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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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    I sincerely advice you to read its generalizations present under section [here](http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Generalization) and also [this](http://mathoverflow.net/questions/15460/variants-of-eisenstein-irreducibility)2011-10-29

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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 $u^2=1$.