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I have a quick question.

Suppose $R$ is a nonzero commutative ring. Is it possible that $|\operatorname{Spec}(R[X])|<\infty$?

Hailie Mathieson
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  • if $R[x]$ is finite... – Don Larynx Nov 29 '13 at 01:48
  • If $R$ is infinite then take a maximal ideal $\mathfrak{m}$ of $R$ and consider the ideals $(\mathfrak{m},x-a)$ where $a\in R$. – Alex Youcis Nov 29 '13 at 01:48
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    @DonLarynx If $R$ is non-zero $R[x]$ is never finite since each of $x^n$ are distinct. – Alex Youcis Nov 29 '13 at 01:49
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    Actually, I guess the geometric answer is this. Take a surjection $R\to k$ where $k$ is a field. Then, one obtains a surjection $R[x]\to k[x]$ and thus a closed embedding $\text{spec } k[x]\to\text{spec } R[x]$. Clearly $\text{spec } k[x]$ is infinite for all $k$ and since $\text{spec } k[x]\to\text{spec } R[x]$ is injective the conclusion follows. – Alex Youcis Nov 29 '13 at 01:51
  • Almost a duplicate of http://math.stackexchange.com/questions/496202/operatornamespeckx-has-infinite-points/496204#496204. –  Nov 29 '13 at 06:08
  • @Alex: Well this "geometric argument" can be easily turned into a purely algebraic one, namely that maximal / prime ideals pull back along surjective ring homomorphisms. And actually you use this, of course. An honest geometric argument would produce points of $\mathbb{A}^1_R$ geometrically. After base change it suffices to consider $\mathbb{A}^1_K$ for a field $K$. And there are many reasons why $\mathbb{A}^1_K$ is infinite, for example otherwise - since it is an irreducible curve - it would just be a point. But we see two $K$-rational points, at least ... – Martin Brandenburg Nov 30 '13 at 11:56
  • @MartinBrandenburg Which is also only geometric until it is pointed out we are just tensoring with my $k$ to reduce to the field case, and then using a (Krull) dimension argument to only need to produce more than one point. Both arguments are only geometric if you believe. – Alex Youcis Nov 30 '13 at 12:11

2 Answers2

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Suppose that $R$ is non-zero. Then, we can find a surjection $R\to k$ for some field $k$, which induces a surjection $R[x]\to k[x]$. From this we obtain a closed embedding $\text{Spec }k[x]\to\text{Spec }R[x]$ and thus it suffices to prove that $\text{Spec }k[x]$ is infinite for $k$ a field.

To do this we can proceed as follows. If $k$ is infinite then $(x-a)\in\text{spec }k[x]$ for all $a\in k$. If $k$ is finite, say $k=\mathbb{F}_q$, then we have infinite elements of $\text{MaxSpec }k[x]$ corresponding to the field extensions $\mathbb{F}_{q^n}$ for every $n\in\mathbb{N}$.

Note that the above actually shows that even $\text{MaxSpec }R[x]$ is infinite if $R\ne 0$.

Alex Youcis
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    +1. To prove that there are infinitely many primes in $k[x]$ you can also imitate Euclid's proof! – Bruno Joyal Nov 29 '13 at 02:11
  • @BrunoJoyal That's true :) Once you're thinking geometrically it's hard not to see those stuck together points. – Alex Youcis Nov 29 '13 at 02:23
  • @AlexYoucis: Nice answer! This maybe a silly question by why does there always exist a surjection $R\rightarrow k$ for some field $k$? – Moss Nov 29 '13 at 02:29
  • @Sebastian Take $m$ a maximal ideal of $R$. The surjection you want is $R \to R/m$. –  Nov 29 '13 at 02:39
  • @Benja: Yes, very good. – Moss Nov 29 '13 at 02:48
  • @Alex Youcis I can not see why you can deduce that "$MaxSpec R[x]$" is infinite if $R≠0$, since your closed embedding may not take maximal ideal to a maximal one. Could you please explain in some detail? – Lao-tzu Jan 10 '14 at 05:56
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Let $J=\operatorname{Rad}(R[X])$ be the Jacobson radical. I claim $R[X]/J$ has infinitely many ideals.

Consider the principal ideal $I=(X+J)$ in $R[X]/J$. I claim that $I^n\supsetneq I^{n+1}$ for all $n\geq 1$. The containment follows immediately from definition. Suppose that $I^n=I^{n+1}$ for some $n$. Then $X^n+J\in I^n=I^{n+1}$, so $X^n+J=pX^{n+1}+J$ for some $p\in R[X]$. This implies $X^n-pX^{n+1}\in J$. Recalling $x\in J$ if $1-xy$ is a unit for all $y$ in the ring (cf. Atiyah & MacDonald Chapter 1), we have that $1-(X^n-pX^{n+1})$ is a unit in $R[X]$. However, $1-(X^n-pX^{n+1})$ is a unit iff the constant coefficient is invertible, and all other coefficients are nilpotent (cf. A&M Chapter 1, Exercises 1,2). But $X^n$ has coefficient $-1$, which is not nilpotent. So necessarily $I^n\supsetneq I^{n+1}$, and we get an infinite descending chain of ideals $I\supsetneq I^2\supsetneq I^3\supsetneq\cdots$.

Now suppose that $R[X]$ has only finitely many prime ideals, and thus only finitely many maximal ideals, since maximal ideals are prime. Enumerate the maximal ideals as $\mathfrak{m}_1,\dots,\mathfrak{m}_n$. Since the maximal ideals are clearly pairwise coprime, the Chinese Remainder Theorem implies $$ R[X]/J=R[X]/\bigcap_{i=1}^n\mathfrak{m}_i\simeq R[X]/\mathfrak{m}_1\times\cdots\times R[X]/\mathfrak{m}_n=:S. $$ It is a standard result that the ideals of $S$ have form $I_1\times\cdots\times I_n$ where $I_i$ is an ideal of $R[X]/\mathfrak{m}_i$. But each $R[X]/\mathfrak{m}_i$ is a field, hence has only trivial ideals. So $S$ has only $2^n$ ideals, a contradiction since $R[X]/J$ has infinitely many ideals. So $R[X]$ must have infinitely many prime ideals, in fact, it must even have infinitely many maximal ideals.

user26857
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Ben West
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