Artin's Algebra Book Problem:Prove that splitting fields of $x^3+ex+6$ over $Q(e)$ and $x^3+\pi x+6$ over $Q(\pi)$ are isomorphic.

Question

Assume that $\pi$ and $e$ are transcendental. Let $K$ be the splitting field of $f(x)=x^{3} + \pi x + 6$ over $F = Q(\pi)$

(a) Prove that $[K : F] = 6$.

(b) Prove that $K$ is isomorphic to the splitting field of $x^3 +ex+6$ over $Q(e)$

It's clear that $f$ has only one real root and also as $\pi$ is transcedental over $Q$ therefore $f$ is irreducible over $Q(\pi)$,which in particular implies that $[K:F]$=6.

Any ideas about part $(b)$ ?

Answer 1

Since $e$ and $\pi$ are transcendental, both $\mathbb Q(e)$ and $\mathbb Q(\pi)$ are isomorphic to the rational function field $\mathbb Q(t)$. If we let $\phi: \mathbb Q(e) \to \mathbb Q(\pi)$ be a field isomorphism, then $\phi$ induces an isomorphism of polynomial rings $\phi: \mathbb Q(e)[x] \to \mathbb Q(\pi)[x]$. Under this isomorphism, we have $\phi(x^3+ex+6)= x^3+\pi x+6$, and in this situation, $\phi$ extends to an isomorphism of the splitting fields of the two polynomials; e.g., see section 13.4 of Dummit and Foote, Abstract Algebra:

Theorem 27. Let $\phi:F \to F'$ be an isomorphism of fields. Let $f(x)\in F[x]$ be a polynomial and let $f'(x) \in F'[x]$ be the polynomial obtained by applying $\phi$ to the coefficients of $f(x)$. Let $E$ be a splitting field for $f(x)$ over $F$ and let $E'$ be a splitting field for $f'(x)$ over $F'$. Then the isomorphism $\phi$ extends to an isomorphism $\sigma: E \to E'$, i.e., $\sigma$ restricted to $F$ is the isomorphism $\phi$.