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Let $K$ be a field, $x=(x_1,\ldots,x_m)$, $y=(y_1,\ldots,y_n)$, $A\!\subseteq\!\mathbb{A}^m_K$, $B\!\subseteq\!\mathbb{A}^n_K$. Does there hold $$I_{K[x,y]}(A\!\times\!B)=\langle\langle I_{K[x]}(A) \cup I_{K[y]}(B)\rangle\rangle?$$ Here $I_{K[x]}(A)=\{f\in K[x]; f(a)\!=\!0\text{ for all }a\!\in\!A\}$ is the vanishing ideal of set $A$, and $\langle\langle\ldots\rangle\rangle$ is the ideal, generated by $\ldots$. The inclusion $\supseteq$ is easy, but I don't see how to show $\subseteq$.

If this equality does not hold, how else then can I prove (knowing $$K[x,y]/\langle\langle\mathfrak{a},\mathfrak{b}\rangle\rangle \,\cong\, K[x]/\mathfrak{a}\otimes_KK[y]/\mathfrak{b}$$ from this post), that there is an isomorphism of $K$-algebras (coordinate rings) $$K[A\!\times\!B] \,\cong\, K[A]\otimes_K K[B]?$$

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Leo
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3 Answers3

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This is well-known and can be found in any some introductions to classical algebraic geometry, in the section about products. Anyway, here is my favorite proof. I assume that $k$ is algebraically closed (otherwise it is wrong).

We have $V(I(A \times B))=A \times B = (A \times \mathbb{A}^n) \cap (\mathbb{A}^m \times B) = V(I(A) \cup I(B))$, thus $I(A \times B) = \sqrt{I(A) + I(B)}$. We have to prove that $I(A) + I(B)$ is a radical ideal, or equivalently that $k[x,y]/(I(A)+I(B))=k[X]/I(A) \otimes k[Y]/I(B)$ is reduced (i.e. $0$ is the only nilpotent element).

Lemma. If $k$ is an algebraically closed field and $R,S$ are reduced $k$-algebras, then $R \otimes_k S$ is reduced.

Proof: A colimit argument shows that we may assume that $R$ is of finite type over $k$. Since $R$ is reduced, the intersection of all prime ideals is $0$, which equals the intersection of all maximal ideals since $R$ is jacobson. This gives an embedding $R \hookrightarrow \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} R/\mathfrak{m}$, where $R/\mathfrak{m}=k$. This induces an embedding of $k$-algebras

$$R \otimes_k S \hookrightarrow (\prod_{\mathfrak{m} \in \mathrm{Spm}(R)} k) \otimes_k S \hookrightarrow \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} (k \otimes_k S) = \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} S.$$ Therefore $R \otimes_k S$ is a subring of a product of reduced algebras, therefore also reduced. $~\square$

If $k$ is not algebraically closed, the Lemma fails, even for fields. In fact, for a polynomial $f \in k[x]$ with splitting field $L$ the tensor product $k[x]/(f) \otimes_k L$ is isomorphic to the product of the algebras $L[x]/(x-\alpha)^{v_{\alpha}}$, where $\alpha$ runs through the roots of $f$ and $v_\alpha$ is its multiplicity. This algebra is reduced iff $v_\alpha=1$ for all $\alpha$ iff $f$ is separable. For example, $\mathbb{F}_p(t) \otimes_{\mathbb{F}_p(t^p)} \mathbb{F}_p(t) = \mathbb{F}_p(t)[x]/(x-t)^p$ is not reduced. The Lemma also fails when $k$ has characteristic zero, but then there is no counterexample for fields.

By the way, the isomorphism $k[A \times B] \cong k[A] \otimes_k k[B]$ holds almost by definition for affine schemes $A,B$. In this context the Lemma translates to the statement that the product of two reduced $k$-schemes is again reduced.

