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How would one show this? is there any test?

say $x^3 - 2x - 5$ over $\mathbb{Q}[\sqrt{2}]$

This is just a made-up example, but I'm interested in the general process and how it differs from showing irreducible over $\mathbb{Q}$.

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since the polynomial degree is 3, it is sufficient to show it has no roots in the field –  Federica Maggioni Apr 26 '13 at 7:18
    
right... so does that involve simple computation of plugging in a + b*sqrt(2) into the polynomial and showing that that can never equal 0? –  tyur43 Apr 26 '13 at 7:19
    
i do agree, except for the word "simple" –  Federica Maggioni Apr 26 '13 at 7:24
    
yeah i realized it becomes quite complicated which is why i would like to hear if there are more general/simpler methods. thanks though. –  tyur43 Apr 26 '13 at 7:31

1 Answer 1

In this case it's easy. Suppose $\theta$ is a root of $x^3-2x-5$. Observe that $x^3-2x-5$ has no rational roots (say by the rational root test) so is irreducible over $\mathbb Q$. Thus $\mathbb Q[\theta]$ has degree $3$. If $\theta\in \mathbb Q[\sqrt{2}]$ then we would have $\mathbb Q[\theta]\subseteq \mathbb Q[\sqrt2]$, thus $[\mathbb Q[\theta]:\mathbb Q]|[\mathbb Q[\sqrt2]:\mathbb Q]$, i.e. $3|2$, a contradiction.

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