Timeline for Maximal ideals of Z[x,y]
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 25, 2012 at 15:38 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| May 22, 2012 at 14:19 | comment | added | Steven Landsburg | Yves Cornulier: Yes, one needs that a field finitely generated (as a ring) over Z is finite, and as you observe this is not entirely trivial. Both my answer and Will Sawin's comment took this for granted. | |
| May 21, 2012 at 21:15 | comment | added | YCor | That every field of characteristic zero is infinitely generated as a ring is true but not a tautology. Sure, it contains $\mathbf{Q}$; sure, $\mathbf{Q}$ is not finitely generated as a ring. If you have in mind that every subfield of a finitely generated field is finitely generated, this is true but this requires some argument. | |
| May 21, 2012 at 13:55 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| May 21, 2012 at 6:30 | comment | added | Martin Brandenburg | It was not obvious to me, neither to anna who asked this question. Each maximal ideal in Z[x,y] intersects Z in a prime ideal, and one has to argue that this prime ideal is not 0. This has been done by Will Sawin above. Perhaps you can just add these "obvious" arguments to your answer so that the reader doesn't have to struggle through our comments. Sorry for yapping.. | |
| May 20, 2012 at 12:58 | comment | added | Steven Landsburg | Martin Brandenburg: But it's clear that any maximal ideal in Z[x,y] contains a prime number p, and this immediately reduces the question to the linked question. I'd thought this was obvious. Are you saying that it's not obvious, or that it's not true? | |
| May 20, 2012 at 7:30 | comment | added | Martin Brandenburg | -1 since this answer doesn't explain what the maximal ideals of $Z[x,y]$ are; the linked question only talks about maximal ideals in polynomial rings over fields. Perhaps Will Sawin can post his comment as a real answer. | |
| May 19, 2012 at 21:17 | comment | added | Will Sawin | One can use the lemma: Let $R$ be a ring finitely generated over $\mathbb Z$. Then a maximal ideal of $R[x]$ is a maximal ideal of $R$ plus a polynomial in $R[x]$ that is irreducible modulo that maximal ideal. Apply to $R=\mathbb Z$, then to $R=\mathbb Z[x]$, etc. Proof: The quotient is some finitely generated field, thus some finite field, so the image of $R$ is a subfield, thus the quotient by a maximal ideal $I$. The field extension is generated by $x$, so it is the quotient of $F[x]$ by the minimal polynomial $m_x$, so it is $R[x]/(I,m_x)$. | |
| May 19, 2012 at 21:16 | vote | accept | Stella | ||
| May 19, 2012 at 20:55 | comment | added | Steven Landsburg | Anna: "$Z$ is not a field". But $Z/pZ$ is. | |
| May 19, 2012 at 20:54 | comment | added | Stella | $Z$ is not field. | |
| May 19, 2012 at 20:47 | history | answered | Steven Landsburg | CC BY-SA 3.0 |