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The version of the Nullstellensatz with which I am familiar states that if $K$ is an algebraically closed field, and $f_1,\dots,f_n\in K[X_1,\dots,X_m]$, then the family $\{f_i\}$ has a common zero iff $\langle f_1,\dots,f_n\rangle\neq K[X_1,\dots,X_m]$.

However, another form I have heard of states that if $K$ is algebraically closed, then the maximal ideals of $K[X_1,\dots,X_m]$ are precisely those of form $(X_1-a_1,\dots,X_m-a_m)$ for some $a_i\in K$.

Can anybody explain how the first form implies the second?

Nastassja
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1 Answers1

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Let $\mathfrak m$ be a maximal ideal. By the Hilbert Basis Theorem, it is generated by some $f_1,\ldots,f_n$. By the first form of the Nullstellensatz that you cite, they have a common zero, say $(a_1,\ldots,a_m)$.

Then the polynomials $f_1,\ldots,f_n,x_1-a_1,\ldots,x_m - a_m$ have a common zero (namely the same point $(a_1,\ldots,a_n)$), and so they generate a proper ideal, which contains $\mathfrak m = (f_1,\ldots,f_n)$. Thus it must equal $\mathfrak m$ (as $\mathfrak m$ is maximal), and so $\mathfrak m \supseteq (x_1-a_1,\ldots,x_m-a_m)$. Since $(x_1-a_1,\ldots,x_m-a_m)$ is clearly maximal, this inclusion is an equality. QED

Matt E
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  • I think that this could well be written without invoking Hilbert's basis theorem, for $f_i$'s are of no use in the proof. – Atom Apr 23 '23 at 19:51