I have just started to try and get my head around extensions, probably a lot of mistakes.

I am trying to solve a problem that goes like this : $ f $ is an irreducible polynomial with rational coefficients, of degree $17$. And $\alpha$ a root. The were three parts to this exercise, first show that $ \alpha $ isn't rational, then compute the degree of $\mathbb Q( \alpha)/ \mathbb Q$ and finally compute the degree of $\mathbb Q ( \alpha^2 + 17)/ \mathbb Q$. Now to prove the second point I used that f is separable (because $char(\mathbb Q)=0$ and $f$ irreducible) and therefore the extension $\mathbb Q (\alpha) / \mathbb Q$ has degree one.

My actual question arised in the third part. I feel like the following reasoning would be correct : $\alpha^2 + 17$ is an element of $\mathbb Q (\alpha) / \mathbb Q$. Therefore $\mathbb Q( \alpha^2 + 17)/ \mathbb Q \subset \mathbb Q (\alpha) / \mathbb Q$. And because the degree of $\mathbb Q (\alpha) / \mathbb Q$ is $1$, $\mathbb Q( \alpha^2 + 17)/ \mathbb Q$ must also have degree one.

Now I feel like this must be BS since I never use that the polynomial has degree 17. But could someone tell me what I should point my nose at ?

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    $\begingroup$ This is hard to read. here is a good tutorial for formatting on this site. $\endgroup$
    – lulu
    Jul 10, 2019 at 9:47
  • $\begingroup$ Sorry about that, it's been a while since i've been here and was writing on my tablet. Had forgotten you can simply use the $ sign --' $\endgroup$
    – Pastudent
    Jul 10, 2019 at 9:50
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    $\begingroup$ The degree of $\mathbb{Q}(\alpha)/\mathbb{Q}$ should equal $\deg f$... $\endgroup$
    – Matt B
    Jul 10, 2019 at 9:52
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    $\begingroup$ If $\Bbb{Q}(\alpha)/\Bbb{Q}$ has degree $1$, then won't $\alpha$ be in $\Bbb{Q}$? $\endgroup$
    – Anurag A
    Jul 10, 2019 at 9:52
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    $\begingroup$ You are right, there are no (non-trivial) intermediate extensions, because the orders of the "step-by-step" extensions multiplied together should equal the order of the "direct" extension $\endgroup$
    – David
    Jul 10, 2019 at 10:09

1 Answer 1


Why is the degree of $\mathbb{Q}(\alpha)/\mathbb{Q}=1$? Shouldn't it be $17$? If it were $1$ then $\mathbb{Q}(\alpha))=\mathbb{Q}$ and the exercise becomes quite simple.

Now, since $a^2+17 \in \mathbb{Q}(\alpha)$, the extension $\mathbb{Q}(\alpha^2+17)/\mathbb{Q}$ should be a divisor of $17$. I think you can conclude something from here

Also, don't worry about that weird $17$, it just means "the first random primer number I came out with"


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