# Showing that $(\mathbb{F}_q[x]/(f_i(x)))^{F_q} = \mathbb{F}_q$

In a proof in my syllabus of a number theory course, they use that

$$(\mathbb{F}_q[x]/(f_i(x)))^{F_q} = \mathbb{F}_q$$

where $f_i(x)$ is irreducible, $F_q$ is the frobenius automorphism of $\mathbb{F}_q$ and $\mathbb{F}_q$ is the finite field of size $q$ (not necessarily prime). The book also says that this can be proven using properties of finite fields. However, I flipped through me algebra book and I didn't see any particular property that I could use to show this.

Could anyone give me some pointers?

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The field of order $q$ , where $q$ is a power of $p$, is the splitting field of $x^q - x$ over $\mathbb{F}_p$ and all its elements are roots; hence every element of the field $\mathbb{F}_q$ is fixed by the $q$-Frobenius automorphism $a\mapsto a^q$.
Conversely, any element of $\mathbb{F}_q[x]/(f_i(x))$ (which is a field extension of $\mathbb{F}_q$) that is fixed by the $q$-Frobenius automorphism must be a root of $x^q-x$. Since this polynomial has at most $q$ roots in this field, and the $q$ elements of $\mathbb{F}_q$ are already among the roots, it follows that the elements of $\mathbb{F}_q$ are the only ones fixed by $a\mapsto a^q$.