Basic Question on Field Theory - Hopefully to Illustrate Several Concepts I am just beginning to study fields and for whatever reason am finding their presentation to be completely baffling - moreso than I think anything I have ever studied. I am reading out of chapter 21 of this free book: http://abstract.ups.edu/download.html. I understand pretty much all the definitions & statements & proofs in isolation but what I find really lacking in my understanding is how to apply any of that to simple computations. Now I understand this is a Q&A site so I have selected one question that I hope might help clear this up for me, 2(g) in Judson's book, page 328:

Find a basis for the following field extension. What is the degree of the extension?
  $\mathbb{Q}(i, \sqrt{2} + i, \sqrt{3} + i)$ over $\mathbb{Q}$.

Now when I look at this a lot of things go through my head. I know this field is the same as the field created by adjoining each of those elements one by one, and that the total degree will be the product of the degree of each of those extensions. In an abstract sense, I know what a basis should look like: all possible products of the bases of each of those extensions.
I'm also fairly certain that $\mathbb{Q}(i, \sqrt{2} + i, \sqrt{3} + i)$ = $\mathbb{Q}(\sqrt{2}+\sqrt{3}, i)$. But does this help me? I can't see how.
Another source of confusion I have is that, say for example in my solution for the above problem I need to find the degree of the extension $\mathbb{Q}(\sqrt{2}+i)$. Well I know that I could find the fact that $\sqrt{2}+i$ is a root of $x^4-2x^2+9$, which has degree 4. I am pretty certain that this is the minimal polynomial, but I see no way of showing that. I keep running into basic misunderstandings like this left and right. Should I instead be approaching finding the degree of that extension from another angle?
If someone could help me get started in this area I would be completely grateful! Alternatively a recommendation for an online treatment of this sort of material would be great, I find Judson's book has few detailed examples that justify the steps - a lot of use words like "obviously" and "easily seen" that are frustrating to a tired newcomer.
 A: One thing you should be aware of is that $\mathbb Q(i,\sqrt2+i,\sqrt3+i)$ is also $\mathbb Q(i,\sqrt2,\sqrt3)$. Recognition of this fact will make all your considerations much more transparent. Thus you should sit down and make sure you understand why it’s true; maybe even give yourself a formal proof.
As for the second question, once you know that $[\mathbb Q(\sqrt2+i)\colon \mathbb Q]=4$, you’re done, but perhaps you have already seen this. But here’s an argument to show that the big field is also $\mathbb Q(\sqrt2,i)$: Look at $(2+1)/(\sqrt2+i)=\sqrt2-i$. Since the left hand side is in the big field, the right hand side is too, and thus their sum is, and $2\sqrt2$ is in your big field, and with only a little more thought, you’re done.
A: So it looks like you understand the definitions of the stuff you're studying but haven't worked out the techniques needed to prove things. This is usually the hard part in learning a new topic. Maybe I can help by thinking about the problem you wrote.
So $\mathbb{Q}( i, \sqrt{2}+i, \sqrt{3}+i)$ is indeed the same as $\mathbb{Q}( i, \sqrt{2}+ \sqrt{3})$. How do we show this? Well the first extension is definitely the same as $\mathbb{Q}( i, \sqrt{2}, \sqrt{3})$ and then $\mathbb{Q}( \sqrt{2}, \sqrt{3})$ is the same as $\mathbb{Q}( \sqrt{2}+ \sqrt{3})$. How can we rigorously prove this latter equivalence? Well, the first field definitely contains the second. Also, $\mathbb{Q}( \sqrt{2}), \mathbb{Q}( \sqrt{3})$ are both degree 2 extensions of $\mathbb{Q}$. Their intersection is $\mathbb{Q}$ since if $a+b \sqrt{2}= c + d \sqrt{3}$ for $b, d $ nonzero, then isolating the $\sqrt{2}$ term, we get $(c-a+d \sqrt{3})^2=2b^2$.The right-hand side is rational, but the left can only be rational if $c-a=0$. But then we have $d \sqrt{3}=b \sqrt{2}$ which is impossible. Thus, $\sqrt{3} \notin \mathbb{Q}( \sqrt{2})$ and so $\mathbb{Q}( \sqrt{2}, \sqrt{3})$ is a degree 4 extension. One can similarly show that $\mathbb{Q}(\sqrt{2}+ \sqrt{3})$ is a degree 4 extension, and so since one contains the other, they are equal.
In any case, if you know that your extension is just $\mathbb{Q}( \sqrt{2}, \sqrt{3}, i)$ then you can easily pin down the degree of the extension. Adjoining the first two square roots is a degree 4 extension as we have seen.But the extension is real and so cannot contain $i$. Thus, adjoining $i$ gives a degree $8$ extension in total. What is a basis? Well it will have $8$ elements. $1, \sqrt{2}, \sqrt{3}, i$ will be in that basis. Also, $\sqrt{6}, \sqrt{2}i, \sqrt{3}i, \sqrt{6}i$. Since we know a basis will have 8 elements, to show this is a basis, we need only show it is a spanning set for our field extension. This is more or less obvious, but a good exercise if you don't see it right away.
