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Let $k$ be an infinite field. Let $d$ be a positive integer. Is it true that the cardinality of the set of maximal ideals of $k[x_1, \dots, x_d]$ is equal to the cardinality of the set of prime ideals of $k[x_1, \dots, x_d]$ and is equal to the cardinality of $k$?

Show your effort: in the case $k$ is algebraically closed, then the maximal ideals are in bijection with the elements of a $d$-dimensional $k$-vector space so the set of maximal ideals does have the same cardinality as $k$. In the case $d=1$, all non-zero prime ideals are maximal ($k[x]$ is a PID) and because there are infinitely many maximal ideals, this addresses the question completely.

I also know that (under ZFC, at least) any algebraic closure of an infinite field has the same cardinality as the field itself but I do not quite see how to apply this here. I think my question is meaningful in ZF as well, so it might be interesting to give a proof in ZF (if one exists).

Community
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jon
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    Every irreducible polynomial in $k[x]$ is the minimal polynomial of finitely many elements of $\bar{k}$. – jgon Jun 06 '19 at 17:11
  • That should tell you that the number of irreducible polynomials in $k[x]$ has the same cardinality as $\bar{k}$, which has the same cardinality as $k$. – jgon Jun 06 '19 at 17:14
  • @jgon I see now, thanks! – jon Jun 06 '19 at 17:15

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As $k$ is an infnite field, then $R=k[x_1,\ldots,x_d]$ has the same cardinality as $k$. The set of finite sequence of elements of $R$ also must have cardinality $|k|$. But $R$ is Noetherian: each ideal is generated by a finite sequence of elements of $R$. The cardinality of the set of all ideals of $R$ is at most $|k|$.

Angina Seng
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  • and to see that the relevant sets have cardinality at least $|k|$ we can consider for example the ideals of the form $(x_1-a_1, \dots, x_d-a_d)$ for $a_1, \dots, a_d\in k$, right? – jon Jun 06 '19 at 17:19
  • Exactly, or even just $(x_1-a,x_2,\ldots,x_d)$. @AsuraPath – Angina Seng Jun 06 '19 at 17:21