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I'm practicing reducing polynomials in different rings, but I'm stuck on a few. I would appreciate any help.

1) I want to factor $7+i$ in $\mathbb Z[i]$. The norm is $50$, so we know it's reducible.

I tried breaking down the norm so it's $(1+2i)^2(1-2i)^2(1+i)(1-i)$ I did $(1+2i)^2 (1+i) = -7 + i$ that's not quite correct though.

2) $x^4 + 2x^3 + x^2 + 2x + 1$ in $\mathbb Q[x]$ I tried to reduce mod $p$ where p$ = 2$ So it's $x^4 + x^2 + 1$ in $F_2$, which IS reducible.

I'm not sure what other prime makes sense though since any other choice won't make the terms disappear

3) $x^8 + x^6 + x^4 + x^2 + 1$ in $\mathbb Q[x]$ Here, I'm having the same issue with picking a suitable prime. But the solution says it's irreducible.

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    Since the norm of $7+i$ is 50 which is even, the gaussian integer $7+i$ is divisible by $1+i$. The other factor is given by $$\frac{1+7}{2}+i\frac{1-7}{2}=4-3i.$$ Thus, $$7+i=(1+i)(4-3i).$$ – 2017-02-15
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    I do suggest you to ask questions one at a time – 2017-02-15
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    You also have the (simpler, in my opinion) factorisation $(2+i)(3-i)$. – 2017-02-15
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    @Bernard, yes, but that's not a factorization into irreducibles, since $3-i$ can be factored further. – 2017-02-15
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    Your comment makes a lot of sense, but quick question. Once an element has an even norm, you know it's divisible by an element with norm 2 automatically? or like how do you reason about that? – 2017-02-15
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    If $z=a+bi$ has an even norm, then either $a$ and $b$ are both even, and life is easy, or they're both odd, and then you can show $z$ is divisible by $1+i$. – 2017-02-16

2 Answers 2

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3) is reducible, and factors as

$$ X^8 + X^6 + X^4 + X^2 + 1 = (X^4 + X^3 + X^2 + X + 1)(X^4 - X^3 + X^2 - X + 1) $$

The easiest way to see this is to note that the roots of this polynomial are tenth roots of unity, and the cyclotomic field $ \mathbf Q(\zeta_{10}) $ has degree $ \varphi(10) = 4 $ over $ \mathbf Q $.

2) is irreducible modulo $ 3 $: its roots all have order $ 10 $, and the order of $ 3 $ modulo $ 10 $ is $ 4 $, so that the roots have degree $ 4 $ over $ \mathbb F_3 $. You already have a good enough hint for 1), so I will not say anything further.

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For $1)$ the comment is excellent. In the Gaussian integers, an element has even norm iff it's divisible by $(1+i)$. I'll address the other two.

For 3) See Starfall's answer or note that it looks like a cyclotomic polynomial of degree $4$ and go directly to town trying to find a polynomial of degree $4$ to multiply $x^4 + x^3 + x^2 + x + 1$ by to get your polynomial.

For $2)$ don't forget that sometimes it's quick to attempt a brute force because Gauss' lemma lets us stay in the integers. Let $P(x) = x^4 + 2x^3 + x^2 + 2x + 1$. $P(-2) = P(0)= 1$ and clearly on either side of that $P(x) > 0$, so checking that $P(-1) \not= 0$ we see $P(x)$ has no linear factor. Now try to write $$P(x) = (ax^2 + bx + c)(dx^2 + ex + f)$$ We see immediately that $a=d=1$ WLOG and $f=c=\pm1$. Then multiplying out we get $$x^4 + (e + b)x^3 + (eb \pm2)x^2 \pm (e+b)x + 1$$ Matching coefficients with $P(x)$ we see that the $-$ of the $\pm$ is impossible so $c=f=1$, and we get the equations $$e+b = 2$$ and $$2+eb = 1$$ The second requires $eb = -1$, so $e=-b$ and the first is impossible. That means $P(x)$ is irreducible over the integers and therefore also over the rationals.

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    The irreducibility of $ P(x+1) $ doesn't imply the irreducibility of $ P(x^2) $. – 2017-02-15
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    Damn good point I was really thinking backwards there. – 2017-02-15