Martin Brandenburg
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  • What books for Classical Algebraic Geometry do you suggest? Hassett, Introduction to Algebraic Geometry, Beltrametti & Carletti & Gallarati, Lectures on Curves, Surfaces and Projective Varieties, Cox & Little & O'Shea, Using Algebraic Geometry, Perrin, Algebraic Geometry, Holme, A Royal Road to Algebraic Geometry, don't have what I need. – Leo Feb 04 '13 at 19:40
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    Ok I agree, it is not so easy to find in the literature. I've also looked at the books by Eisenbud-Harris, Ueno, Griffiths-Harris, Hartshorne, Shafarevich, Gathmann, but nowhere it is proved that $I+J \subseteq k[x,y]$ is a radical ideal when $I \subseteq k[x]$ and $J \subseteq k[y]$ are radical ideals. Sometimes it is argued that $\sqrt{I+J}$ is prime, which of course suffices to prove that (classical defined) affine varieties have products. But this does not suffice to prove that the coordinate ring is just the tensor product, it only shows that it is the tensor product mod nilpotents. – Martin Brandenburg Feb 04 '13 at 23:56
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    Anyway, a more modern account can be found in the book by Görtz-Wedhorn, Proposition 5.49. It is essentially the proof I gave above. I haven't even found it in EGA ... – Martin Brandenburg Feb 04 '13 at 23:57
  • Thank you for your help. The Görtz-Wedhorn book looks really good. Just one more question, does there hold $K[A\sqcup B]\cong K[A]\times K[B]$, i.e. does $K[-]$ send coproducts to products and vice versa? Do we need $K$ to be algebraically closed for both statements? – Leo Feb 05 '13 at 00:43
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    Yes $k[A \sqcup B] = k[A] \times k[B]$ holds for all affine schemes $A,B$. And $k[A \times B] = k[A] \otimes_k k[B]$ holds for all affine schemes, since $k[-]$ has a left adjoint. But in the context of classical algebraic geometry, where everything is reduced, one has to assume that $k$ is algebraically closed. – Martin Brandenburg Feb 05 '13 at 00:54
  • @MartinBrandenburg I am only a beginner in algebraic geometry - what do you mean by "A colimit argument shows..." ? –  Feb 06 '13 at 14:42
  • @MartinBrandenburg, may I know what goes wrong when $k$ is not algebraically closed? Nullstellensatz fails for sure, but the vanishing ideal claim still seems to hold. –  Feb 07 '13 at 07:37
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At the request of Martin Brandenburg, here is the result which may help you:

Let $k$ be a field and $V\subseteq \Bbb{A}_k^n$, $W\subseteq \Bbb{A}_k^m$ affine algebraic sets ( the proof I believe will also work if $V,W$ are affine varieties). Then $$I(V \times W )= I(V) + I(W).$$ By $I(V)$ we mean now the extension of the ideal $I(V)$ in the polynomial ring $k[x_1,\ldots,x_{m+n}]$.

Proof: See the discussion in my question here.

Using this result we can now prove the following theorem.

Theorem (Coordinate ring of a product): Let $V,W$ be as before. Then as $k$ - algebras we have $$k[V \times W] \cong k[V] \otimes_k k[W].$$

Proof: Let us write $I = \mathcal{I}(V)$ and $J = \mathcal{I}(W)$ and define $R = k[x_1,\ldots,x_n]$ and $S = k[x_{n+1},\ldots,x_{m+n}]$. Then by the lemma above, we have that $\mathcal{I}(V \times W) = I^e + J^e$ where the superscript denotes ideal extension in the ring $T = R \otimes_k S$, the polynomial algebra in $m+n$ variables. Now by the usual extension of scalars process we have $$I^e + J^e = I \otimes_k S +R \otimes_k J.$$ Thus to prove the theorem we need to prove that

$$\frac{T}{I \otimes_k S +R \otimes_k J} \cong \frac{R}{I} \otimes_k \frac{S}{J}.$$

The only tricky part in the proof is to construct a well-defined $k$ - algebra homomorphism $$f : \frac{R}{I} \times \frac{S}{J} \to \frac{ T}{I \otimes_k S + R \otimes_k J}$$

We can define one by declaring that $f$ sends $(\bar{a},\bar{b}) $ to $a\otimes b \mod{(I \otimes_k S + R \otimes_k J)}$. Is this well defined? Suppose $(\bar{a},\bar{b}) = (\bar{c},\bar{d})$. Then notice that $$\begin{eqnarray*} a \otimes b - c\otimes d &=& a \otimes b + c\otimes b - c\otimes b - c\otimes d \\ &=& (a-c) \otimes b + c \otimes (b-d)\\ &=& 0 \mod{(I \otimes_k S+ R \otimes_k J)}\end{eqnarray*}$$ because $a -c \in I$ and $b-d \in J$. Thus we see that $f$ is well-defined and leave the rest of the details for your to fill in, including constructing an inverse map. If you have any questions, I can add more details.

