# Field extensions, stupid question

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 ?

• This is hard to read. here is a good tutorial for formatting on this site.
– lulu
Jul 10, 2019 at 9:47
• 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 $sign --' Jul 10, 2019 at 9:50 • The degree of$\mathbb{Q}(\alpha)/\mathbb{Q}$should equal$\deg f$... Jul 10, 2019 at 9:52 • If$\Bbb{Q}(\alpha)/\Bbb{Q}$has degree$1$, then won't$\alpha$be in$\Bbb{Q}\$? Jul 10, 2019 at 9:52
• 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 Jul 10, 2019 at 10:09

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"