# Proof that a Polynomial is Irreducible

In my abstract algebra class, we were given some polynomials to prove irreducible or not over the rationals. I have come up with a proof for this one, but I'm not sure if it's valid.

The polynomial: $$x^4 + 2x^3 + x^2 + x + 1$$

I've started by taking the modulus $2$ of all the coefficients to get:

$$x^4 + x^2 + x + 1$$There's a theorem in my book that says if you take the modulus using a prime and the new polynomial has the same degree and is irreducible, than the previous polynomial is also irreducible.

So to prove $x^4 + x^2 + x + 1$ is irreducible:

If it is reducible it can be written as $f(x)g(x)$. But because it has a constant term, both $f$ and $g$ must have a constant term. But this also implies neither $f$ nor $g$ can have an $x^3$ term because their product does not have an $x^3$ term. But since the product has an $x^4$ term, they both must have $x^2$ terms. Also, since the product has an $x$ term, at least one of $f$ and $g$ must have an $x$ term. But then the $x$ term times the $x^2$ term in the other would make an $x^3$ term, which the product does not have, so $f$ and $g$ cannot exist.

Is this proof solid? Is there a more elegant way?

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There appears to be a typo,\ (or error), the first polynomial has root $\, x = -1\$ (which remains a root mod $\,2).\ \$ – Bill Dubuque Jun 16 '14 at 3:50
The product f(x)g(x) not having an x^3 term does not imply f,g do not have an x^3 term. Nor does it having an x^4 term imply f,g both have x^2 terms. – blue Jun 16 '14 at 3:53
Both $f$ and $g$ could have cubic terms which are eliminated under some sort of cancellation when they're multiplied. – user61527 Jun 16 '14 at 3:55
Did you try the rational roots test? – Eric Towers Jun 16 '14 at 3:57
Is it possible that there is a typo, and the polynomial is really $x^4 - 2x^3 + x^2 + x + 1$? Because this one is reducible modulo $2$, but irreducible modulo $3$. – Andreas Caranti Jun 16 '14 at 7:54

## 1 Answer

Unfortunately, $x^4+x^2+x+1$ is not irreducible in the two element field, since $x=1$ is a root.

And the polynomial you started with is not irreducible over the rationals, since $x=-1$ is a root. Thus the polynomial $x^4+2x^3+x^2+x+1$ is divisible by $x+1$.

Remark: We have $x^4+2x^3+x^2+x+1=(x+1)(x^3+x^2+1)$. The polynomial $x^3+x^2+1$ is irreducible over the rationals. For a polynomial of degree $2$ or $3$ is irreducible over a field $F$ if and only if it has no root in that field. And by the Rational Roots Theorem, the polynomial $x^3+x^2+1$ has no rational roots.

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How is $x = 1$ a root of $x^4+x^2+x+1$? – Ivo Beckers Jun 16 '14 at 8:16
@IvoBeckers, modulo $2$. – Andreas Caranti Jun 16 '14 at 8:35