$ be a field, $X$ and $Y$ two locally Noetherian $k$-preschemes such that $X{\times }_{k}Y$ is locally Noetherian. Suppose in addition that $X$ and $Y$ are integral. Then:

(i) $X{\times }_{k}Y$ is without embedded associated prime cycle; each of the irreducible components of $X{\times }_{k}Y$ dominates $X$ and $Y$, and the set of these components is in bijective correspondence with the set of irreducible components of $\mathrm{Spec}(R(X){\otimes }_{k}R(Y))$ (in other words, the set of minimal prime ideals of $R(X){\otimes }_{k}R(Y)$), where $R(X)$ and $R(Y)$ are the fields of rational functions of $X$ and $Y$ respectively.

(ii) If a maximal point $z$ of $X{\times }_{k}Y$ is identified with a minimal prime ideal $\mathfrak{p}$ of $R(X){\otimes }_{k}R(Y)$, the local ring ${\mathcal{O}}_{X{\times }_{k}Y,z}$ is isomorphic to the ring of fractions $(R(X){\otimes }_{k}R(Y){)}_{\mathfrak{p}}$. In particular, if $R(X)$ or $R(Y)$ is separable over $k$, $X{\times }_{k}Y$ is reduced.

(iii) If in addition $X$ and $Y$ are locally of finite type over $k$, every irreducible component of $X{\times }_{k}Y$ has dimension $\dim (X)+\dim (Y)$.

Assertion (iii) follows from (4.2.1.2), taking into account (i) and (ii). To prove (ii), one may evidently restrict to the case where $X=\mathrm{Spec}(A)$, $Y=\mathrm{Spec}(B)$ are affine, $A$ and $B$ being thus integral rings of respective fields of fractions $K=R(X)$ and $L=R(Y)$; assertion (i) shows that every minimal prime ideal $\mathfrak{q}$ of $A{\otimes }_{k}B$ is the trace on $A{\otimes }_{k}B$ of a minimal prime ideal $\mathfrak{p}$ of $K{\otimes }_{k}L$; since $K{\otimes }_{k}L$ is a ring of fractions of $A{\otimes }_{k}B$, the isomorphy of the rings $(A{\otimes }_{k}B{)}_{\mathfrak{q}}$ and $(K{\otimes }_{k}L{)}_{\mathfrak{p}}$ follows from $(0_I,\mathrm{1.2.5})$.

Finally, if for example $R(Y)$ is separable over $k$, one knows that the ring $R(X){\otimes }_{k}R(Y)$ is reduced (Bourbaki, Alg., chap. VIII, §7, n° 3, th. 1); the same therefore holds for the local rings of the maximal points of $X{\times }_{k}Y$. One deduces from this that $X{\times }_{k}Y$ is reduced (3.2.1).

Proposition (4.2.5).

Let $k$ be a field, $X$ and $Y$ locally Noetherian $k$-preschemes, $\mathcal{F}$ (resp. $\mathcal{G}$) a quasi-coherent ${\mathcal{O}}_X$-Module (resp. ${\mathcal{O}}_Y$-Module). Let $(Z_{\lambda }^{\prime })$ (resp. $(Z_{\mu }^{\prime \prime })$) be the family of associated prime cycles of $\mathcal{F}$ (resp. $\mathcal{G}$), and let us again denote by $Z_{\lambda }^{\prime }$ (resp. $Z_{\mu }^{\prime \prime }$) the reduced sub-prescheme of $X$ (resp. $Y$) having $Z_{\lambda }^{\prime }$ (resp. $Z_{\mu }^{\prime \prime }$) as underlying space. Then, if $Z_{\lambda }^{\prime }{\times }_k{Z}_{\mu }^{\prime \prime }$ is locally Noetherian, the irreducible components $Z_{\lambda ,\mu ,\nu }$ of $Z_{\lambda }^{\prime }{\times }_k{Z}_{\mu }^{\prime \prime }$ dominate $Z_{\lambda }^{\prime }$ and $Z_{\mu }^{\prime \prime }$, and $(Z_{\lambda ,\mu ,\nu })$ is the family of distinct associated prime cycles of $\mathcal{F}{\otimes }_k\mathcal{G}$.

It suffices to apply (4.2.4) to the product $\mathrm{Spec}(k(x)){\times }_k\mathrm{Spec}(k(y))$.

In particular:

Corollary (4.2.6).

Let $k$ be a field, $X$, $Y$ two locally Noetherian $k$-preschemes such that $X{\times }_{k}Y$ is locally Noetherian. Let $(Z_{\lambda }^{\prime })$ (resp. $(Z_{\mu }^{\prime \prime })$) be the family of reduced sub-preschemes of $X$ (resp. $Y$) having as underlying spaces the irreducible components of $X$ (resp. $Y$). Then the irreducible components $Z_{\lambda ,\mu ,\nu }$ of $Z_{\lambda }^{\prime }{\times }_k{Z}_{\mu }^{\prime \prime }$ dominate $Z_{\lambda }^{\prime }$ and $Z_{\mu }^{\prime \prime }$, and $(Z_{\lambda ,\mu ,\nu })$ is the family of irreducible components of $X{\times }_{k}Y$.

Indeed, one may restrict to the case where $X$ and $Y$ are reduced (I, 5.1.8); the irreducible components of $X$ (resp. $Y$) are then the associated prime cycles of ${\mathcal{O}}_X$ (resp. ${\mathcal{O}}_Y$) (3.2.1). Apply (4.2.5) to $\mathcal{F}={\mathcal{O}}_X$ and $\mathcal{G}={\mathcal{O}}_Y$, noting that by definition ${\mathcal{O}}_X{\otimes }_k{\mathcal{O}}_Y={\mathcal{O}}_{X{\times }_{k}Y}$; the corollary follows from the fact that ${\mathcal{O}}_X{\otimes }_k{\mathcal{O}}_Y$ has no embedded associated prime cycles, since this holds for ${\mathcal{O}}_X$ and ${\mathcal{O}}_Y$ by hypothesis (4.2.3).

We apply the preceding results to the case where $Y$ is the spectrum of an extension $K$ of $k$:

Proposition (4.2.7).

Let $k$ be a field, $X$ a $k$-prescheme, $K$ an extension of $k$ such that $X{\otimes }_{k}K$ is locally Noetherian, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module, $x^{\prime }$ a point of $X{\otimes }_{k}K$, $x$ its image in $X$.

