2

I was trying to prove that $x^4+3x^3+3x^2-5$ is irreducible over $\mathbb{Q}$ but I have troubles understanding the solution. The solution reads:

Consider the polynomial mod 2: $x^4+3x^3+3x^2-5 \equiv x^4+x^3+x^2+1 = (x+1)(x^3+x+1)$. Hence if $x^4+3x^3+3x^2-5$ was reducible over $\mathbb{Q}$, it must be a product of a linear polynomial and a cubic polynomial. But it is easy to check that $x^4+3x^3+3x^2-5$ have no rational root by rational root theorem and hence it has no linear factors and the result follows.

I do not understand the statement "if $x^4+3x^3+3x^2-5$ was reducible over $\mathbb{Q}$, it must be a product of a linear polynomial and a cubic polynomial".

For example, let $f(x)=(x^2+1)(x^2+2)$ is reducible over $\mathbb{Q}$, and $f(x) \equiv (x+1)^2x^2 \equiv (x+1)(x^3+x^2)\pmod{2}$. Yet $f$ still has no linear factor. Am I missing something here?

Thanks.

Ishigami
  • 2,202

1 Answers1

3

First of all, by Gauss's lemma, irreducibility over $\mathbb{Q}$ is here equivalent to irreducibility over $\mathbb{Z}$. There are no rational roots, so the only possibility is that the polynomial is a product of two monic polynomials in $\mathbb{Z}[x]$ of degree $2$, say:

$x^4+3x^3+3x^2-5=(x^2+ax+b)(x^2+cx+d)$

But now reducing the polynomial mod $2$ we get that its mod $2$ reduction is also a product of two (not necessary irreducible) polynomials of degree $2$. But that is a contradiction, because the reduction mod $2$ has an irreducible factor of degree $3$, and such factorization is unique.

Note that in your other example with $(x^2+1)(x^2+2)$ you don't get that contradiction, because in that case the reduction mod $2$ is indeed a product of two polynomials of degree $2$. (not irreducible ones, but it doesn't matter)

Mark
  • 43,582