Let $F$ be a finite field. If $f, g \in F[x]$ are irreducible polynomials of the same degree, show that they have the same splitting field.
I tried this problem by induction on the degree of the polynomials, but I couldn't get anything. Is there another way to approach this? Could someone help me on this? Thanks in advance!
I found this post and it is related to this question:
How to deduce the irreducible factors of $x^4 +1$ in $\Bbb F_3$.
On
I suppose that what’s below is just the same as the answer of @LordSharkTheUnknown, but maybe it’s different enough to give some help.
Let $F$ have cardinality $q$, and let the common degree of $f$ and $g$ be $d$. Then the field $K_f$ gotten by adjoining a root of $f$ is of degree $d$ over $F$, and consequently of cardinality $q^d$. But the multiplicative group of $K_f$, call it $K_f^\times$, is a cyclic group of order $q^d-1$, contained in the zero-set of $X^{q^d-1}-1$ (in some fixed algebraic closure of $F$), and therefore equal to this set, since they have the same cardinality. Thus $K_f$ is the zero-set of $X^{q^d}-X$. Same goes for $K_g$, the field gotten by adjoining a root of $g$ to $F$.
This not only proves that if $\alpha$ and $\beta$ are roots of $f$, then both generate the same field, so that $K_f$ is normal, similarly for $K_g$; but also it proves that $K_f=K_g$, which is what you asked.
This is standard theory in finite fields.
Every finite extension of finite fields is Galois with cyclic Galois group. This comes from the theory of the Frobenius automorphism. If $|F|=q$ and $f$ has degree $d$ then $L=F[x]/(f(x))$ is the splitting field and has order $q^d$. Then every element of $L$ satisfies $a^{q^d}-a=0$ so $L$ is also the splitting field of $x^{q^d}-x$. Thus the splitting fields of $f$ and $g$ are splitting fields of $x^{q^d}-x$. A polynomial's splitting field is unique up to isomorphism; this is another foundational result in Galois theory.