(i) Let $(Z_{\lambda }^{\prime })$ be the family of reduced sub-preschemes of $X$ having as underlying spaces the associated prime cycles of $\mathcal{F}$; then the irreducible components $Z_{\lambda }$ of $Z_{\lambda }^{\prime }{\otimes }_{k}K$ are the associated prime cycles of $\mathcal{F}{\otimes }_{k}K$, and $Z_{\lambda }$ dominates $Z_{\lambda }^{\prime }$; in addition, for a $Z_{\lambda }$ to be embedded, it is necessary and sufficient that $Z_{\lambda }^{\prime }$ be so.

(ii) For $x$ to belong to an embedded associated prime cycle of $\mathcal{F}$, it is necessary and sufficient that $x^{\prime }$ belong to an embedded associated prime cycle of $\mathcal{F}{\otimes }_{k}K$; for $\mathcal{F}$ to be without embedded associated prime cycle, it is necessary and sufficient that the same hold for $\mathcal{F}{\otimes }_{k}K$.

(iii) The indices $\lambda$ such that $x\in Z_{\lambda }^{\prime }$ are the same as those for which there exists an index $\mu$ such that $x^{\prime }\in Z_{\lambda ,\mu }$. In particular, if $x^{\prime }$ belongs to only one associated prime cycle of $\mathcal{F}{\otimes }_{k}K$, $x$ belongs to only one associated prime cycle of $\mathcal{F}$.

(iv) If $X$ is locally of finite type over $k$, one has $\dim (Z_{\lambda })=\dim (Z_{\lambda }^{\prime })$.

We note that the hypothesis implies that $X$ itself is locally Noetherian (2.2.13 and 2.2.14); assertion (i) follows from (4.2.5) and from the proof of (4.2.3), for $\mathcal{G}={\mathcal{O}}_Y$ with $Y=\mathrm{Spec}(K)$; (ii) and (iii) follow from (i) and from (2.3.5); finally, (iv) is a special case of (4.2.4, (iii)).

Corollary (4.2.8).

If $X$ is locally of finite type over $k$, the set of dimensions of the associated prime cycles is the same for $\mathcal{F}$ and $\mathcal{F}{\otimes }_{k}K$; the set of dimensions of the irreducible components is the same for $X$ and $X{\otimes }_{k}K$.

Proposition (4.2.9).

Suppose the hypotheses of (4.2.5) are satisfied, and suppose in addition that $\mathcal{F}$ and $\mathcal{G}$ are coherent. Let $({\mathcal{F}}_{\lambda }{)}_{\lambda \in L}$ and $({\mathcal{G}}_{\mu }{)}_{\mu \in M}$ be irredundant decompositions of $\mathcal{F}$ and $\mathcal{G}$ respectively; for every pair $(\lambda ,\mu )\in L\times M$, let $({\mathcal{H}}_{\lambda ,\mu ,\nu }{)}_{\nu \in S(\lambda ,\mu )}$ be a reduced irredundant decomposition of ${\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu }$, where $S(\lambda ,\mu )=Ass({\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu })$ (3.2.5). Then $({\mathcal{H}}_{\lambda ,\mu ,\nu })$ (for all triples $(\lambda ,\mu ,\nu )$) is an irredundant decomposition of $\mathcal{F}{\otimes }_k\mathcal{G}$; it is reduced if $({\mathcal{F}}_{\lambda })$ and $({\mathcal{G}}_{\mu })$ are.

Note that $\mathcal{F}{\otimes }_k\mathcal{G}$ and each of the ${\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu }$ are coherent, and each of the ${\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu }$ is identified with a quotient of $\mathcal{F}{\otimes }_k\mathcal{G}$; each of the ${\mathcal{H}}_{\lambda ,\mu ,\nu }$ is identified by definition with a coherent quotient of ${\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu }$, hence also with a coherent quotient of $\mathcal{F}{\otimes }_k\mathcal{G}$. The family of supports of the ${\mathcal{H}}_{\lambda ,\mu ,\nu }$ is locally finite, since $Supp({\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu })$ is contained in the underlying space of $Supp({\mathcal{F}}_{\lambda }){\times }_{k}Supp({\mathcal{G}}_{\mu })$ (where $Supp({\mathcal{F}}_{\lambda })$ and $Supp({\mathcal{G}}_{\mu })$ denote the corresponding reduced closed sub-preschemes), and for given $\lambda$ and $\mu$, the family of supports of the ${\mathcal{H}}_{\lambda ,\mu ,\nu }$ is locally finite by hypothesis; our assertion therefore follows from the fact that the families $(Supp({\mathcal{F}}_{\lambda }))$ and $(Supp({\mathcal{G}}_{\mu }))$ are locally finite. To prove that $({\mathcal{H}}_{\lambda ,\mu ,\nu })$ is an irredundant decomposition of $\mathcal{F}{\otimes }_k\mathcal{G}$, it suffices therefore to prove that the canonical homomorphism $\mathcal{F}{\otimes }_k\mathcal{G}\to {\oplus }_{\lambda ,\mu ,\nu }{\mathcal{H}}_{\lambda ,\mu ,\nu }$ is injective; now, it is composed of the homomorphisms $\mathcal{F}{\otimes }_k\mathcal{G}\to ({\oplus }_{\lambda }{\mathcal{F}}_{\lambda }){\otimes }_k({\oplus }_{\mu }{\mathcal{G}}_{\mu })\to {\oplus }_{\lambda ,\mu }{\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu }\to {\oplus }_{\lambda ,\mu ,\nu }{\mathcal{H}}_{\lambda ,\mu ,\nu }$; the last of these homomorphisms is injective by definition, and the same holds for the first, which is none other than the tensor product of the canonical injective homomorphisms $\mathcal{F}\to {\oplus }_{\lambda }{\mathcal{F}}_{\lambda }$, $\mathcal{G}\to {\oplus }_{\mu }{\mathcal{G}}_{\mu }$ (recall that $\mathcal{F}$ and $\mathcal{G}$ are flat over $k$). Finally, if $({\mathcal{F}}_{\lambda })$ and $({\mathcal{G}}_{\mu })$ are reduced, one may suppose that $L=Ass(\mathcal{F})$ and $M=Ass(\mathcal{G})$; the fact that the ${\mathcal{H}}_{\lambda ,\mu ,\nu }$ then form a reduced decomposition follows from the fact that the $Ass({\mathcal{F}}_{\lambda }{\otimes }_k{\mathcal{G}}_{\mu })$ form a partition of $Ass(\mathcal{F}{\otimes }_k\mathcal{G})$ by virtue of (4.2.2) (cf. (3.2.5)).

