# Polynomial of degree $5$ reducible over $\mathbb Q(\sqrt 2)$

Give an example of an irreducible monic polynomials of degree (a) $4$; (b) $5$ in $\mathbb Z[x]$ that is reducible over $\mathbb Q(\sqrt 2)$, or prove that none exists.

I managed to find a polynomial of degree $4$: $x^4+1=(x^2-\sqrt 2x+1)(x^2+\sqrt 2x+1)$, which is irreducible over $\mathbb Z$ (or equivalently $\mathbb Q$) because it has no rational roots by the rational root test, and cannot have a quadratic factor with rational coefficients because $\mathbb R[x]$ is a UFD, and the two quadratic factors have irrational coefficients.

I've been trying to construct an example for degree $5$, but it seems it's impossible. How can I prove it?

• If $p$ is irreducible over $\Bbb F$, then what can you say about the degree of the splitting field of $p$ over $\Bbb F$? – Travis Jul 19 '18 at 1:11
• It must be $5$ if $\deg p=5$. – user437309 Jul 19 '18 at 1:17
• If $f(x)$ has degree $5$, is irreducible over $\mathbb{Q}$ and reducible over $\mathbb{Q}(\sqrt{2})$, then: (1) It cannot have a root in $\mathbb{Q}(\sqrt{2})$, because the elements of this have minimal polynomials of at most degree $2$. (2) It must factor in a polynomial of degree $2$ and a polynomial of degree $3$. Then adjoining the roots of these two factor you get an extension of $\mathbb{Q}$ of degree dividing $2^3\cdot 3$. On the other hand, since $f$ is irreducible, extending by one of its roots gives an extension of degree $5$. Hence the splitting field of $f$ is multiple of $5$. – user574889 Jul 19 '18 at 1:19
• @user437309 On your comment above: Not $5$ exactly, rather divisible by $5$. – user574889 Jul 19 '18 at 1:21