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How can I prove that every integral domain has at least one prime ideal? I don't know if I'm overthinking it, but maybe I am. I know how to prove it for something like $\Bbb Z$, but I'm not sure how to prove it for the general case.

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Hint: What specific assumption about integral domains pertains to the ideal $0$?

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  • $\begingroup$ That the characteristic of an integral domain is either 0 or prime? I'm not sure... But every ring has 0 and itself as ideals $\endgroup$ – Alice Aug 24 '13 at 0:52
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    $\begingroup$ @Alice The special assumption is that if $ab = 0$, then $a = 0$ or $b = 0$. So suppose that you have ideals $A$ and $B$ with $AB \subset \{0\}$; can $A$ and $B$ be nonzero? $\endgroup$ – user61527 Aug 24 '13 at 0:57
  • $\begingroup$ I don't think so? Because one of them has to be zero, right? $\endgroup$ – Alice Aug 24 '13 at 1:12
  • $\begingroup$ @Alice Exactly. $\endgroup$ – user61527 Aug 24 '13 at 1:28
  • $\begingroup$ OH OH wow I feel silly. I get it now, thankyou! $\endgroup$ – Alice Aug 24 '13 at 1:39
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As the other answer has suggested, $\{0\}$ is in fact a prime ideal for an integral domain.

But this is a sort of bizarre question as any ring has at least one prime ideal. In particular, one can use a Zorn's Lemma argument to show that any ring has a maximal ideal, and then one can prove that maximal ideals are prime. To see the second part, note that if $\mathfrak{M}$ is not prime, then there are $a, b$ so that $ab \in \mathfrak{M}$ but neither $a \in \mathfrak{M}$ nor $b \in \mathfrak{M}$. But then $(a+\mathfrak{M})(b+\mathfrak{M}) \subset \mathfrak{M}$ so at least one of these ideals is proper and both properly contain $\mathfrak{M}$, which contradicts maximality of $\mathfrak{M}$.

I suppose the advantage in an integral domain is that one doesn't need the axiom of choice to find a prime ideal.

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  • $\begingroup$ The trivial ring $\{0\}$ is a ring with no prime ideals, if you call such a thing a ring. I personally do not. In any case, your answer certainly works for rings with unity. $\endgroup$ – RghtHndSd Aug 24 '13 at 2:11
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    $\begingroup$ @rghthndsd: Dear rghthndsd, It is common in commutative algebra to regard the trivial ring $\{0\}$ as a ring with unity, in which one happens to have $0 = 1$ (in fact it is the unique ring with unity in which this equation holds). (One advantage of this is that the empty scheme is an affine scheme, namely the Spec of $\{0\}$.) So I would rather say that Alexander's example works for non-trivial rings with unity. Of course this is a fairly minor point ... . Regards, $\endgroup$ – Matt E Aug 24 '13 at 3:10

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