Corollary (4.2.10).

Under the hypotheses of (4.2.7), suppose in addition that $\mathcal{F}$ is coherent and let $({\mathcal{F}}_{\lambda }{)}_{\lambda \in L}$ be an irredundant decomposition of $\mathcal{F}$; for every $\lambda \in L$, let $({\mathcal{H}}_{\lambda ,\mu }{)}_{\mu \in Ass({\mathcal{F}}_{\lambda }{\otimes }_{k}K)}$ be a reduced irredundant decomposition of ${\mathcal{F}}_{\lambda }{\otimes }_{k}K$. Then $({\mathcal{H}}_{\lambda ,\mu })$ is an irredundant decomposition of $\mathcal{F}{\otimes }_{k}K$; it is reduced if $({\mathcal{F}}_{\lambda })$ is.

One applies (4.2.9) with $Y=\mathrm{Spec}(K)$, $\mathcal{G}={\mathcal{O}}_Y=K$, $M$ reduced to a single element.

4.3. Reminders on tensor products of fields

For the convenience of the reader, we recall here certain properties of tensor products of fields which we shall use in the following numbers; for the proofs, we refer to [1].

(4.3.1)

Recall that an extension $L$ of a field $k$ is called primary if the largest separable algebraic extension of $k$ contained in $L$ is $k$ itself.

Proposition (4.3.2).

Let $K$, $L$ be two extensions of a field $k$. If $L$ is a primary extension of $k$, then $\mathrm{Spec}(L{\otimes }_{k}K)$ is irreducible, and if $\zeta$ is its generic point, $k(\zeta )$ is a primary extension of $K$. Conversely, if for every finite separable extension $K$ of $k$, $\mathrm{Spec}(L{\otimes }_{k}K)$ is irreducible, $L$ is a primary extension of $k$.

See the proof in [1], pp. 14-03 to 14-06.

Corollary (4.3.3).

If $k$ is separably closed (i.e. if its algebraic closure is radicial over $k$), then, for any two extensions $K$, $L$ of $k$, $\mathrm{Spec}(L{\otimes }_{k}K)$ is irreducible, and conversely.

This follows at once from (4.3.2).

Corollary (4.3.4).

Let $L$ be an extension of a field $k$, $L_s$ the largest separable algebraic extension of $k$ contained in $L$ (sometimes called the separable algebraic closure of $k$ in $L$); suppose $L_s$ is of finite degree over $k$, and let $K$ be a Galois extension of $k$ containing $L_s$. Then $\mathrm{Spec}(L{\otimes }_{k}K)$ has $[L_s:k]$ irreducible components of which it is the sum, and the residue fields at the generic points of these components are primary extensions of $K$.

Indeed, one has $L{\otimes }_{k}K=L{\otimes }_{L_s}(L_s{\otimes }_{k}K)$ and one knows that $L_s{\otimes }_{k}K$ is isomorphic to a product of $[L_s:k]$ fields isomorphic to $K$ (Bourbaki, Alg., chap. VIII, §7, n° 3, cor. 2 of th. 1); hence $L{\otimes }_{k}K$ is isomorphic to a product of $[L_s:k]$ rings isomorphic to $L{\otimes }_{L_s}K$; since $L$ is a primary extension of $L_s$, the conclusion follows from (4.3.2).

Proposition (4.3.5).

Let $K$, $L$ be two extensions of a field $k$. If $L$ is a separable extension of $k$, the ring $L{\otimes }_{k}K$ is reduced. Conversely, if for every finite radicial extension $K$ of $k$, the ring $L{\otimes }_{k}K$ is reduced, $L$ is a separable extension of $k$.

For the proof, see Bourbaki, Alg., chap. VIII, §7, n° 3, th. 1.

Corollary (4.3.6).

If $k$ is perfect, then, for any two extensions $K$, $L$ of $k$, $K{\otimes }_{k}L$ is reduced, and conversely.

This follows at once from (4.3.5).

Corollary (4.3.7).

Let $L$ be a separable extension of $k$ and let $K$ be an extension of $k$ such that $K$ or $L$ is finite over $k$; then the residue fields of the semi-local ring $L{\otimes }_{k}K$ are separable extensions of $K$.

One knows in fact that $L{\otimes }_{k}K$ is the direct composite of these residue fields $E_i$; for every extension $K^{\prime }$ of $K$, $L{\otimes }_k{K}^{\prime }$ is therefore isomorphic to the direct composite of the rings $E_i{\otimes }_K{K}^{\prime }$; since $L{\otimes }_k{K}^{\prime }$ is reduced, the $E_i$ are separable over $K$ by virtue of (4.3.5).

Corollary (4.3.8).

If $K$ and $L$ are two finite separable extensions of $k$, $K{\otimes }_{k}L$ is the product of a finite number of separable extensions of $k$.

Proposition (4.3.9).

If $k$ is algebraically closed, then, for any two extensions $K$, $L$ of $k$, $L{\otimes }_{k}K$ is integral, and conversely.

This is a consequence of (4.3.3) and (4.3.6), a perfect and separably closed field being algebraically closed.