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    Concerning the second part: It is not tricky at all, when you use the Yoneda Lemma - the proof is just one line. – Martin Brandenburg Feb 07 '13 at 16:38
  • @MartinBrandenburg As a beginner, the proof is going to be low-level. But I am very interested in the Yoneda Lemma now. I don't have to keep defining maps all over the place and check well-definiteness! –  Feb 07 '13 at 16:44
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Let us assume that $K$ is algebraically closed.

If we have two $K$-algebras $C$ and $D$, there are canonical morphisms $$ C\overset{\alpha}{\longrightarrow} C\otimes_KD\overset{\beta}{\longleftarrow} D $$ defined by $\alpha(c)=c\otimes 1_D$ and $\beta(d)=1_C\otimes d$. If you are given two morphisms of $K$-algebras $\phi:C\to E$ and $\psi:D\to E$ to a third $K$-algebra $E$, then the universal property of tensor product (in the category of $K$-algebras) tells you that there exists a unique morphism of $K$-algebras $q:C\otimes_KD\to E$ such that $\phi=q\circ\alpha$ and $\psi=q\circ\beta$. In other words, in the category of $K$-algebras, $C\otimes_KD$ together with the arrows $\alpha$ and $\beta$ satisfies the universal property above.

Now, there is a duality of categories between the category of $K$-algebras of finite type and the category of affine varieties. This duality is given by the functor Spec. So take $$ \textrm{Spec (universal property of } C\otimes_KD),$$ after having assumed that $C$ and $D$ are of finite type (hence of the kind $C:=K[A]$ and $D:=K[B]$ as in your question). The universal property becomes a new universal property in the category of affine varieties, and now the universal object is $\textrm{Spec}(C\otimes_KD)$ with the two projections (the images $\textrm{Spec}(\alpha)$ and $\textrm{Spec}(\beta)$). But this is the universal property of $\textrm{Spec }C\times_K\textrm{Spec }D$, so that $$ \textrm{Spec}(C\otimes_KD)\cong \textrm{Spec }C\times_K\textrm{Spec }D=A\times_KB. $$ Now, to recover $K[A\times_KB]\cong K[A]\otimes_KK[B]=:C\otimes_KD$ it is enough to go through the duality in the other direction: since the inverse of Spec is $K[-]$ (the functor "global sections", or "take the coordinate ring of"), just take global sections of the last displayed formula to get what you want.

[Note that the identity you wrote between the ideals follows now as an obvious corollary.]

Brenin
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  • For the category of affine schemes in $\mathbb{A}^n_K$ and the category of finitely generated reduced (commutative unital) $K$-algebras to be equivalent, need $K$ be algebraically closed? Furthermore, I think I've read that the first category is also equivalent to the category of finitely generated reduced (commutative unital) rings. Is this true? – Leo Feb 04 '13 at 14:06
  • For the algebraically closed hypothesis, you are right. I will correct my answer. And for the rest, as far as I know, it depends whether your definition of variety includes reducedness. I think the right correspondences are: rings, with affine schemes; and (reduced) $K$-algebras of finite type, with (reduced) affine varieties over $K$. – Brenin Feb 04 '13 at 14:37
  • I don't think that abstract nonsense can answer the question (only the second one for affine schemes). Even for schemes it has a content (see my answer). The question was obviously asked for varieties in the classical sense, which are therefore reduced. Then one has to show that there is a product of varieties (more generally reduced schemes, doesn't really matter). – Martin Brandenburg Feb 04 '13 at 14:56
  • Dear Martin, thank you for your comment. Can you please explain to me why the equivalence between reduced algebras of finite type and affine varieties (in the classical sense, i.e. reduced) does not "include" the result that the ideal of the product is itself radical? Sorry to bother you but I want to understand this point. – Brenin Feb 05 '13 at 16:07
  • No problem. A priori, the coproduct in the category of reduced $k$-algebras (of finite type) is given by $(A,B) \mapsto (A \otimes_k B)_{\mathrm{red}}$. Here, $\mathrm{red}$ means that we mod out all nilpotent elements. This is abstract nonsense, but you can verify it directly. By the equivalence, this correspondce to the product of affine algebraic sets. But without further work it does not simply to $A \otimes_k B$ (as I have argued). – Martin Brandenburg Feb 07 '13 at 11:49