The Stacks project

Proof. By the Hilbert Nullstellensatz (Theorem 10.34.1) we know the residue field $\kappa = \kappa (\mathfrak m)$ is a finite extension of $k$. Denote $\alpha _ i \in \kappa $ the image of $x_ i$. Denote $\kappa _ i = k(\alpha _1, \ldots , \alpha _ i) \subset \kappa $, $i = 1, \ldots , n$ and $\kappa _0 = k$. Note that $\kappa _ i = k[\alpha _1, \ldots , \alpha _ i]$ by field theory. Define inductively elements $f_ i \in \mathfrak m \cap k[x_1, \ldots , x_ i]$ as follows: Let $P_ i(T) \in \kappa _{i-1}[T]$ be the monic minimal polynomial of $\alpha _ i $ over $\kappa _{i-1}$. Let $Q_ i(T) \in k[x_1, \ldots , x_{i-1}][T]$ be a monic lift of $P_ i(T)$ (of the same degree). Set $f_ i = Q_ i(x_ i)$. Note that if $d_ i = \deg _ T(P_ i) = \deg _ T(Q_ i) = \deg _{x_ i}(f_ i)$ then $d_1d_2\ldots d_ i = [\kappa _ i : k]$ by Fields, Lemmas 9.7.7 and 9.9.2.

We claim that for all $i = 0, 1, \ldots , n$ there is an isomorphism

By construction the composition $k[x_1, \ldots , x_ i] \to k[x_1, \ldots , x_ n] \to \kappa $ is surjective onto $\kappa _ i$ and $f_1, \ldots , f_ i$ are in the kernel. This gives a surjective homomorphism. We prove $\psi _ i$ is injective by induction. It is clear for $i = 0$. Given the statement for $i$ we prove it for $i + 1$. The ring extension $k[x_1, \ldots , x_ i]/(f_1, \ldots , f_ i) \to k[x_1, \ldots , x_{i + 1}]/(f_1, \ldots , f_{i + 1})$ is generated by $1$ element over a field and one irreducible equation. By elementary field theory $k[x_1, \ldots , x_{i + 1}]/(f_1, \ldots , f_{i + 1})$ is a field, and hence $\psi _ i$ is injective.

This implies that $\mathfrak m = (f_1, \ldots , f_ n)$. Moreover, we also conclude that

Hence $(f_1, \ldots , f_ i)$ is a prime ideal. Thus

is a chain of primes of length $n$. The lemma follows. $\square$

Proof. By Lemma 10.114.1 all localizations $k[x_1, \ldots , x_ n]_{\mathfrak m}$ at maximal ideals are regular local rings of dimension $n$. Hence we conclude by Lemma 10.110.8. $\square$

Proof. By Proposition 10.114.2 any local ring of $k[x_1, \ldots , x_ n]$ is regular. Hence all local rings are Cohen-Macaulay, see Lemma 10.106.3. The local rings at maximal ideals have dimension $n$ hence every maximal chain of primes in $k[x_1, \ldots , x_ n]$ has length $n$, see Lemma 10.104.3. Hence every maximal chain of primes between $(0)$ and $\mathfrak p$ has length $\text{height}(\mathfrak p)$, see Lemma 10.104.4 for example. Putting these together leads to the assertion of the lemma. $\square$

Proof. Write $S = k[x_1, \ldots , x_ n]/\mathfrak p$. By Proposition 10.114.2 and Lemma 10.114.3 all the maximal chains of primes in $S$ (which necessarily end with a maximal ideal) have length $n - \text{height}(\mathfrak p)$. Thus this number is the dimension of $S$ and of $S_{\mathfrak m}$ for any maximal ideal $\mathfrak m$ of $S$. $\square$

Proof. Let $X = \bigcup _{i \in I} Z_ i$ be the decomposition of $X$ into its irreducible components. There are finitely many of them (see Lemmas 10.31.3 and 10.31.5). Let $I' = \{ i \mid x \in Z_ i\} $, and let $T = \bigcup _{i \not\in I'} Z_ i$. Then $U = X \setminus T$ is an open subset of $X$ containing the point $x$. The number (2) is $\max _{i \in I'} \dim (Z_ i)$. For any open $W \subset U$ with $x \in W$ the irreducible components of $W$ are the irreducible sets $W_ i = Z_ i \cap W$ for $i \in I'$ and $x$ is contained in each of these. Note that each $W_ i$, $i \in I'$ contains a closed point because $X$ is Jacobson, see Section 10.35. Since $W_ i \subset Z_ i$ we have $\dim (W_ i) \leq \dim (Z_ i)$. The existence of a closed point implies, via Lemma 10.114.4, that there is a chain of irreducible closed subsets of length equal to $\dim (Z_ i)$ in the open $W_ i$. Thus $\dim (W_ i) = \dim (Z_ i)$ for any $i \in I'$. Hence $\dim (W)$ is equal to the number (2). This proves that (1) $ = $ (2).

Let $\mathfrak m \supset \mathfrak p$ be any maximal ideal containing $\mathfrak p$. Let $x_0 \in X$ be the corresponding point. First of all, $x_0$ is contained in all the irreducible components $Z_ i$, $i \in I'$. Let $\mathfrak q_ i$ denote the minimal primes of $S$ corresponding to the irreducible components $Z_ i$. For each $i$ such that $x_0 \in Z_ i$ (which is equivalent to $\mathfrak m \supset \mathfrak q_ i$) we have a surjection

Moreover, the primes $\mathfrak q_ i S_\mathfrak m$ so obtained exhaust the minimal primes of the Noetherian local ring $S_{\mathfrak m}$, see Lemma 10.26.3. We conclude, using Lemma 10.114.4, that the dimension of $S_{\mathfrak m}$ is the maximum of the dimensions of the $Z_ i$ passing through $x_0$. To finish the proof of the lemma it suffices to show that we can choose $x_0$ such that $x_0 \in Z_ i \Rightarrow i \in I'$. Because $S$ is Jacobson (as we saw above) it is enough to show that $V(\mathfrak p) \setminus T$ (with $T$ as above) is nonempty. And this is clear since it contains the point $x$ (i.e. $\mathfrak p$). $\square$

Proof. This is a special case of Lemma 10.114.5. $\square$

Proof. The equivalence of the two statements follows from Lemma 10.24.3. Let $\mathfrak m \subset S$ be a maximal ideal. Every maximal chain of primes in $S_{\mathfrak m}$ has the same length equal to $\dim (S_{\mathfrak m})$, see Lemma 10.104.3. Hence, the dimension of the irreducible components passing through the point corresponding to $\mathfrak m$ all have dimension equal to $\dim (S_{\mathfrak m})$, see Lemma 10.114.4. Since $\mathop{\mathrm{Spec}}(S)$ is a Jacobson topological space the intersection of any two irreducible components of it contains a closed point if nonempty, see Lemmas 10.35.2 and 10.35.4. Thus we have shown that any two irreducible components that meet have the same dimension. The lemma follows easily from this, and the fact that $\mathop{\mathrm{Spec}}(S)$ has a finite number of irreducible components (see Lemmas 10.31.3 and 10.31.5). $\square$