4.4. Irreducible preschemes and connected preschemes over an algebraically closed field

(4.4.1)

Let $k$ be a field, $X$ a $k$-prescheme, $K$ an extension of $k$. Since the morphism $\mathrm{Spec}(K)\to \mathrm{Spec}(k)$ is faithfully flat, quasi-compact, and universally open (2.4.9), the same holds for the projection morphism $p:X{\otimes }_{k}K\to X$ (2.2.13). It therefore follows from (2.3.5) that every irreducible component of $X{\otimes }_{k}K$ dominates an irreducible component of $X$, and (since $p$ is surjective) every irreducible component of $X$ is dominated by an irreducible component of $X{\otimes }_{k}K$; one thus deduces from $p$ a surjective map from the set of irreducible components of $X{\otimes }_{k}K$ to the set of irreducible components of $X$. Likewise, it follows from the fact that $p$ is continuous and surjective that every connected component of $X$ is the image by $p$ of a union of connected components of $X{\otimes }_{k}K$, whence a surjective map from the set of connected components of $X{\otimes }_{k}K$ to the set of connected components of $X$. One concludes from this that every irreducible (resp. connected) component $Z^{\prime }$ of $X{\otimes }_{k}K$ is contained in a unique set of the form $p^{-1}(Z)$, where $Z$ is an irreducible (resp. connected) component of $X$; for every sub-prescheme of $X$ having $Z$ as underlying space (and still denoted $Z$), $Z^{\prime }$ is an irreducible (resp. connected) component of $Z{\otimes }_{k}K$. In particular, if $X{\otimes }_{k}K$ has a finite number $n$ (resp. $n^{\prime }$) of irreducible (resp. connected) components (which will be the case when $X$ is of finite type over $k$, since then $X{\otimes }_{k}K$ is of finite type over $K$, hence Noetherian), the number of irreducible (resp. connected) components of $X$ is $\le n$ (resp. $\le n^{\prime }$); to say that it is equal to $n$ (resp. $n^{\prime }$) means that for every reduced sub-prescheme $Z$ of $X$ having as underlying space an irreducible (resp. connected) component of $X$, $Z{\otimes }_{k}K$ is irreducible (resp. connected) (I, 5.1.8); the number of irreducible (resp. connected) components of $X{\otimes }_{k}K$ is therefore independent of $K$ if and only if for every irreducible (resp. connected) component $Z$ of $X$, $Z{\otimes }_{k}K$ is irreducible (resp. connected) for every extension $K$ of $k$.

In particular, if $X{\otimes }_{k}K$ is irreducible (resp. connected), the same holds for $X$ (which already follows from the fact that $p$ is surjective). Likewise, if $X{\otimes }_{k}K$ is reduced (resp. integral), the same holds for $X$ since $p$ is faithfully flat (2.1.13).

In this number and the next, we shall examine more closely what can be said of the irreducible (resp. connected) components of $X{\otimes }_{k}K$ when $K$ varies; what precedes already shows us that the number of these components is an increasing function of $K$.

Lemma (4.4.2).

Let $X^{\prime }$, $X$ be two topological spaces, $f:X^{\prime }\to X$ a continuous map. Suppose that the following conditions are satisfied:

(i) $f$ is surjective open (resp. such that the topological space $X$ is canonically identified with the quotient of $X^{\prime }$ by the equivalence relation defined by $f$).

(ii) For every $x\in X$, $f^{-1}(x)$ is irreducible (resp. connected).

Then for $X^{\prime }$ to be irreducible (resp. connected), it is necessary and sufficient that $X$ be so.

The assertion relative to connectedness is none other than Bourbaki, Top. gén., chap. 1, 3rd ed., §11, n° 3, prop. 7. Let us prove the assertion relative to irreducibility; the condition being trivially necessary since $f$ is surjective, let us prove that it is sufficient. Let $X_1^{\prime }$, $X_2^{\prime }$ be two closed parts of $X^{\prime }$ such that $X^{\prime }=X_1^{\prime }\cup X_2^{\prime }$, and denote by $X_i$ $(i=1,2)$ the set of $x\in X$ such that $f^{-1}(x)\subset X_i^{\prime }$; one therefore has $X-X_i=f(X^{\prime }-X_i^{\prime })$, and since $f$ is open, one concludes that $X_i$ is a closed part of $X$ $(i=1,2)$. On the other hand, for every $x\in X$, $f^{-1}(x)$ is irreducible by hypothesis, and is the union of the two closed parts $X_1^{\prime }\cap f^{-1}(x)$ and $X_2^{\prime }\cap f^{-1}(x)$; one of these two parts must therefore be equal to $f^{-1}(x)$, which means that one must have $x\in X_1$ or $x\in X_2$; one therefore has $X=X_1\cup X_2$, and since $X$ is irreducible by hypothesis, one deduces from this $X=X_1$ or $X=X_2$, hence $X^{\prime }=X_1^{\prime }$ or $X^{\prime }=X_2^{\prime }$.

Remark (4.4.3).

If the topology of $X$ is not the quotient of that of $X^{\prime }$ by the equivalence relation defined by $f$, it can happen that $X$ and all the fibres $f^{-1}(x)$ are irreducible, without $X^{\prime }$ being connected: one has an example by taking for $X$ an irreducible scheme (for example the affine line $\mathrm{Spec}(k[T])$ over an algebraically closed field $k$), considering a closed point $x_0$ of $X$ and taking for $X^{\prime }$ the sum space of $x_0$ and the open subspace $X-x_0$ of $X$. Likewise, if one replaces the hypothesis "$f$ open" by "$f$ closed" (and even "$f$ proper") (which however implies that the topology of $X$ is then quotient of that of $X^{\prime }$ by the relation defined by $f$), it can happen that $X$ and all the fibres $f^{-1}(x)$ are irreducible without $X^{\prime }$ being so. Take again for $X$ the affine line over $k$, denote by $S$ the product $X{\times }_k\mathbf{P}$, where $\mathbf{P}$ is the projective line over $k$; if $x_0$ is a closed point of $X$, $t_0$ a closed point of $\mathbf{P}$, $p:S\to X$, $q:S\to \mathbf{P}$ the projections, denote by $X^{\prime }$ the reduced sub-prescheme having as underlying space the closed set $p^{-1}(x_0)\cup q^{-1}(t_0)$, and take for $f$ the restriction to $X^{\prime }$ of $p$; $f$ is proper but $X^{\prime }$ is not irreducible.

Theorem (4.4.4).

Let $k$ be an algebraically closed field, $X$ a $k$-prescheme. If $X$ is irreducible (resp. connected), the same holds for $X{\otimes }_{k}K$ for every extension $K$ of $k$.

One has seen indeed (4.4.1) that the morphism $p:X{\otimes }_{k}K\to X$ is faithfully flat and open; in order to apply the lemma (4.4.2), it therefore suffices to verify that the fibres $p^{-1}(x)$ are irreducible for every $x\in X$. But the $k(x)$-prescheme $p^{-1}(x)$ is isomorphic to $\mathrm{Spec}(k(x){\otimes }_{k}K)$ (I, 3.6.2), hence integral by virtue of the hypothesis on $k$ and of (4.3.9).

Corollary (4.4.5).

