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Let $k$ be an algebraically closed field. Nullstellensatz states that the maximal ideals of the polynomial ring $R=k[X_1,\dots,X_n]$ are precisely those of the form $\langle X_1-a_1,\dots,X_n-a_n\rangle$, with $(a_1,\dots,a_n)\in k^n$.

What if we are working with infinitely many variables, say $\{X_i\}_{i\in I}$, being $I$ an infinite set? Again, the ideals of the form $\bigl\langle\{X_i-a_i: i\in I\}\bigr\rangle$, with $(a_i)_{i\in I}\subseteq k$ are maximal (even if $k$ is not algebraically closed), but what about the converse?

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Let $k$ be a field and let $K$ be an extension of $k$ of degree larger than $1$. Let $R=k[X_i, i\in K]$ be a polynomial ring with one variable per element of $K$. There is a $k$-linear ring homomorphism $\phi:R\to K$ which maps $X_i$ to $i$ for all $i\in K$, and it is surjective. It follows that the kernel of $\phi$ is maximal. Clearly, it is not of the form you mentioned: indeed, the quotient of $R$ by all ideals of that form is $1$-dimensional as a $k$-algebra.

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  • $\begingroup$ Notice that even if $k$ is algebraically closed this gives examples: it is enough to take $K=k(x)$, a field of rational functions, for example. $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '13 at 22:11
  • $\begingroup$ Very ingenious, Mariano: bravo! $\endgroup$ – Georges Elencwajg Oct 29 '13 at 7:48
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If $|k| > |I|$, then any maximal ideal of $k[\{x_i\}_{i \in I}]$ is of the standard form, see MO/41262. The "cheap proof" of the Nullstellensatz works. The example by Mariano shows that this doesn't hold if $|k|=|I|$.

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