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Why is $p(x)=x^2+t\in\mathbb{F}_2(t)[x]$ irreducible?

The only argument I can think of is that $\sqrt t$ doesn't seem like a rational function, but I tried to prove this and got stuck when I had to square a polynomial...is this the way to prove this claim, or is there another argument ?

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3 Answers

up vote 4 down vote accepted

My algebra is rusty, so hopefully I don't make any embarassing mistake:

$\mathbb{F}_2[t]$ is an integral domain where the ideal $\langle t \rangle$ generated by $t$ is maximal, thus prime.

Then, by the Generalized Eisenstein Criterion the polynomial $x^2+t$ is irreducible over both $\mathbb{F}_2[t]$ and its field of fractions.

P.S. You actually don't even need to employ the GEC. All you need is the Gauss Lemma, which sais that since $\mathbb{F}_2[t]$ is an UFD, then $x^2+t$ is irreducible over $\mathbb{F}_2(t)$ if and only if $x^2+t$ is irreducible over $\mathbb{F}_2[t]$. The last follows from Eisenstein, but Eisenstein is overkill here :)

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I can't figure why $\langle t \rangle$ is maximal... –  Belgi May 6 '12 at 16:12
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@Belgi Isn't $\mathbb{F}_2[t]/ \langle t \rangle \sim \mathbb{F}_2$? You can also prove directly that $\langle t \rangle$ is prime. To see this you can probably apply the First Isomorphism Theorem to $\phi : \mathbb{F}_2[t] \to \mathbb{F}_2 \,;\, \phi(P)=P(0)$. –  N. S. May 6 '12 at 16:14
    
@ZevChonoles are you sure ? What about $\langle 4 \rangle$ contained in $\langle 2 \rangle$ in $\mathbb{Z}$ –  Belgi May 6 '12 at 16:16
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@Belgi $\langle 4 \rangle$ is not prime. He said that any prime ideal is maximal. BTW, you don't even need that $\langle t \rangle$ is maximal, you only need it is prime. And keep in mind that as Zev mentioned, $\mathbb{F}_2[t]$ is PID, so it is enough to prove that $t$ is an irreducible element here. –  N. S. May 6 '12 at 16:16
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You could try the following: If $p$ is reducible, it will neccesarily reduce to a pair of linear factors. Furthermore, as $p$ is monic, one can assume that the factors are monic as well. Thus, if $p$ were reducible, $p = (x+a)(x+b)$. By comparison of coefficients, $a + b = 0$ and $ab = t$, so $b = -a$ and thus $-a^2=t$. Now, $a = q_1/q_2$, where $q_1, q_2$ are polynomials in $t$, so $-q_1^2/q_2^2 = t$, or equivalently, $-q_1^2 = t q_2^2$. But since $\deg q_1^2$ is even and $\deg t q_2^2$ is odd, no such factorization can exist. This is, essentially, a rigourous version of your argument.

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Just a thought. Say that $p(x) = x^2 + t \in \mathbb{F}_2(t)[x]$ is reducible. Then there is a rational function $\frac{q(x)}{d(x)}$ with $q(x), d(x) \in \mathbb{F}_2(t)$ such that $$\begin{align} \left(\frac{q(x)}{d(x)}\right)^2 &= -t = t\quad \Rightarrow \\ q(x)^2 &= t d(x)^2. \end{align} $$ Now the polynomial with coefficients in the field $\mathbb{F}_2(t)$ on the left hand side has even degree (being the square of a polynomial), but the polynomial on the right hand side has odd degree, so we have a contradiction.

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