Is the ring of the polynomials on a finite algebraic set always a product of local rings?

Question

A proposition from Fulton's book 'Algebraic Curves':

Let $I$ be an ideal in $k[X_1, ..., X_n]$ ($k$ algebraically closed), and suppose $V(I) = \{ P_1, \ldots, P_N \}$ is finite. Let $\mathcal{O}_i = \mathcal{O}_{P_i}(\mathbb{A}^n)$. Then there is a natural isomorphism of $k[X_1, \ldots, X_n] \mathbin{/} I$ with $\prod_{i = 1}^{N} \mathcal{O}_i \mathbin{/} I\mathcal{O}_i$.

But if $k$ is not algebraically closed, is it still true that, if $V(I)$ is finite, $k[X_1, \ldots, X_n] \mathbin{/} I$ is a direct product of local rings? And if so, then some factors are most likely given by localizations at the points of $V(I)$, whereas other factors are not; how to describe these latter factors?

The reason for this question is the observation that field extensions of $k$ are fields and thus local rings. For instance, $\mathbb{Q}[X] \mathbin{/} (X^2 + 1)$ is a field, but $V(I)$ is empty and $\mathbb{Q}$ is not algebraically closed. Hence, by the Chinese remainder theorem, it seems to me that, for any $f$ in $k[X]$, $k[X] \mathbin{/} (f)$ is the direct product of local rings anyway. Is it at least that correct?

Answer 1

If you upgrade to schemes and take $V(I)$ to be the closed subscheme $\operatorname{Spec} k[X_1,\cdots,X_n]/I \subset \operatorname{Spec} k[X_1,\cdots,X_n] = \Bbb A^n_k$, then your claim that $V(I)$ finite implies a natural isomorphism between $k[X_1,\cdots,X_n]/I$ and $\prod \mathcal{O}_i/I\mathcal{O}_i$ is still true. Without upgrading to schemes or enforcing the requirement that $k$ is algebraically closed, you cannot guarantee this result will still hold.

In the case of a scheme, the reason is that the former is the global sections of the structure sheaf on $V(I)$, while the latter is the product of the stalks, and on any discrete space these two things are naturally isomorphic. The sheaf condition gives that the global sections are the product of sections over each open set $\{p_i\}$, while the fact that the $\{p_i\}$ are open means they're the final object among open sets containing $p_i$ and thus the stalk at $p_i$ is the value of the sheaf on $\{p_i\}$.

Answer 2

Let $B:=k[x_1,..,x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal. Let $X:=Spec(A)$ with $A:=B/I$. By the Noether normalization lemma it follows there is a finite set of elements $y_1,..,y_d \in A$ with $d=dim(X)$ and the property that the sub-ring $k[y_1,..,y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1,..,y_d] \subseteq A$ is an integral extension of rings. If $d=0$ it follows from the same lemma the ring extension $k \subseteq A$ is integral and in particular it follows $A$ is an artinian ring. In particular it follows $A\cong R_1 \prod \cdots \prod R_l$ is a direct product of artin local rings $R_i$ with $\mathfrak{p_i} \subseteq R_i$ the maximal ideal (this is Atiyah-Macdonald's book Thm 8.7). Hence $Spec(A) \cong \prod Spec(R_i)$ is a disjoint union of the schemes $U_i:=Spec(R_i)$. We may write $X:=\{ \mathfrak{p_1}, \ldots , \mathfrak{p_l}\}$ as a disjoint union of $l$ distinct points, with $\mathfrak{p_i} \subseteq R_i$ the unique maximal ideal in the local ring $R_i$. Since $U_i \subseteq X$ is an open subscheme it follows there are isomorphisms $\mathcal{O}_{X,\mathfrak{p_i}} \cong \mathcal{O}_{U_i,\mathfrak{p_i} } \cong (R_i)_{\mathfrak{p_i}} \cong R_i$ since $R_i$ is a local ring and every element in $R_i - \mathfrak{p_i}$ is a unit. Here $\mathcal{O}_{U_i}$ is the restriction of $\mathcal{O}_X$ to the open set $U_i$. Hence for any base field $k$ it follows from Atiyah-Macdonalds (AM) book "Commutative Algebra" that your ring $A:=B/I$ decompose as a finite direct sum of Artinian local rings $R_i$. At the level of sheaves you get the following formulas:

