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The problem (from D&F) is to prove that the ring $\mathbb{Z}[x]$ is not isomorphic to the ring $\mathbb{Q}[x]$.I can't come up to any idea, so I'm asking for a hint.

Invincible
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    In one of the rings the identity element has the property that to each positive integer $m$ there is an element $r$ of the ring such that $m\cdot r=1$. – Jyrki Lahtonen Apr 11 '17 at 12:35
  • Can you prove that $\Bbb{Z}$ is not isomorphic to $\Bbb{Q}$? That might be a good first step. Jyrki's hint will help here as well. – badjohn Apr 11 '17 at 12:39

2 Answers2

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Hint: What do you know about ideals in those rings? With respect to ideals, one of the rings is _______, while the other is not.

Ennar
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    Hint or not, that's too vague. – quid Apr 11 '17 at 12:44
  • @quid, well this one really is hard to balance. Any more details is a complete answer, is it not? I'd appreciate help, downvote is hardly useful. – Ennar Apr 11 '17 at 12:46
  • @quid, is it better now? – Ennar Apr 11 '17 at 12:50
  • Slightly. Yet, at the very least one could stress "What do you know about the ideals in Q[X]?" This is more likely to put somebody on the right track. In addition I do not think that giving away the keyword principal ideal would amount to a complete answer at that level. Anyway it's rather moot as the Q is closed as a dupe and there are numerous answers there. – quid Apr 11 '17 at 12:52
  • @Ennar, polynomial ring over field is always P.I.D, but I don't know how to prove that Z[x] is not. – Invincible Apr 11 '17 at 12:53
  • @Vladislav, a standard way is to write explicitly an ideal in $\Bbb Z[x]$ that is not principal. Try to come up with simple polynomials $p,q$ such that $(p,q)$ is not principal. You can find an answer in the linked question, or can ask me again if you can't find it. – Ennar Apr 11 '17 at 12:56
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Hint: Assume, for contradiction, that there exists a ring isomorphism $\varphi \colon \mathbb{Z}[x] \rightarrow \mathbb{Q}[x]$. Since $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ have identities (in fact the same identities), $\varphi (1) = 1$. Since $\varphi$ is surjective, $\ldots$

Mark Twain
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