I had a question, I was just wondering about something.
So there's a question in my textbook that asks for the degree of this field extension for:$$\mathbb{Q}(i, \sqrt2 + i, \sqrt3 + i)$$
I can break this into towers of extensions: So $$[\mathbb{Q}(i, \sqrt 2 + i, \sqrt 3 + i): \mathbb{Q}(\sqrt 2, i)]* [\mathbb{Q}(i, \sqrt 2): \mathbb{Q}(i)]*[\mathbb{Q}(i): \mathbb{Q}]$$
So for me, it's clear that $[\mathbb{Q}(i, \sqrt 2): \mathbb{Q}(i)] = 2$ (since $x^2 - 2$ is irreducible in $\mathbb{Q}(i)$ )
and also $[\mathbb{Q}(i): \mathbb{Q}] = 2.$ (since $x^2 + 1$ is irreducible in $\mathbb{Q}$)
My issue here is trying to find $[\mathbb{Q}(i, \sqrt 2 + i, \sqrt 3 + i): \mathbb{Q}(\sqrt 2, i)]$.
I'm not sure what I'm supposed to do, my gut is telling me to check if $x^2 - 3$ is irreducible in $\mathbb{Q}(\sqrt 2, i)$. So if I'm supposed to do that, do I just show that $\sqrt 3 = a + bi + c\sqrt 2 + di\sqrt 2 \notin \mathbb{Q}(\sqrt 2, i)$?
Any hints would be helpful.
Note that your field is the same as $\mathbf{Q}(\sqrt2,\sqrt3,i)$
Now write the (real) field $\mathbf{Q}(\sqrt2,\sqrt3)$ as a simple extension $\mathbf{Q}(\sqrt2+\sqrt3)$ (easily verified). Now $x^2+1$ remains irreducible over this degree four field.