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I think $X^7-(4+i)\in\mathbb{Q}(i)[X]$ is irreducible (simply because I don't know how to go about factoring it). Would it suffice to show that it is irreducible over $\mathbb{Z}[i]$?

If so, I can consider \begin{align*} X^7-(4+i)+\langle i\rangle=X^7-1+\langle i\rangle\in\dfrac{\mathbb{Z}[i]}{\langle i\rangle}[X]\cong\mathbb{Z}[X], \end{align*} where $\langle i\rangle$ is the prime ideal (since $\mathbb{Z}[i]/\langle i \rangle\cong\mathbb{Z}$ is a domain) generated by $i$. Am I completely off my rocker?

Michael Hardy
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user31415926535
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    You made a very bad error. In $\mathbf Z[i]$ the element $i$ is a unit, so $\mathbf Z[i]/(i)$ is zero rather than $\mathbf Z$. The ideal $(i)$ in $\mathbf Z[i]$ is the whole ring, not a prime ideal. This situation is not at all like $\mathbf Z[x]/(x) \cong \mathbf Z$. – KCd Dec 24 '14 at 16:26
  • I see. We do have though $\mathbb{Z}[i]/(4+i)\cong\mathbb{Z}$, yes? – user31415926535 Dec 24 '14 at 16:32
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    Not by a long shot: $\mathbf Z[i]/(4+i)$ is finite. You really need to review modular arithmetic in the Gaussian integers. – KCd Dec 24 '14 at 17:10
  • Another error was in typesetting: writing $$ instead of $\langle i\rangle$. I fixed it. – Michael Hardy Dec 24 '14 at 18:39

2 Answers2

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Hint: I think you might be expected to notice that $4+i$ is prime in $\mathbb Z[i]$ and to think about whether you can adapt Eisenstein's Criterion to this slightly different case.

Mark Bennet
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  • I see how $4+i$ is prime. Since $4+i\in<4+i>$ and $4+i\not\in<(4+i)^2>=<15+8i>$, by Eisenstein's we conclude irreducibility. Yes? In general, is it sufficient to show irreducibility in $\mathbb{Q}$ adjoin anything by showing irreducibility in $\mathbb{Z}$ adjoin the same thing? In general, this is because the field of fractions of $\mathbb{Z}$ is $\mathbb{Q}$? – user31415926535 Dec 24 '14 at 14:14
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    @Brandon: No, that idea does not work in general. Consider in $\mathbf Q(\sqrt{5})[x]$ the quadratic polynomial $x^2-x-1$. It is irreducible in $\mathbf Z[\sqrt{5}][x]$ but it factors in $\mathbf Q(\sqrt{5})[x]$. This is related to the issue of rings being integrally closed or not. Unlike $\mathbf Z[\sqrt{5}]$, the ring $\mathbf Z[i]$ is integrally closed. – KCd Dec 24 '14 at 16:21
  • @Brandon KCd's comment illustrates what I mean about thinking about this problem. If you study algebra you will find ideas like "integrally closed" often given as abstract definitions, but they were developed by thinking about problems which arise when you try to extend useful things you know about $\mathbb Z$ and $\mathbb Q$ to extensions. – Mark Bennet Dec 24 '14 at 17:48
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Let $R$ be a GCD domain and $K$ its field of fractions. A non-constant polynomial $f\in R[X]$ is irreducible if and only if it is primitive and irreducible in $K[X]$; see here.

Fortunately $\mathbb Z[i]$ is a GCD domain (it's even a Euclidean domain), so $X^7-(4+i)$ is irreducible in $\mathbb Q(i)[X]$ iff $X^7-(4+i)$ is irreducible in $\mathbb Z[i][X]$. Now you can use the Eisenstein's criterion for the prime ideal $\mathfrak p=(4+i)$. (Why is this a prime ideal? Well, this is proved e.g. here: $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$. In fact, $\mathbb Z[i]/(4+i)\simeq\mathbb Z/17\mathbb Z$ which is a finite field.)

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