Let $k$ be an algebraically closed field, $X$ a $k$-prescheme, $K$ an extension of $k$, $p:X{\otimes }_{k}K\to X$ the canonical projection. If $Z$ is an irreducible (resp. connected) part of $X$, $p^{-1}(Z)$ is irreducible (resp. connected); in particular, if X_0 is an irreducible component (resp. the connected component) of $X$ containing $Z$, $p^{-1}(X_0)$ is an irreducible (resp. connected) component of $X{\otimes }_{k}K$ containing $p^{-1}(Z)$.

The second assertion follows from (4.4.1) and from the first applied by replacing $Z$ by X_0; to prove the first, let $X^{\prime }$ be the underlying space of $X{\otimes }_{k}K$, $Z^{\prime }=p^{-1}(Z)$; the

equivalence relation $R$ defined by $p$ on $X^{\prime }$ is open, hence the same holds for the relation $R_{Z^{\prime }}$ it induces on the saturated part $Z^{\prime }$, and $Z$ is identified with the quotient space $Z^{\prime }/R_{Z^{\prime }}$ (Bourbaki, Top. gén., chap. I, 3rd ed., §5, n° 2, prop. 4); one may therefore apply to $Z$ and $Z^{\prime }$ the lemma (4.4.2), whence the first assertion of (4.4.4).

Corollary (4.4.6).

With the hypotheses and notations of (4.4.5), the map $Z\mapsto p^{-1}(Z)$ is a bijection of the set of irreducible (resp. connected) components of $X$ onto the set of irreducible (resp. connected) components of $X{\otimes }_{k}K$, whose inverse bijection is $Z^{\prime }\mapsto p(Z^{\prime })$.

4.5. Geometrically irreducible and geometrically connected preschemes

Proposition (4.5.1).

Let $k$ be a field, $X$ a $k$-prescheme, $\Omega$ an algebraically closed extension of $k$. Let $n$ (resp. $n^{\prime }$) be the cardinal of the set of irreducible components (resp. of connected components) of $X{\otimes }_k\Omega$; this number is independent of the algebraically closed extension $\Omega$ chosen. In addition, for every extension $K$ of $k$, the cardinal of the set of irreducible (resp. connected) components of $X{\otimes }_{k}K$ is $\le n$ (resp. $\le n^{\prime }$).

Indeed, two algebraically closed extensions $\Omega$, ${\Omega }^{\prime }$ of $k$ can always be considered as sub-extensions of a same extension $E$ of $k$; it suffices then to apply (4.4.6) to $X{\otimes }_k\Omega$ and $X{\otimes }_k{\Omega }^{\prime }$ and to the common extension $E$ of $\Omega$ and ${\Omega }^{\prime }$ to prove the first assertion. The second is obtained by taking for $\Omega$ an algebraically closed extension of $K$ and using (4.4.1).

Definition (4.5.2).

The cardinal $n$ (resp. $n^{\prime }$) of (4.5.1) is called the geometric number of irreducible components (resp. of connected components) of $X$ (relative to $k$). If $n=1$ (resp. $n^{\prime }=1$) one says that $X$ is a geometrically irreducible (resp. geometrically connected) $k$-prescheme.

One thereby recovers a definition given for a particular case in (III, 4.3.4). By virtue of (4.5.1), the following properties are equivalent:

a) $X$ is geometrically irreducible (resp. geometrically connected).

b) For one algebraically closed extension $\Omega$ of $k$, $X{\otimes }_k\Omega$ is irreducible (resp. connected).

c) For every extension $K$ of $k$, $X{\otimes }_{k}K$ is irreducible (resp. connected).

(4.5.3)

Let $X$ be a $k$-prescheme, $Z$ a locally closed part of $X$. If one again denotes by $Z$ any of the sub-preschemes of $X$ having $Z$ as underlying space (I, 5.2.1), it follows from (I, 5.1.8) that for an extension $K$ of $k$, the number of irreducible (resp. connected) components of $Z{\otimes }_{k}K$ is independent of the sub-prescheme with underlying space $Z$ that one has chosen; thus one defines the geometric number of irreducible components (resp. of connected components) of $Z$ as that of any sub-prescheme of $X$ having $Z$ as underlying space.

Proposition (4.5.4).

Let $X$, $Y$ be two $k$-preschemes, $f:X\to Y$ a surjective $k$-morphism. If $X$ is geometrically irreducible (resp. geometrically connected), the same holds for $Y$.

Indeed, for every extension $K$ of $k$, $f_{(K)}:X_{(K)}\to Y_{(K)}$ is surjective (I, 3.5.2), and $X_{(K)}$ being irreducible (resp. connected) by hypothesis, the same holds for $Y_{(K)}$.

Definition (4.5.5).

One says that a morphism $f:X\to Y$ of preschemes is irreducible (resp. connected) if, for every $y\in Y$, the $k(y)$-prescheme $f^{-1}(y)$ is geometrically irreducible (resp. geometrically connected).

Proposition (4.5.6).

(i) Let $X$ be a $k$-prescheme, $K$ an extension of $k$. Then the geometric number of irreducible (resp. connected) components of $X_{(K)}$ relative to $K$ is equal to the geometric number of irreducible (resp. connected) components of $X$ relative to $k$. In particular, for the $K$-prescheme $X_{(K)}$ to be geometrically irreducible (resp. geometrically connected), it is necessary and sufficient that the $k$-prescheme $X$ be so.

(ii) Let $f:X\to Y$, $g:Y^{\prime }\to Y$ be two morphisms, and put $X^{\prime }=X_{(Y^{\prime })}=X{\times }_Y{Y}^{\prime }$, $f^{\prime }=f_{(Y^{\prime })}:X^{\prime }\to Y^{\prime }$. If $f$ is irreducible (resp. connected), the same holds for $f^{\prime }$. The converse is true if $g$ is surjective.

(i) If $\Omega$ is an algebraically closed extension of $K$, one has $X{\otimes }_k\Omega =X_{(K)}{\otimes }_K\Omega$, whence the assertion.

(ii) For every $y^{\prime }\in Y^{\prime }$, if one puts $y=g(y^{\prime })$, one has $f^{\prime -1}(y^{\prime })=f^{-1}(y){\otimes }_{k(y)}k(y^{\prime })$ (I, 3.6.4), hence the two assertions follow from (i).

Proposition (4.5.7).

