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Show that $x^5-(3+i)x+2$ is irreducible in $(\mathbb{Z}[i])[x]$.

$\mathbb{Z}[i]$ is a UFD and hence $(\mathbb{Z}[i])[x]$ is a UFD. So they are integral domains. Thus I can use Eisenstein's Criterion here.

$2=(1+i)(1-i)$ and these factors are irreducibles, and hence primes. Let $P=((1+i))$.

$-(3+i)=-(1+i)(2-i)$. So $-(3+i),2\in P$. But $P^2=(2i)$ and $2\in P^2$ too. So I cannot use Eisenstein's Criterion. Am I missing something here? Or else what other methods I can use here in order to show the irreducibility?

Extremal
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2 Answers2

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Sorry for being stupid earlier, substitute $x$ by $x-i$ and use $1+2i$, a factor of $5$ and conjugate of $2-i$ as your prime.

Rene Schipperus
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  • Why I want to substitute for $x$? – Extremal Jan 01 '16 at 17:58
  • Its a standard trick. It wont change the reducibility and the resulting poly satisfies Eisenstein. – Rene Schipperus Jan 01 '16 at 18:00
  • I have never heard that trick. I am wondering how can I prove that it wont change the reducibility. – Extremal Jan 01 '16 at 18:06
  • If $p(x-i)=q(x)r(x)$ then $p(x)=q(x+i)r(x+i)$ – Rene Schipperus Jan 01 '16 at 18:09
  • Well spotted. I used the other factor of $5$ and came up with a different argument. I had to fire up Mathematica to check your calculation, because I thought (due to my argument) that the polynomial can have zeros mod five, but it CAN (as you spotted) have a zero modulo the other factor of five :-) I apologize for first doubting the correctness of your calculation. – Jyrki Lahtonen Jan 01 '16 at 18:53
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In addition to Eisenstein's criterion an often useful tool in the study of a polynomial with (algebraic) integer coefficients is to reduce everything modulo a suitable prime. If the reduced polynomial can be shown to be irreducible over the quotient ring (it is a field actually), then the original polynomial must have irreducible as well.

First an example. We can tell that the cubic $p(x)=x^3-x+1$ is irreducible over the field $\Bbb{F}_3=\Bbb{Z}/3\Bbb{Z}$. This is because $a^3\equiv a\pmod3$ for all integers $a$, so $p(x)$ has no zeros in $\Bbb{F}_3$. Therefore it has no linear factors and, being cubic, it must be irreducible.

So for example a polynomial like $$ x^3-3333x^2-1000x+1234 $$ is automatically irreducible in $\Bbb{Z}[x]$ because modulo three it is congruent to $x^3-x+1$. (By Gauss Lemma it is then also irreducible in $\Bbb{Q}[x]$.)

On with your quintic. Hopefully you know that polynomials $x^p-x-a$, $a=1,2,\ldots,p-1$ are all irreducible in $\Bbb{F}_p[x]$. If you don't, then read this - the question has been asked umpteen times. Anyway, this result generalizes the above $p=3$ irreducibility result from my example, and comes in handy here.

Consider the prime $p=2+i$ of $\Bbb{Z}[i]$. We easily see that $\Bbb{Z}[i]/(p)\cong\Bbb{Z}/5\Bbb{Z}= \Bbb{F}_5.$ But modulo $p$ your polynomial is congruent to $$ x^5-(3+i)x+2\equiv x^5-x+2\pmod{2+i}. $$ So after reduction modulo $p$ the polynomial is irreducible over $\Bbb{Z}[i]/(p)$. Therefore it must have been irreducible over $\Bbb{Z}[i]$ to begin with.

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Jyrki Lahtonen
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