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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?

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    $\begingroup$ This looks like the correct statement to me $\endgroup$
    – Mike
    Commented Apr 18, 2020 at 2:17

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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 $F$.

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