# Is every integral finitely generated reduced $k$-algebra a domain?

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?

• The canonical morphism $k \to A$ is very rarely integral. "Integral" here means an integral domain (so "reduced" is redundant). – Qiaochu Yuan Dec 1 '20 at 19:11
• Now that you mention it, it was kind of obvious. $\mathbb{A}^n_k$ itself is irreducible, with coordinated ring $k[X_1, \dots, X_n]$ that definitely isn't integral over $k$. – user480840 Dec 1 '20 at 19:20
• You were looking at Gortz and Wedhorn, weren't you? I had the exact same question! – Mehta Mar 16 at 9:07

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 $$k$$.