$\mathcal{O}_X \cong \mathcal{O}_{U_1} \oplus \cdots \oplus \mathcal{O}_{U_l}$

and

$ \mathcal{O}_{X,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{X,\mathfrak{p_l}} \cong \mathcal{O}_{U_1,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{U_l,\mathfrak{p_l}} \cong \mathcal{O}_{U_1}(U_1) \oplus \cdots \oplus \mathcal{O}_{U_l}(U_l) .$

These formulas answers the following questions:

Question 1: "But if k is not algebraically closed, is it still true that, if V(I) is finite, k[X1,…,Xn]/I is a direct product of local rings?"

Answer 1: Yes, this is true by the above argument.

Question 2: "And if so, then some factors are most likely given by localizations at the points of V(I), whereas other factors are not; how to describe these latter factors?"

Answer 2: All factors are given as localizations at points by the above argument.

Consider your example with $A:=\mathbb{Q}[x,y]/(x^2+y^2-3)$ and let $C:=Spec(A)$. By Zorn's Lemma (if you accept this) there is a maximal ideal $\mathfrak{m} \subseteq A$, and the residue field $\kappa(\mathfrak{m})$ is a finite extension of $\mathbb{Q}$. Hence $C$ - which as a set is the set of prime ideals in $A$ - is non-empty. If $k$ is the algebraic closure of $\mathbb{Q}$ and you pass to the fiber product $C\otimes k$ you get a curve. By Hartshorne Prop I.1.13 it follows $dim(C\otimes k)=1$ since it is defined by a hypersurface in $k[x,y]$.

Let $k$ be any field and let $A:=k[x]$ be the polynomial ring in the variable $x$. Let $f(x) \in A$ be any polynomial. We may write $f(x)=\prod_{i=1}^l p_i(x)^{d_i}$ where $p_i(x)$ is an irreducible polynomial and $d_i \geq 1$ is an integer for every $i$. Here $p_i(x)$ and $p_j(x)$ are distinct for $i\neq j$. Let $B_i:=k[x]/(p_i(x)^{d_i})$ and let $B:=A/(f(x))$. There is a canonical isomorphism

$\phi: B \rightarrow B_1\oplus \cdots \oplus B_l$

defined by

$\phi(x + (f(x))) :=(x+(p_1(x)^{d_1}),..,x+(p_l(x)^{d_l}))$.

This is Prop 1.10 in the AM-book. The ideals $(p_i(x)^{d_i})$ and $(p_j(x)^{d_j})$ are coprime for $i\neq j$. Since the ideals $(p_i(x))$ are maximal, it follows the rings $B_i$ are Artinian local rings, hence the decomposition given by $\phi$ decompose the Artinian ring $B$ into a direct sum of Artinian local rings $B_i$. Hence in this case Prop 1.10 in AM implies Thm 8.7 in the AM book (the "stucture theorem on Artin rings").

Question 3: "And is there an higher-level explanation why it (apparently, not quite sure) works in one variable, that is, k[X]/f is a direct product of local rings for any f∈k[X], even if k is not algebraically closed, and even if V(f) is empty?"

Here you must understand the fact that any polynomial $f(x) \in k[x]$ factors as a product of powers of irreducible polynomials, and that an irreducible polynomial defines a maximal ideal in $k[x]$. Moreover if $p(x)$ is an irreducible polynomial, it follows the ring $k[x]/((p(x)^d)$ for $d\geq 1$ is an Artinian ring. Then you must read AM, Chapter1 where it is proved that the map $\phi$ is an isomorphism of rings.