I am having a lot of trouble figuring out which ideals are maximal in rings that aren't fields. I read that maximal ideals of $\mathbb{Z}[x]$ are of the form $(p,f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in $\mathbb{Z}[x]$ which is irreducible modulo $p$.
$\\$I am trying to apply this to figuring out if the ideals generated by $x+1$ and $x^2+x+1$ are maximal in $\mathbb{Z}$. I've been told to consider the ideals $(2,x+1)$ and $(2, x^2+x+1)$ respectively, but I don't get where the two comes from. Can you pick any prime number to check this condition? Any explanation of how to proceed would be appreciated. Thanks
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Vinny Chase
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$2$ is a prime and $x+1$ and $x^2+x+1$ are both irreducible modulo in $\mathbb{Z}_2[X]$. So it fits the description. – Henno Brandsma Mar 19 '18 at 17:32
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@HennoBrandsma so you can use any prime number? – Vinny Chase Mar 19 '18 at 17:32
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I think so. But if you use $p$ the polynomial should be irredicible over $\mathbb{Z}_p[X]$ instead. – Henno Brandsma Mar 19 '18 at 17:33
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@HennoBrandsma I think then that these would both be maximal, but I've also found this question though which would seem to contradict that: https://math.stackexchange.com/questions/1591558/nonconstant-polynomials-do-not-generate-maximal-ideals-in-mathbb-zx – Vinny Chase Mar 19 '18 at 17:34
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These ideals have two generators, not $1$, so what's the contradiction? – Henno Brandsma Mar 19 '18 at 17:35
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The example uses $2$ as it's the easiest to check irreducibility of $p(x)$ for, I suppose. – Henno Brandsma Mar 19 '18 at 17:36
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See here for a proof of your statement. – Henno Brandsma Mar 19 '18 at 17:38
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@HennoBrandsma Ok I see that. Thank you. Thats actually where I got the theorem from in the first place. This may be really dumb since I am teaching myself this course, but I still don't understand why though the ideal (2,x+1) is the same as (x+1) – Vinny Chase Mar 19 '18 at 17:43
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$(2,x+1)$ is not the same as $(x+1)$. $x+3$ is in the former but not in the latter. – Henno Brandsma Mar 19 '18 at 19:17
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@HennoBrandsma So why can we just switch to it to show that (x+1) is maximal in $\mathbb{Z}$? – Vinny Chase Mar 19 '18 at 19:34
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@HennoBrandsma why is $M_{1}$ is a proper ideal of $\mathbb{Q}[x]$ in the link you provided? – Delong Mar 19 '18 at 22:30
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We show that $(x+1)$ is not a maximal ideal in $\mathbb{Z}[x]$. Since $(x+1)\subset (2,x+1)\subset \mathbb{Z}[x]$, we just need to show that $(x+1)\neq (2,x+1)$ and $(2,x+1)\neq \mathbb{Z}[x]$. The first statement holds because $2\notin (x+1)$. The second statement is true by the result you said in the first paragraph.
Delong
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Thank you! So similarly for $(x^2+x+1)$, its not maximal because there exists this other ideal $(2,x^2+x+1)$ that contains it but isn't equal to all of $\mathbb{Z}[x]$? – Vinny Chase Mar 19 '18 at 22:30
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@VinnyChase Yes. It is similar for the other one, and use the definition of maximal ideal. – Delong Mar 19 '18 at 22:32