Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.?

share|improve this question
    
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
1  
I sincerely advice you to read its generalizations present under section here and also this –  Iyengar Oct 29 '11 at 17:50
add comment

1 Answer 1

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 $u^2=1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.