I'm studying some classical algebraic geometry, and the book I'm using says that the contravariant functor from the category of algebraic affine sets ($k$ algebraically closed) to the category of finitely generated reduced $k$-algebraic that associates to $X \subseteq \mathbb{A}^n_k$ to the coordinate ring $k[X]$ is an equivalence of categories. Ok, no problem with this. But then, the book says that the restriction of this functor to irreducible algebraic sets gives an equivalence between the opposed category of irreducible algebraic sets and integral finitely generated reduced $k$-algebras. Since the coordinates ring of an irreducible algebraic set is always integral domain, this would imply that integral finitely generated reduced $k$-algebras are integral domains. Is this the case? Here, I'm assuming that integral $k$-algebras are algebras such that the canonical morphism $k \to A$ is integral. How can this last equivalence of categories arise?
Is every integral finitely generated reduced $k$-algebra a domain
algebraic-geometrycommutative-algebra
Related Solutions
1) Given a field $k$ and two domains $A,B$ over $k$, their tensor product $A\otimes _k B$ will be a domain as soon as one of the domains is a separable $k$-algebra and one of them (maybe the same!) has a fraction field which is a primary extension of $k$.
This is probably the most general possible result and is proved in Bourbaki's Algebra, Chapter v, ยง17, Corollary to Proposition 1.
(You can also find it on page 203 of Jacobson's Lectures in Abstract Algebra: III Theory of fields and Galois theory.)
Here separable means universally reduced and an extension of fields $K/k$ is said to be primary if the algebraic closure of $k$ in $K$ is purely inseparable over $k$.
These are quite advanced concepts in field theory but the good news is that for an algebraically closed field $k$ every algebra is separable and every extension field is primary, so that indeed for an algebraically closed field $k$, the $k$-algebra $A\otimes_k B$ is a domain as soon as $A$ and $B$ are domains.
2) For a general (non algebraically closed) field $k$, if $X,Y$ are the affine varieties associated to the $k$-algebras of finite type $A,B$ without zero-divisors, the irreducibility of $X\times Y$ does not imply that $A\otimes_k B$ is a domain.
For example, suppose $p$ is a prime integer. Then for $k=\mathbb F_p(t)$ ($t$ an indeterminate) and $A=k(\sqrt [p]t)$, we have $A\otimes_k A=\frac {A[X]}{\langle X^p-t\rangle}=\frac {A[X]}{\langle (X-\sqrt [p]t)^p\rangle }$, a ring with non-zero nilpotent elements (for example the class of $X-\sqrt [p]t$) which is thus certainly not a domain although the corresponding "variety" (or rather scheme) is irreducible (since its underlying topological space has just one point!).
Edit (September 28th, 2014)
As noticed by @lee in the comments, a non irreducible example follows from the isomorphism $\mathbb C\otimes_ \mathbb R \mathbb C= \mathbb C \times \mathbb C$ .
So I don't think that your teacher's remark is correct: it is not clear that the algebra corresponding to $X\times Y$ is the ring $A\otimes_k B$ rather than its reduction $(A\otimes_k B)_{\mathrm{red}}$ (obtained by killing the nilpotents: $(A\otimes_k B)_{\mathrm{red}}=A\otimes_k B/\mathrm{Nil}(A\otimes_k B)$).
This tends to show that, despite what your teacher claims, you cannot replace the hard algebra in 1) by elementary topological considerations of irreducibility: Milne himself proves the result (for an algebraically closed field) purely algebraically although, believe me, he knows what irreducibility means!
You only need to use the fact that the rings you're are working with are finitely generated. In the first example, note that by the definition of $P$, $k[T_1,\ldots,T_d][X]/(P(X))$ is just a finitely generated subring of $K(A)$. The second case is almost the same. Consider the generators of $k$-algebra $A$, pick $R$ for each one of them and finally take a product of all $R$'s.
Best Answer
It seems you confuse the notions "integral scheme" and "integral ring extension".
A scheme $(X,\mathcal{O}_X)$ is integral (this is Hartshorne's definition) iff for any open subset $U\subseteq X$ it follows the ring $\mathcal{O}_X(U)$ is an integral domain.
If $X:=\mathrm{Spec}(A)$ is an affine scheme it follows (this must be proved) $X$ is integral iff $A$ is an integral domain.
One implication is trivial. If $X:=\mathrm{Spec}(A)$ and $A$ is an integral domain you must prove that $\mathcal{O}_X(U)$ is an integral domain for any open subset $U \subseteq X$, and this follows from the fact that there is an inclusion $\mathcal{O}_X(U) \subseteq K(A)$ where $K(A)$ is the quotient field. Since $K(A)$ is a field, it follows any sub ring of $K(A)$ is an integral domain, in particular it follows $\mathcal{O}_X(U)$ is an integral domain.
Let $B:=k[x_1,...,x_n]$ be a polynomial ring in $n$ variables over a field $k$ and let $I \subseteq B$ be an ideal. Let $X:=\mathrm{Spec}(B/I)$. We may view $X$ as a closed sub-scheme of affine $n$-space $\mathbb{A}^n_k:=\mathrm{Spec}(B)$. It follows $\mathcal{O}_X(X)=A$ where $A:=B/I$. Noether normalization lemma says there is a finite set of elements $y_1,...,y_d \in A$ with $d=\dim(X)$ and the ring $k[y_1,...,y_d]\subseteq A$ generated by the elements $y_i$ is a non-trivial polynomial ring if $d \geq 1$. It follows the elements $y_i$ are transcendental over the field $k$ and do not satisfy a monic polynomial relation over $k$ in general, hence the ring extension $k \subseteq A$ is not an integral ring extension if $d \geq 1$. If $k \subseteq A$ is an integral ring extension it follows $\dim(X)=0$. It follows $A$ is an Artinian ring and there is a direct sum decomposition $A\cong A_1\oplus \cdots \oplus A_l$ with $A_i$ an Artinian local ring. Hence in this case $A$ is an integral domain iff $k \subseteq A$ is a finite field extension. $A$ is reduced iff $A$ is a finite direct product of finite field extensions of $k$.