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 ?
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"