I am trying to find out whether the following two examples of field extensions could possibly exist(they are related):
Two different irreducible monic polynomials $f(x), g(x) ∈ \Bbb Q[x]$ such that the fields $\Bbb Q[x]/<f(x)>$ and $\Bbb Q[x]/<g(x)>$ are isomorphic.
A degree $3$ extension of $\Bbb Q$ which is not isomorphic to one of the form $Q(\sqrt[3]{a})$.
My intution is that the first example does not exist and the second does exist. But I am really not sure. Thanks so much.
The first one does exist. To roughly see why let $\mathbb Q[x]/\langle f(x)\rangle$ be one field extension of $\mathbb Q$, then consider the minimal polynomials of $x+1,x+2,x+3,\dots$. At least one of those will be different from $f(x)$. For an actual example:
$$\mathbb Q[x]/\langle x^2+x+1\rangle\cong\mathbb Q[x]/\langle x^2+3\rangle$$ are isomorphic. To see why recall that a primitive cube root of unity $(-1\pm\sqrt{-3})/2$ has minimal polynomial $x^2+x+1$.
Or there's a really trivial example
$$\mathbb Q[x]/\langle x\rangle \cong\mathbb Q[x]/\langle x-1\rangle.$$
For the second one we turn to automorphism groups. A degree $3$ extension of the form $\mathbb Q(\sqrt[3]{a})$ has trivial automorphism group. To see why note that we can always assume $\sqrt[3]{a}$ is real, then the other two cube roots of $a$ are complex. So no Galois extension of degree $3$ is of the form $\mathbb Q(\sqrt[3]{a})$. Looking at Wikipedia I see that $\mathbb Q[x]/\langle x^3+x^2-2x-1\rangle$ is a Galois extension, and I guess to prove it you should show that $\mathbb Q[x]/\langle x^3+x^2-2x-1\rangle$ has $3$ roots of $x^3+x^2-2x-1$.
Edit: $x^3+x^2-2x-1$ is irreducible with discriminant $49$ and therefore its splitting field is Galois of degree $3$. Theorem 2.1. here: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf