# Every irreducible polynomial over $\mathbb F_p$ has a root in $\mathbb F_{p^{\deg f}}$ [duplicate]

I found the following question in my Galois theory book:

Let $$F$$ be a field with $$|F|=p^2$$ for some prime $$p$$. Show that $$a^2=5$$ for some $$a\in F$$, and generalize this statement.

My supposed proof is this: Let $$a$$ be a root of $$x^2-5$$, and suppose it is not already in $$\mathbb F_p$$. Then $$\mathbb F_p(a)$$ is a degree $$2$$ extension and so by uniqueness of fields of a certain order, $$F=\mathbb F_{p^2}=\mathbb F_p(a)$$ . Q.E.D.

This seems to me like it is a bit too easy and we didn't really do any work, and if this were to work, then we can generalise it to

Let $$|F|=p^n$$, and suppose $$f$$ is an irreducible polynomial over $$\mathbb F_p$$ of degree $$n$$. Then it has a root in $$F$$.

To me this sounds a bit too good to be true, but since I'm relatively new to field theory I'm not too sure. Is this correct?

• This looks like the correct statement to me
– Mike
Commented Apr 18, 2020 at 2:17

Your proof and conjecture are correct. To elaborate a little bit further, if $$f(x)$$ is an irreducible polynomial of degree $$n$$ over $$\mathbb{F}_p$$, then $$\mathbb{F}_p[x]/(f(x)))$$ is a field with $$p^n$$ elements.
As you mention, finite fields of the same order are unique up to isomorphism. Therefore, there is an isomorphism $$\phi: \mathbb{F}_p[x]/(f(x)) \to F$$. Viewing $$f$$ as a polynomial in $$\mathbb{F}_p[x]/(f(x))$$ we see that $$\overline{x}$$ (the coset containing $$x$$) is a zero in the quotient $$\mathbb{F}_p[x]/(f(x))$$. Since the $$\phi$$ fixes $$\mathbb{F}_p$$, we see that $$\phi(\overline{x}) \in F$$ is a zero of $$f$$ in $$F$$.