Let $f:X\to Y$ be a surjective irreducible (resp. connected) morphism, $g:Y^{\prime }\to Y$ a morphism, and put $X^{\prime }=X{\times }_Y{Y}^{\prime }$. If in addition $Y^{\prime }$ is irreducible (resp. connected), and $f$ universally open (resp. flat and quasi-compact, or universally open, or universally closed), then $X^{\prime }$ is irreducible (resp. connected).

The properties of $f$ that one considers being all stable under base change, one may restrict to the case where $Y^{\prime }=Y$; it then suffices to apply (4.4.2) (taking into account (2.3.12)).

Corollary (4.5.8).

Let $X$, $Y$ be two $k$-preschemes.

(i) If $X$ is geometrically irreducible (resp. geometrically connected) and $Y$ irreducible (resp. connected), then $X{\times }_{k}Y$ is irreducible (resp. connected).

(ii) If $X$ and $Y$ are geometrically irreducible (resp. geometrically connected), the same holds for $X{\times }_{k}Y$.

(i) The structure morphism $X\to \mathrm{Spec}(k)$ is surjective and universally open (2.4.9) and in addition irreducible (resp. connected); it therefore suffices to apply (4.5.7).

(ii) If $\Omega$ is an algebraically closed extension of $k$, one has $(X{\times }_{k}Y{)}_{(\Omega )}=X_{(\Omega )}{\times }_{\Omega }{Y}_{(\Omega )}$ (I, 3.3.10); by hypothesis $X_{(\Omega )}$ and $Y_{(\Omega )}$ are geometrically irreducible (resp. geometrically connected) $\Omega$-preschemes (4.5.6) and it therefore suffices to apply (i).

Proposition (4.5.9).

Let $X$ be a $k$-prescheme. The following conditions are equivalent:

a) $X$ is geometrically irreducible (in other words, for every extension $K$ of $k$, $X{\otimes }_{k}K$ is irreducible).

b) For every finite separable extension $K$ of $k$, $X{\otimes }_{k}K$ is irreducible.

c) $X$ is irreducible, and if $x$ is its generic point, $k(x)$ is a primary extension of $k$.

It is clear that a) implies b). To see that b) implies c), consider a finite separable extension $K$ of $k$; if $p:X{\otimes }_{k}K\to X$ is the canonical projection, the

maximal points of $X{\otimes }_{k}K$ are those of the fibre $p^{-1}(x)=\mathrm{Spec}(k(x){\otimes }_{k}K)$ ((2.3.4) and $(0_I,\mathrm{2.1.8})$); to say that this fibre is irreducible for every finite separable extension $K$ of $k$ amounts to saying that $k(x)$ is a primary extension of $k$, by virtue of (4.3.2). Conversely, if c) is satisfied, the same reasoning (taking into account (4.3.2)) shows that for every extension $K$ of $k$, $X{\otimes }_{k}K$ is irreducible, hence c) implies a).

Corollary (4.5.10).

Let $X$ be an irreducible $k$-prescheme, $x$ its generic point, $k^{\prime }$ the separable algebraic closure of $k$ in $k(x)$, k'' a Galois extension of $k$ (of finite degree or not) containing $k^{\prime }$. Suppose $k^{\prime }$ finite over $k$; then the irreducible components of $X{\otimes }_k{k}^{\prime \prime }$ are geometrically irreducible; their number is equal to $[k^{\prime }:k]$ and is also the geometric number of irreducible components of $X$.

As in (4.5.9), one is reduced to considering the maximal points of the fibre $\mathrm{Spec}(k(x){\otimes }_k{k}^{\prime \prime })$; by virtue of (4.3.4), they are $[k^{\prime }:k]$ in number and the residue fields at these points (which are also those of $X{\otimes }_k{k}^{\prime \prime }$) are primary extensions of k'', which, taking into account (4.5.9), proves the corollary.

Corollary (4.5.11).

Let $X$ be a $k$-prescheme; suppose that $X$ has only a finite number of maximal points $x_i$ $(1\le i\le r)$ and that for each $i$, the separable algebraic closure $k_i^{\prime }$ of $k$ in $k(x_i)$ is of finite degree over $k$. Then there exists a finite separable extension $L$ of $k$ such that the irreducible components of $X{\otimes }_{k}L$ are geometrically irreducible; their number, equal to ${\Sigma }_i[k_i^{\prime }:k]$, is the geometric number of irreducible components of $X$.

The maximal points of $X{\otimes }_{k}L$ are those of the various fibres $\mathrm{Spec}(k(x_i){\otimes }_{k}L)$ by virtue of (2.3.4) and $(0_I,\mathrm{2.1.8})$ applied to the irreducible components of $X$ and to their inverse images in $X{\otimes }_{k}L$. It therefore suffices to take for $L$ a finite Galois extension of $k$ containing all the $k_i^{\prime }$ and to apply (4.5.10).

One will note that (4.5.11) applies in particular to every $k$-prescheme $X$ of finite type over $k$, for $X$ is then Noetherian, hence has only a finite number of irreducible components, and each of the $k(x_i)$ is an extension of finite type of $k$, hence so is the algebraic closure of $k$ in $k(x_i)$ ([1], p. 6-06, lemma 5).

Remarks (4.5.12).

(i) The notions defined in (4.5.2) depend on the base field $k$. For brevity, and when it will be necessary to mention the base field, one may say "$k$-irreducible" (resp. "$k$-connected") in place of "geometrically irreducible (resp. connected) relative to $k$". One will note that this terminology, completely natural in the context of schemes, is exactly opposite to that of Weil: in that author, "$k$-irreducible" (resp. "$k$-connected", "$k$-normal", etc.) designates an intrinsic property of a prescheme $X$, independent of the field $k$ taken as "base field" (in other words, of the chosen morphism $X\to \mathrm{Spec}(k)$); we express these properties by saying simply that $X$ is irreducible (resp. connected, normal, etc.). Weil says on the other hand "absolutely irreducible", "absolutely connected", "absolutely normal", etc., for the corresponding notions relative to the prescheme $X{\otimes }_k\Omega$, where $\Omega$ is a suitable algebraically closed extension of $k$. This terminology is explained by the different point of view in which that author places himself: for him, an "algebraic variety" $V$ is given first as a

geometric object over an algebraically closed field $\Omega$; the datum of a smaller "field of definition" $k$ (i.e., of a scheme over $k$ defining $V$ by extension of the base field to $\Omega$) is for him an additional and, to a certain extent, secondary structure, so that his qualification of "absolute" or "intrinsic" on one side, "relative to $k$" on the other, is opposite to ours (¹). We shall therefore avoid in the sequel using the qualifier "absolute", which may lead to confusion, and when we use the abbreviated terminology introduced above (in conflict with received terminology), we shall refer to (4.5.12) to avoid all ambiguity.

