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.



Generators of a maximal ideal of $k[X_1,\cdots,X_n]$
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 [Commutative algebra](http://alpha.math.uga.edu/%7Epete/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.


