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Say I have a quotient ring $\mathbb Z/4\mathbb{Z}[x]/(x^4 + 2x + 1)$.

What is the process for finding the maximal ideas of this ring?

Martin Brandenburg
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csss
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    You don't mean $\mathbb{Z}_4=\lim_n \mathbb{Z}/4^n$ or $\mathbb{Z}[\frac{1}{4}]$, but rather $\mathbb{Z}/4 \mathbb{Z}$. Therefore I've changed the notation. – Martin Brandenburg Nov 13 '13 at 18:10
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    @MartinBrandenburg $\mathbb{Z}_n$ is an established notation for the ring of integers modulo $n$, one of many. – TBrendle Nov 13 '13 at 18:15
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    @MartinBrandenburg Please, no need for abusive language. It was clear from context what was meant. Your edit was condescending and unnecessary, as is your reply to me. – TBrendle Nov 14 '13 at 04:38

3 Answers3

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I assume you mean $\mathbb{Z}_4[x]$, the ring of polynomials over the integers modulo $4$. Let $p(x)= x^4 + 2x+1$. We seek the maximal ideals of $R=\mathbb{Z}_4[x]/(p(x))$.

First, notice that $p(3)=0$, so $p(x)$ is divisible by $x+1$. We have $p(x)=(x+1)(x^3 + 3 x^2 + x + 1)$. The second factor, $q(x) = x^3 + 3 x^2 + x + 1$, is irreducible mod 4 because it is of degree $3$ but has no roots mod 4.

We know that a polynomial ring modulo a polynomial is a field if the modulus is prime and the polynomial being modded out is irreducible. Using the isomorphism theorems, you can show that $R/(\overline{q(x)}, 2) \cong \mathbb{Z}_2[x] / (\overline{q(x)})$ and $R/(\overline{x+1}, 2) \cong \mathbb{Z}_2[x] /( x+1)$. Now, $\mathbb{Z}_2[x] / (\overline{q(x)})$ is not a field because $q(1)\equiv 0 \pmod{2}$ implies $q(x)$ is not irreducible mod 2 (in fact it is divisible by $x-1$). However, $\mathbb{Z}_2[x] /( x+1)$ is clearly a field. Therefore of these two ideals, only $(\overline{x+1}, 2)$ is a maximal ideal of $R$. On the other hand, by the isomorphism theorems, the ideals of $R$ correspond to the ideals of $\mathbb{Z}_4[x]$ containing $(p(x))$, which are obtained by the factorization of $p(x)$ and of the modulus $4$, so those were the only two candidates for maximal ideals, and so we are done.

TBrendle
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  • I'm not sure how you are able to say $q(x)$ is irreducible because it is of degree 3 but has no roots? – csss Nov 13 '13 at 17:23
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    @csss If a degree 3 polynomial factors then one of the factors must be degree 1. – TBrendle Nov 13 '13 at 17:25
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    @TBrendle: Be careful, $\mathbb{Z}_4[x]/(x+1)$ has non-zero zero-divisors, hence it is not a field; you must add a new generator, 2, to your ideal. – arienda Nov 13 '13 at 17:26
  • I.e. the factors must either be $(x + 3), (x + 2), (x + 1), (x + 0)$, yet $q(3), q(2), q(1), q(0)$ never equal $0$. Is that it? – csss Nov 13 '13 at 17:30
  • The question is about maximal ideals not prime ideals. The correct statements on this topic are actually "$R/I$ integral domain $\Leftrightarrow I$ prime" and "$R/I$ field $\Leftrightarrow I$ maximal." – anon Nov 13 '13 at 17:39
  • That is correct, but you know the original version had an incorrect statement that you now edited in response to my comment. Furthermore, when the question is to find maximal ideals, one does not generically respond with "here is how to find the prime ideals, because the maximal ideals are somewhere among them..." (although in nice cases, max and prime ideals coincide). – anon Nov 13 '13 at 18:14
  • @anon if you don't like my answer, feel free to post your own :) I tried to keep it as elementary as possible so everyone could understand. – TBrendle Nov 13 '13 at 18:22
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If you know about the prime ideals of $\Bbb Z[x]$, and you can recognize this ring as $\Bbb Z[x]/(4,x^4+2x+1)$, then the maximal ideals (among the prime ideals) will be $(p, f(x))$ where $p$ is a prime divisor of $4$ and $f(x)$ mod $p$ is an irreducible polynomial over $\Bbb F_p$ dividing $x^4+2x+1$.

So we've got just $2$ as a candidate for $p$, and then we are looking to see how $x^4+1\pmod{2}$ factors. Luckily, this is easy: $x^4+1=(x+1)^4$. So, you've got one maximal ideal: $(2,x+1)$

Community
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rschwieb
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A slight variation of the answer by rschwieb. Since the element $2$ of $R=\mathbb Z_4[x]/(x^4 + 2x + 1)$ is nilpotent, it is contained in every prime ideal. Therefore asking for the prime ideals or $R$ is equivalent to asking for the prime ideals of $R/2R\cong \mathbb Z_2[x]/(x^4 + 1)$. Now $\mathbb Z_2[x]$ is a PID, so the non-zero prime ideals are generated by irreducible polynomials. Since $x^4+1=(x+1)^4$ has only one irreducible factor, the only prmie ideal containing it is generated by that factor$~(x+1)$.

Concluding, $R$ has just one prime ideal, generated by $2$ and $x+1$; it is of course also maximal.

Marc van Leeuwen
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