Non-isomorphic field extensions of $\mathbb{Q}$

Question

I'm having a little bit of a problem with the following question:

Show that there do not exist two irreducible polynomials $a(x)$ and $b(x)$ in $\mathbb{Q}[x]$ of degrees 6 and 7 respectively which have isomorphic splitting fields.

Here's what I've thought of so far.

Let $\mathbb{K_a}$ and $\mathbb{K}_b$ denote the splitting fields of $a$ and $b$ respectively. Then $\mathcal{Gal}(\mathbb{K_a}/\mathbb{Q})\leq S_6$ and $\mathcal{Gal}(\mathbb{K}_b/\mathbb{Q})\leq S_7$.

Now, I somehow want to use the structure of $S_6$ and $S_7$ to make the conclusion, but am not too sure how. Any ideas?

Note: This isn't homework or classwork.

Answer 1

(1) If $K/F$ is Galois then $[K:F]={\rm Gal}(K/F)$.

(2) If $K/F$ is the splitting field of $f\in F[x]$ and $n=\deg f$, then $G={\rm Gal}(K/F)$ permutes the roots of $f$ in $K$ and is determined by this permutation, hence $G\hookrightarrow S_n$.

(3) If $K/B/F$ then $[K:F]=[K:B][B:F]$ is divisible by $[B:F]$.

(4) For any irreducible $f\in F[x]$, its rupture fields $K/F$ have degree $[K:F]=\deg f$.

Combining these results tells you that if $a(x),b(x)\in\Bbb Q[x]$ are irreducible of degrees six and seven respectively, then $a(x)$'s splitting field has degree dividing $6!$ whereas $b(x)$'s splitting field has degree a multiple of $7$; therefore it is not possible for their degrees to be equal, precluding isomorphism.

Answer 2

Recall that we can build the splitting field of an irreducible polynomial $p(x)$ by adjoining roots of the polynomials one at a time. f $K_0=\mathbb Q \subset K_1 \subset K_2 \subset ... \subset K_n$ is a tower of fields obtained by adjoining roots one by one, and $K_n$ is the splitting field of $p(x)$, we have that $[K_n:\mathbb Q]=[K_n:K_{n-1}]\cdot [K_{n-1}:K_{n-2}]\cdots [K_1:K_0].$ Note also that $[K_{i+1}:K_i] < [K_1:K_0]$ for all $i>1$.

The first extension $K_1$, obtained by adjoining any root of $p(x)$, is isomorphic to $\mathbb Q[x]/\langle p(x) \rangle.$ The dimension of this extension over $\mathbb Q$ is precisely $\deg p$.

Do you see how to apply this to finish your problem?