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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.

  • It looks better if you use \langle and \rangle instead of < and > in this context. – Greg Martin Mar 25 '17 at 09:52
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    As stated, you could take $f(x)=x$ and $g(x)=x-1$ for the first bullet point. Do you mean to exclude linear polynomials? In any case, I believe $f(x)=x^3-3x+1$ and $g(x)=x^3-3x^2+1$ are examples for the first one, and both generate a field extension that satisfies the second bullet point as well; but without knowing what context you're seeing these problems in (for example, do you know Galois theory?), these might or might not be the simplest examples. – Greg Martin Mar 25 '17 at 10:00
  • For the second point check http://math.stackexchange.com/questions/598285/degree-3-galois-extension-of-mathbbq-not-radical – DonAntonio Mar 25 '17 at 10:12
  • @GregMartin I have not learnt Galois Theory yet. May I ask for a simpler example for the second bullet point? –  Mar 25 '17 at 12:43

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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

  • I have not learnt Galois Theory yet. May I ask for a simpler example for the second bullet point? –  Mar 25 '17 at 12:44
  • @ButterMath I can't think of any easy examples. –  Mar 25 '17 at 14:47
  • @ButterMath I take that back. The link in DonAntonio's comment has an easy example. –  Mar 25 '17 at 14:48