# Find an irreducible polynomial in this case, if it exists

Is there a second degree polynomial $$P(x)\in\mathbb{Q[x]}$$ so that it is irreducible in $$\mathbb{Q}$$ but reducible in $$\mathbb{Q[i]}\in\Bbb{C}$$?

I know that an irreducible polynomial cannot be factored into the product of two non-constant polynomials and it depends on where the coefficients are taken from. What can I say in this case? Any help or hint is much appreciated.

• A polynomial of degree $2$ is reducible over a field if and only if it has a root in that field. (Do you understand why?). If you know that, it should be really easy to find an example.
– Mark
Jun 8 at 15:51
• Thank you for the comment. No, unfortunately I don't know why since this is the first time we are doing polynomial over rings or fields. And also, since Q isn't a field, could you please explain me also the connection here? Jun 8 at 15:55
• Why $\mathbb{Q}$ isn't a field? It is. Anyway, if a polynomial $f\in F[x]$ is reducible then it is a product $f=gh$ of two polynomials of lower degree. But if the degree of $f$ is $2$ then the polynomials $g,h$ must have degree $1$. So $g$ has the form $g(x)=ax+b$ for some $a,b\in F$ with $a\ne 0$. So $\frac{-b}{a}\in F$ is a root of $g$, and hence a root of $f$.
– Mark
Jun 8 at 16:03
• So if a monic 2nd degree polynomial is reducible, it has a monic factor of degree 1. That gives a root. Jun 8 at 16:04

A second degree polynomial is reducible in any field means the roots of the polynomial exist in that Field. So you need a second degree polynomial which is irreducible in $$\mathbb{Q[x]}$$ but in $$\mathbb{Q[i]}$$. So overall you just need to find a equations whose roots are not rational but either real or complex . So you can find many examples over it. Very simple example i can give you is
$$X^2+ 1\in\mathbb{Q[x]}$$ but irreducible there.
it's reducible in $$\mathbb{C[x]}$$ as into $$(x-i)(x+i)$$ .