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Let $R$ be a commutative ring. Clearly the kernel of $h$ is a prime ideal whenever $h : R \rightarrow \mathbb{Z}$ is a ring homomorphism. But is the converse true: does every prime ideal arise as kernel of a homomorphism into $\mathbb{Z}$?

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    $\begingroup$ Since the quotient group of an ideal is an integral domain iff the ideal is prime, this translates to "Given an integral domain $R$, is there always a homomorphism into $\Bbb Z$ with trivial kernel?" $\endgroup$
    – Arthur
    Oct 1, 2012 at 8:02
  • $\begingroup$ @Arthur, yes, I had commutative rings in mind. I've edited the question accordingly. $\endgroup$ Oct 1, 2012 at 8:08
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    $\begingroup$ As @Arthur's comment shows, is this wrong, since a homomorhpism $\mathbb Z[X] \to \mathbb Z$ can't have trivial kernel ... so the ideal $(Y) \subseteq \mathbb Z[X,Y]$ is prime but arises not as a kernel of a homomorphism to $\mathbb Z$. $\endgroup$
    – martini
    Oct 1, 2012 at 8:09

3 Answers 3

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Let $R$ be a field of uncountable cardinality. For a minimal counterexample, let $R = \mathbb{F}_2$.

The correct salvage is that every prime ideal arises as the kernel of a homomorphism into some integral domain (in fact, into some field). It shouldn't be possible to say anything stronger than this.

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    $\begingroup$ "Into some integral domain" and "into some field" are equivalent statements, since any integral domain is contained in its field of fractions. Just as a side comment. $\endgroup$
    – Arthur
    Oct 1, 2012 at 8:13
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    $\begingroup$ Yes, but they're not identical statements; it takes a tiny bit of work to show that they're equivalent, and if I had said "surjective homomorphism" then they wouldn't be equivalent. I only mentioned integral domains because $\mathbb{Z}$ is an integral domain, not a field. $\endgroup$ Oct 1, 2012 at 8:14
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    $\begingroup$ Because $R/\mathfrak p$ is an integral domain into which there is a surjective homomorphism with exactly $\mathfrak p$ as kernel, but it's a field only if $\mathfrak p$ is maximal, right? $\endgroup$
    – Arthur
    Oct 1, 2012 at 8:17
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    $\begingroup$ @Marc: if something doesn't exist for cardinality reasons, then it really doesn't exist. It's a fun proof technique to use because it rarely applies, but when it does, it really applies. $\endgroup$ Oct 1, 2012 at 8:24
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    $\begingroup$ Only a minor comment: The first two sentences being in the same line makes them seem related, but they are not. Do you mind separating them (e.g. by an 'or')? $\endgroup$ Oct 1, 2012 at 9:04
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The converse is very, very false. For instance if $R$ is a ring of nonzero characteristic (think $\mathbb Z/n\mathbb Z$ with $n>0$), there aren't any homomorphismes $R\to\mathbb Z$ at all, yet $R$ will always (assuming the axiom of choice) have at least one prime ideal. Simple counterexample $\mathbb Z/2\mathbb Z$ and its zero ideal, or (if you don't like zero ideals) $\mathbb Z/4\mathbb Z$ and its ideal $\{\overline0,\overline2\}$.

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    $\begingroup$ @HagenvonEitzen Below "false" there is "wrong", "really wrong" and then "not even wrong" :P $\endgroup$
    – rschwieb
    Oct 1, 2012 at 12:46
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If $h: R \rightarrow \mathbb{Z}$ is a ring homomorphism, then $h$ is also a homomorphism of (additive) abelian groups. If $P$ is the kernel of $h$, then $R/P$ is isomorphic to the image of $h$, which is a subgroup of $\mathbb{Z}$. And a subgroup of $\mathbb{Z}$ is either trivial or isomorphic to $\mathbb{Z}$.

So if $P$ is a prime ideal of $R$ (prime ideals are by definition proper, so $R/P$ is never trivial) such that $R/P$ is not isomorphic to $\mathbb{Z}$ as an abelian group, then $P$ cannot possibly be the kernel of a homomorphism $R \rightarrow \mathbb{Z}$. There are plenty of prime ideals which fit this description. If $R$ is $\mathbb{Z}$ itself, and $P = \mathbb{Z} p$ for some prime number $p$, then $R/P$ is finite!

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