I am trying to show that the splitting field for an irreducible quadratic polynomial over Q\ \mathbb{Q} is of the form Q(√D)\ \mathbb{Q}(\sqrt{D}) where DD is a square-free integer not equal to 00 or 11.

If we let f(x)=ax2+bx+c∈Q[x]f(x) = ax^2+bx+c \in \mathbb{Q}[x] be irreducible in Q\mathbb{Q} then the roots are

x=−b±√b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.

Then using field operations we can say that the splitting field of ff is Q(√b2−4ac)\mathbb{Q}(\sqrt{b^2-4ac}). Since ff is irreducible over Q⟹\ \mathbb{Q} \implies no roots in Q⟹b2−4ac≠\mathbb{Q} \implies b^2-4ac \neq a square (including 00 and 11).

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

Any help would be appreciated!

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If b2−4acb^2 – 4ac contains a square, i.e. b2−4ac=d2eb^2 – 4ac = d^2e, then Q(√b2−4ac)=Q(√e)\mathbb{Q}(\sqrt{b^2 – 4ac}) = \mathbb{Q}(\sqrt{e}).

– ec92

2 days ago

@ec92 thanks, just figured this out for myself 🙂

– Joseph

2 days ago

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