## Generators of a maximal ideal of k[X1,…,Xn]

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Hi. Given $k$ an algebraically closed field, I know that that a maximal ideal $\mathfrak{m}$ of $A = k[X_1,\cdots,X_n]$ is just a $\langle X_1-a_1,\cdots,X_n-a_n \rangle$ (Nullstellensatz). Knowing that, it seems intuitive that $\mathfrak{m}$ can not be generated by less than $n$ elements. Is that true ? In that case, how can I show that ?

(Actually, I need that, just after using Nakayama's lemma, to show that $dim_{A/\mathfrak{m}=k} \mathfrak{m}/\mathfrak{m}^2$ is greater than $n$.)

(I've seen things that might be relevant like "Krull height theorem" but I think such things take place in a more general context and I have some difficulties, first to understand them, and second to adapt them...)

Thank you.

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## 4 Answers

Since you mentioned Krull's height theorem (= the generalized principal ideal theorem) and having difficulty applying it, I thought you or someone else might appreciate seeing how this works: it is quite straightforward.

The generalized principal ideal theorem is as follows: let $R$ be a Noetherian ring and $I$ a proper ideal of $R$ which can be generated by $n$ elements. Let $\mathfrak{p}$ be a prime ideal which is minimal among all primes containing $I$. Then $\mathfrak{p}$ has height at most $n$, that is, there do not exist prime ideals $\mathfrak{p}_0,\ldots,\mathfrak{p}_n$ such that

$\mathfrak{p}_0 \subsetneq \mathfrak{p_1} \subsetneq \ldots \subsetneq \mathfrak{p_n} \subsetneq \mathfrak{p}$.

[For a deduction of this from Krull's Principal Ideal Theorem, see e.g. Theorem 96 on p. 70 of http://math.uga.edu/~pete/integral.pdf.]

Let us apply this with $R = k[x_1,\ldots,x_n]$ and the ideal $I = \langle x_1 - a_1,\ldots,x_n - a_n \rangle$. $I$ is itself a maximal -- hence prime -- ideal, since $R/I \cong k$. Thus the generalized principal ideal theorem simply says that $I$ cannot be generated by fewer elements than its height. But its height is certainly at least $n$. No geometry is needed here: just define $\mathfrak{p}_0 = 0$ and for $1 \leq i \leq n$, $\mathfrak{p}_i = \langle x_1 - a_1,\ldots,x_i - a_i \rangle$.

Finally, a comment: I did not use that $k$ was algebraically closed per se but only worked with maximal ideals of this particular form. On the other hand, it is still true over an arbitrary field $k$ that every maximal ideal of $k[x_1,\ldots,x_n]$ has height $n$ and can be generated by $n$ elements (and no fewer, by Krull's theorem): see Corollary 130 on p. 83 of the document linked to above.

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Thanks a lot, that exactly answers my question ! – Laurent May 25 2010 at 18:22

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The classes of the $X_i - a_i$ are easily seen to be a basis for $\mathfrak{m}/\mathfrak{m}^2$.

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The dimension is exactly n: you are counting linear polynomials modulo terms of higher order.

The geometrical interpretation is that intersecting n - 1 hypersurfaces with a common point must give an algebraic set with a component of dimension at least one. If you insist on a commutative algebra approach, you are going to have to study those theorems ...

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Trying to write down something along the lines you were looking for, I would argue that over an algebraically closed field the set of common zeros of $k < n$ polynomials has codimension at most $k$ and so cannot be a single point $(a_1,\ldots,a_n)$. You can use Krull theorem indeed (e.g. check out Ravi Vakil's notes: http://math.stanford.edu/~vakil/725/class14.pdf), or argue more along the 19th century and prove by some sort of elimination that the set of common zeros depend on at least $n-k$ parameters...

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