$\# V(I) \le \text{dim}_K(K[X_1, \ldots, X_n]/I)$ Let $K$ be an algebraically closed field and $I \subseteq K[X_1, \ldots, X_n]$ be a zero-dimensional ideal, i.e. the set $V(I)$ is finite. This is equivalent to $\text{dim}_K (K[X_1, \ldots, X_n]/I)&lt;\infty$. But the proof I know for this equivalence does not highlight any relation between $V(I)$ and $\text{dim}_K (K[X_1, \ldots, X_n]/I)$. I found recently that, in fact, one has the following inequality:
$$\# V(I) \le \text{dim}_K (K[X_1, \ldots, X_n]/I)$$
Here $\# V(I)$ denotes the number of points in $V(I)$. I am trying to prove this result. I know that the dimension of $K[X_1, \ldots, X_n]/I$ over $K$ is precisely the number of monomials from $K[X_1, \ldots, X_n]$ which do not belong to the initial ideal of $I$ (the classes of these monomials modulo $I$ form a $K$-basis of the vector space $K[X_1, \ldots, X_n]/I$). How can I relate these monomials with the number of solutions $(x_1, \ldots, x_n)$ of the set of equations $\{f(x_1, \ldots, x_n) = 0 \ | \ f\in I \}$?
At @KReiser's suggestion, I'll post the solution of this problem. Let me denote the ring $K[X_1, \ldots, X_n]$ with $R$.
The key observation is the following: Since $K$ is algebraically closed, all the maximal ideals of $R$ are exactly ones of the form $(X_1-c_1, \ldots. X_n-c_n)$, with $c_i \in K$, so there is a bijection between the elements of $V(I)$ and the maximal ideals of $R$ which contain $I$, namely $(a_1, \ldots, a_n) \rightarrow (X_1-a_1, \ldots. X_n-a_n)$.
Now $R/I$ is a finite dimensional $K$-algebra, and as is shown here using CRT, the number of maximal ideals of $R/I$ (which is the number of maximal ideals of $R$ which contain $I$) is less than or equal to $\text{dim}_K(R/I)$.
Here you get an approach to the problem over any field or Artinian ring:
Isomorphism of product of local ring of variety
I copy the post given in the above link here:
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 &quot;Commutative Algebra&quot; 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: &quot;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?&quot;
Answer 1: Yes, this is true by the above argument.
Question 2: &quot;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?&quot;
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 &quot;stucture theorem on Artin rings&quot;).
Question 3: &quot;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?&quot;
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.
Here the copied post ends.
This is an answer to your question: In your case you may write
$$D1.\text{   }A \cong \oplus_{i=1}^l R_i$$
where $R_i$ is a local ring with maximal ideal $\mathfrak{p}_i$. It follows
$$dim_k(A) :=d = \sum_{i=1}^l dim_k(R_i)$$
hence $l \leq d:=dim_k(A)$$.
Question: &quot;I found recently that, in fact, one has the following inequality:
$$\# V(I)≤dim_k(k[x_1,…,x_n]/I)$$
I am trying to prove this result.&quot;
Answer: Here I interpret $\# V(I)$ as the &quot;number of points&quot; in the scheme $V(I)$. By the above argument it follows the number of maximal ideals in $A$ is less than or equal to $dim_k(A)$. Here $k$ is any field. Hence the general case (over any field) can be proved using methods from Atiyah-Macdonalds book &quot;Commutative Algebra&quot;. In fact:
It seems to me the methods introduced above proves the result for any Artinian ring $A$ over any field $k$. Let $V(I) \subseteq Spec(A[x_1,..,x_n]):=\mathbb{A}^n_A$ and let $B:=A[x_i]/I$.  It follows
$$ \# V(I) \leq length_A(B)$$
where $length_A(B)$ is the &quot;length&quot; of $B$ as left $A$-module.
Formula $D1$ gives
$$D1.\text{   }B\cong R_1 \oplus \cdots \oplus R_l$$
where $R_i$ is a local ring, and each $R_i$ is an $A$-algebra. We get
$$length_A(B) = length_A(\oplus_{i=1}^l R_i)=\sum_{i=1}^l length_A(R_i).$$
Hence $\# V(I) \leq length_A(B)$.