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I am trying to show that the splitting field for an irreducible quadratic polynomial over $\ \mathbb{Q}$ is of the form $\ \mathbb{Q}(\sqrt{D})$ where $D$ is a square-free integer not equal to $0$ or $1$.

If we let $f(x) = ax^2+bx+c \in \mathbb{Q}[x]$ be irreducible in $\mathbb{Q}$ then the roots are

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Then using field operations we can say that the splitting field of $f$ is $\mathbb{Q}(\sqrt{b^2-4ac})$. Since $f$ is irreducible over $\ \mathbb{Q} \implies$ no roots in $\mathbb{Q} \implies b^2-4ac \neq$ a square (including $0$ and $1$).

So I've shown that at least $D$ cannot be a square - but I am stuck on how to prove the stronger condition that $D$ must be square-free.

Any help would be appreciated!

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    $\begingroup$ If $b^2 - 4ac$ contains a square, i.e. $b^2 - 4ac = d^2e$, then $\mathbb{Q}(\sqrt{b^2 - 4ac}) = \mathbb{Q}(\sqrt{e})$. $\endgroup$ – ec92 Oct 21 '16 at 18:02
  • $\begingroup$ @ec92 thanks, just figured this out for myself :) $\endgroup$ – Joe Oct 21 '16 at 18:03

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