Is $x^4+15x^3+7$ irreductible on $\mathbb{Q}[x]$?

Question

I have to discuss wether the polynomial $f(x)=x^4+15x^3+7$ is irreductible on $\mathbb{Q}[x]$.

To show this I will use a combination between the Gauss Criterion and the Modular criterion, which states that

Let $I$ be an ideal of $R$ and $f(x)\in R[x]$ a monic polynomial. If $f(x)$ has no trivial decompositions on $R/I[x]$, then $f(x)$ is irreductible on $R[x]$.

Using 1 we see that if $f(x)$ is irreductible on $\mathbb{Z}[x]$, so is on $\mathbb{Q}[x]$. To apply this second criterion we choose $R=\mathbb{Z}$ and $I=3\mathbb{Z}$ such that $[f(x)]=x^4+7\in\mathbb{Z}/3\mathbb{Z}$. We can see that $[f(x)]$ has no roots on $\mathbb{Z}/3\mathbb{Z}$, thus being irreductible. By virtue of the Modular criterion we get that $f(x)$ is also irreductible.

Is my reasoning correct? Have I commited any error? Is it true? How would you solve this problem?

Answer 1

The commenters are correct; although no roots exist for $x^4+1\equiv0\bmod3$, the polynomial could have quadratic factors. Indeed, the polynomial is also $\equiv x^4+4\bmod3$ and then Sophie Germain's Identity immediately gives $(x^2-2x+2)(x^2+2x+2)$ (or $(x^2+x-1)(x^2-x-1)$).

To prove the nonexistence of quadratic factors over the rationals you can use another modulus. As the cubic term disappears $\bmod 5$ you may want to explore that ring. A quadratic factorization of $x^4+2\bmod 5$ must have one of the forms

$(x^2+ax+1)(x^2-ax+2)$ or

$(x^2+ax-1)(x^2-ax-2)$

If you can prove that neither candidate holds, you can proceed to the conclusion. Incidentally, $\bmod 5$ also admits no linear factor.

Answer 2

I give a proof that follows a personal idea I exposed in a congress several years ago.

First at all, it is easy to discard a linear factor so if $f(x)=x^4+15x^3+7$ would be reducible we would have $f(x)=(x^2+ax+b)(x^2+cx+d)$.

Now each time $f(n)$ is a rational prime for $n\in\mathbb Z$, one of the supossed quadratic factors should be equal to $\pm1$. It follows that $f(n)$ cannot be a prime for many values of $n$.

Since $f(n)$ is prime for $0,\pm1,\pm2,-3,5$, giving respectively the seven primes $7,23,-7,53,-7,43,557$, we have proved that $f(x)$ is irreducible.

NOTE.-We can calculate the maximum number of values $\pm1$ can have a polynomial but for this particular problem it is not necessary.

Answer 3

You don't need to go to $\mathbb{Z}_3$, the polynomial is already irreducible over $\mathbb{Z}_2$. Indeed, $\overline{f}(x)=x^4+x^3+1$ has no root in $\mathbb{Z}_2$, ruling out a linear factor. We can also check it is not divisible by the only irreducible quadratic $x^2+x+1$, so it must be irreducible.

Alternatively, if you have a degree $n$ monic polynomial $f\in \mathbb{Z}[x]$ with non-negative coefficients, you can check if the coefficient at $x^{n-1}$ is greater than sum of all the other coefficients. If so, then it is irreducible over integers and rationals. This is known as Perron's irreducibility criterion. In this case $15>1+7$ and the result follows.