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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 $\Bbb F_3$.

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  • $\begingroup$ Induction is not the good strategy. $\endgroup$ – Lubin Oct 26 '17 at 3:07
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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.

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  • $\begingroup$ Don't you want to mean $a^{q^d+1}-a=0$? Because $a^{q^d}=1$ or am I wrong? $\endgroup$ – user425181 Oct 23 '17 at 19:02
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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.

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  • $\begingroup$ Roots of $X^{q^d-1}-1$ distinct? Sure: derivative is $-X^{q^d-2}$, relatively prime to the original. $\endgroup$ – Lubin Oct 26 '17 at 3:12

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