Lemma 10.114.1. Let $\mathfrak m$ be a maximal ideal in $k[x_1, \ldots , x_ n]$. The ideal $\mathfrak m$ is generated by $n$ elements. The dimension of $k[x_1, \ldots , x_ n]_{\mathfrak m}$ is $n$. Hence $k[x_1, \ldots , x_ n]_{\mathfrak m}$ is a regular local ring of dimension $n$.

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

$\psi _ i : k[x_1, \ldots , x_ i] /(f_1, \ldots , f_ i) \cong \kappa _ i.$

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

$k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ i) \cong \kappa _ i[x_{i + 1}, \ldots , x_ n].$

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

$(0) \subset (f_1) \subset (f_1, f_2) \subset \ldots \subset (f_1, \ldots , f_ n) = \mathfrak m$

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

Comment #2237 by David Savitt on

Why not use part (2) of [00GB] and induction on n to do this in just a few lines?

Comment #2272 by on

OK, I think I see what you mean, but I am too lazy to change this now.

Comment #7783 by Zhenhua Wu on

This result got me to think that $\mathop{\mathrm{ht}}P$, the height of a prime ideal $P$ of $k[x_1,...,x_n]$, might be the same as $\mu(P)$, the number of minimal generators of $P$. A counter example is described in this link https://math.stackexchange.com/questions/96691/height-and-minimal-number-of-generators-of-an-ideal. In short, $P:=(xz−y^2,x^3−yz,z^2−x^2y)⊂\mathbb{C}[x,y,z]$ is a prime ideal with $\mathop{\mathrm{ht}}P=2$ and $\mu(P)=3$. I suggest that we add this example after this tag.

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