# Proving that a polynomial is irreducible over a field extension of Q

Suppose I have the field $\mathbb Q[\sqrt d]$ where d is some square free positive integer.

How can I prove that a polynomial with integer coefficients is irreducible over this field?

And what if the field is something like $\mathbb Q[\sqrt d_1, \sqrt d_2]$ both $d_1, d_2$ square free.?

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Do you have any specific polynomial in mind? –  lhf Oct 29 '11 at 17:27
Yeah. Suppose $x^2 - 7$. But I would also like to know some general techniques. –  Mohan Oct 29 '11 at 17:32
Yeah.I know. But Eisenstein's crterion is to prove irreducibility over $\mathbb Z$. How do I use it to prove irreducibility over $\mathbb Q[\sqrt d]$? –  Mohan Oct 29 '11 at 17:37
I sincerely advice you to read its generalizations present under section here and also this –  Iyengar Oct 29 '11 at 17:50

For $x^2-7$, it is easy because it is reducible iff it has a root. Now if $(a+b\sqrt d)^2=7$ then $a^2+b^2 d=7$ and $2ab=0$. If $b=0$, we get $a^2=7$, which is impossible because $\sqrt 7$ is irrational. If $a=0$, we get $b^2 d = 7$ or $u^2 d = 7 v^2$, if $b=u/v$ with $u,v$ coprime integers. Then $v^2$ divides $d$ and so $v^2=1$, because $d$ is square free. This implies that $d=7$ and $u^2=1$.

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+1 for elaborating. –  user124471 Aug 29 '14 at 12:09