(ii) The proposition (4.5.9) gives a "birational" criterion (in other words, depending only on the residue field at the generic point) for a $k$-prescheme $X$ to be geometrically irreducible. There is no analogous criterion for $X$ to be geometrically connected. For example, take $X=\mathrm{Spec}(\mathbf{R}[[T,U]]/(T^2+U^2))$ ($T$, $U$ indeterminates); $X$ is an integral $\mathbf{R}$-scheme, and its field of rational functions $k(x)$ is $\mathbf{C}((T))$, as one easily verifies; $X{\otimes }_{\mathbf{R}}\mathbf{C}$ decomposes into two irreducible components (the two "isotropic lines") which have one point in common (the maximal ideal image in $\mathbf{C}[[T,U]]/(T^2+U^2)$ of the maximal ideal $(T)+(U)$ of $\mathbf{C}[[T,U]]$); hence $X{\otimes }_{\mathbf{R}}\mathbf{C}$ is connected. Let $X^{\prime }$ be the (open) complement of the closed point of $X$ corresponding to the maximal ideal $(T)+(U)$ of $\mathbf{R}[[T,U]]$; then $X^{\prime }{\otimes }_{\mathbf{R}}\mathbf{C}$ is not connected, although $X$ and $X^{\prime }$ have the same field of rational functions.

Proposition (4.5.13).

Let $X$, $Y$ be two $k$-preschemes, $f:Y\to X$ a $k$-morphism. Suppose $Y$ geometrically connected and non-empty, and $X$ connected. Then $X$ is geometrically connected.

Let $\Omega$ be an algebraic closure of $k$, $X^{\prime }=X_{(\Omega )}$, $Y^{\prime }=Y_{(\Omega )}$, $f^{\prime }=f_{(\Omega )}:Y^{\prime }\to X^{\prime }$; it must be shown that $X^{\prime }$ is connected. Let $U^{\prime }$ be a non-empty open and closed part of $X^{\prime }$; note that the morphisms $p:X^{\prime }\to X$ and $q:Y^{\prime }\to Y$ are open (2.4.10) and closed since $\Omega$ is an algebraic extension of $k$ (II, 6.1.10). Hence $U=p(U^{\prime })$ is non-empty open and closed in $X$, and consequently equal to $X$. Since $Y$ is non-empty, one concludes (I, 3.4.7) that $V^{\prime }=f^{\prime -1}(U^{\prime })$ is not empty; moreover, it is an open and closed part of $Y^{\prime }$, and the latter space is connected by hypothesis; hence $V^{\prime }=Y^{\prime }$. One deduces from this that $U^{\prime }=X^{\prime }$, since otherwise the same reasoning applied to the open and closed set $X^{\prime }-U^{\prime }$, would show that $f^{\prime -1}(X^{\prime }-U^{\prime })=Y^{\prime }$, which is absurd since $Y^{\prime }$ is non-empty.

Corollary (4.5.13.1).

(i) Let $X$, $Y$ be two $k$-preschemes, $f:Y\to X$ a $k$-morphism. If $Y$ is geometrically connected and non-empty, the connected component X_0 of $X$ containing $f(Y)$ is geometrically connected.

(ii) Let $X$ be a $k$-prescheme. If $Y$ is an irreducible component of $X$ which is geometrically irreducible, then the connected component X_0 of $X$ containing $Y$ is geometrically connected.

(i) One may suppose $Y$ reduced, so that $f$ factors as $f:Y\to X_0\to X$, X_0

(¹) Zariski's point of view is already closer to ours, and his terminology is not generally in conflict with that introduced here.

denoting the reduced sub-prescheme of $X$ having X_0 as underlying space (I, 5.2.2); it suffices to apply (4.5.13) to $g$.

(ii) By considering a prescheme having $Y$ as underlying space, and noting that $Y$ is a fortiori geometrically connected, it suffices to apply (i) to the canonical injection $Y\to X$.

Corollary (4.5.14).

Let $X$ be a $k$-prescheme. If $x$ is a point of $X$ such that $k(x)$ is a primary extension of $k$ (in particular if $x$ is a rational point of $X$), the connected component of $x$ in $X$ is geometrically connected.

It suffices to apply (4.5.13.1) to $Y=\mathrm{Spec}(k(x))$ and to the canonical morphism $Y\to X$, taking into account (4.5.9), which implies that $Y$ is geometrically irreducible.

Proposition (4.5.15).

Let $X$ be a $k$-prescheme, $x$ a point of $X$, $k^{\prime }$ the separable algebraic closure of $k$ in $k(x)$. Suppose the following conditions satisfied:

(i) $X$ is connected.

(ii) $k^{\prime }$ is a finite extension of $k$.

Then the geometric number of connected components of $X$ is $\le [k^{\prime }:k]$, and if k'' is a finite Galois extension of $k$ containing $k^{\prime }$, the connected components of $X{\otimes }_k{k}^{\prime \prime }$ are geometrically connected.

One will note that condition (ii) is satisfied if $X$ is locally of finite type over $k$.

There exist finite Galois extensions k'' of $k$ containing $k^{\prime }$ by virtue of condition (ii). Put $X^{\prime \prime }=X{\otimes }_k{k}^{\prime \prime }$; the projection morphism $p:X^{\prime \prime }\to X$ is faithfully flat and finite, hence closed (II, 6.1.10) and consequently (2.3.6, (ii)) the image by $p$ of every connected component $X_{\alpha }^{\prime \prime }$ of X'' is equal to $X$; $X_{\alpha }^{\prime \prime }$ therefore contains a point $x_{\alpha }\in p^{-1}(x)$. But $p^{-1}(x)$ has a number of points equal to the number of irreducible components of $\mathrm{Spec}(k(x){\otimes }_k{k}^{\prime \prime })$ (I, 3.4.9), that is $[k^{\prime }:k]$ (4.3.4), and a fortiori the number of connected components of X'' is $\le [k^{\prime }:k]$. For every $x_{\alpha }\in p^{-1}(x)$, $k(x_{\alpha })$ is a primary extension of k'' by virtue of (4.3.5) and of (I, 3.4.9). One may therefore apply to $x_{\alpha }$ and X'' the corollary (4.5.14), which proves that all the connected components of X'' are geometrically connected.

Corollary (4.5.16).

Let $X$ be a $k$-prescheme, $(X_{\alpha })$ the family of its connected components, and for every $\alpha$, let $x_{\alpha }\in X_{\alpha }$. Suppose the following conditions satisfied:

(i) The family $(X_{\alpha })$ is finite.

(ii) For every $\alpha$, the separable algebraic closure $k_{\alpha }^{\prime }$ of $k$ in $k(x_{\alpha })$ is a finite extension of $k$.

Then the geometric number of connected components of $X$ is at most ${\Sigma }_{\alpha }[k_{\alpha }^{\prime }:k]$, and there exists a finite separable extension k'' of $k$ such that all the connected components of $X{\otimes }_k{k}^{\prime \prime }$ are geometrically connected.

Noting that the connected components of $X$ are open by virtue of condition (i), it suffices to apply to each of the sub-preschemes of $X$ induced on the open sets $X_{\alpha }$ the result of (4.5.15); to have an extension k'' answering the question, it suffices to take a finite Galois extension of $k$ containing all the $k_{\alpha }^{\prime }$.

Corollary (4.5.17).

Suppose the $k$-prescheme $X$ contains a point $x$ such that the separable algebraic closure of $k$ in $k(x)$ is finite over $k$. For $X$ to be geometrically connected, it is necessary and sufficient that for every finite separable extension $K$ of $k$, $X{\otimes }_{k}K$ be connected.

Remark (4.5.18).

We shall see in §8 (8.4.5) that the conclusion of (4.5.17) is still valid if, instead of supposing that the condition of the statement is satisfied, one supposes that $X$ is quasi-compact.

Proposition (4.5.19).

Let $X$ be a $k$-prescheme, $Z$ a part of $X$, $k^{\prime }$ an algebraically closed extension of $k$, $X^{\prime }=X_{(k^{\prime })}$, $p:X^{\prime }\to X$ the canonical projection. Suppose that $Z^{\prime }=p^{-1}(Z)$ is contained in a single irreducible component $X_0^{\prime }$ of $X^{\prime }$. Then one has $X_0^{\prime }=p^{-1}(X_0)$, where $X_0=p(X_0^{\prime })$ is an irreducible component of $X$ containing $Z$, and in addition X_0 is geometrically irreducible. If moreover $X_0^{\prime }$ is the only irreducible component of $X^{\prime }$ meeting $Z^{\prime }$, then X_0 is the only irreducible component of $X$ meeting $Z$.

To show that $X_0^{\prime }=p^{-1}(X_0)$, we shall apply the

Lemma (4.5.19.1).

Let $T$, $T^{\prime }$ be two preschemes, $p:T^{\prime }\to T$ a morphism; consider the prescheme $T^{\prime \prime }=T^{\prime }{\times }_T{T}^{\prime }$, and let $p_1:T^{\prime \prime }\to T^{\prime }$, $p_2:T^{\prime \prime }\to T^{\prime }$ be the canonical projections. For a part $U^{\prime }$ of $T^{\prime }$ to be of the form $p^{-1}(U)$, where $U\subset T$, it is necessary and sufficient that one have $p_1^{-1}(U^{\prime })=p_2^{-1}(U^{\prime })$.

The structure morphism $q:T^{\prime \prime }\to T$ is equal to $p\circ p_1$ and to $p\circ p_2$ by definition, hence the relation is necessary. Conversely, suppose it satisfied; let $u^{\prime }$ be a point of $U^{\prime }$, $t^{\prime }$ a point of $p^{-1}(p(u^{\prime }))$; since $p(u^{\prime })=p(t^{\prime })$, there exists a point $t^{\prime \prime }\in T^{\prime \prime }$ such that $p_1(t^{\prime \prime })=u^{\prime }$, $p_2(t^{\prime \prime })=t^{\prime }$ (I, 3.4.7); hence $t^{\prime \prime }\in p_1^{-1}(U^{\prime })=p_2^{-1}(U^{\prime })$ by hypothesis, and consequently $t^{\prime }=p_2(t^{\prime \prime })\in U^{\prime }$, which proves the lemma.

This lemma being established, to show that $X_0^{\prime }=p^{-1}(X_0)$, let us form therefore the product $X^{\prime \prime }=X^{\prime }{\times }_X{X}^{\prime }$ (relative to the morphism $p:X^{\prime }\to X$); if $p_1$ and $p_2$ are the canonical projections of X'' onto $X^{\prime }$, it is therefore a matter of showing that $p_1^{-1}(X_0^{\prime })=p_2^{-1}(X_0^{\prime })$. Now, if one puts $S=\mathrm{Spec}(k)$, $S^{\prime }=\mathrm{Spec}(k^{\prime })$, observe that if $S^{\prime \prime }=S^{\prime }{\times }_S{S}^{\prime }=\mathrm{Spec}(k^{\prime }{\otimes }_k{k}^{\prime })$, one may also write $X^{\prime \prime }=X^{\prime }{\times }_{S^{\prime }}{S}^{\prime \prime }$. If $q^{\prime \prime }:X^{\prime \prime }\to S^{\prime \prime }$ is the structure morphism, it suffices to prove that for every $s^{\prime \prime }\in S^{\prime \prime }$, one has $p_1^{-1}(X_0^{\prime })\cap q^{\prime \prime -1}(s^{\prime \prime })=p_2^{-1}(X_0^{\prime })\cap q^{\prime \prime -1}(s^{\prime \prime })$. If one puts $T=\mathrm{Spec}(k(s^{\prime \prime }))$, $U=X^{\prime \prime }{\times }_{S^{\prime \prime }}T=q^{\prime \prime -1}(s^{\prime \prime })$, and if $v:U\to X^{\prime \prime }$ is the canonical projection, it is therefore a matter of proving that $U_1=v^{-1}(p_1^{-1}(X_0^{\prime }))$ and $U_2=v^{-1}(p_2^{-1}(X_0^{\prime }))$, which are irreducible components of $U$ (4.4.5), are identical. One has the diagram