Finding determinant of higher degree number fields.

Question

I know I am forgetting something but I am not sure what. Could somebody please point me out in the real direction? I just want an explanation of the concept because I feel like I am confusing myself with all this and I am tired of getting myself confused with different incomplete explanation from various materials(I say incomplete because most just assume you have the maturity to catch their missing steps and as a non math student I find that very challenging.)

Given $K = \mathbb{Q}(\sqrt{5},\sqrt{-1})$, how does one find its discriminant?

From Wikipedia they say, let $K = \mathbb{Q}(\sqrt{-15})$ of discriminant $-15$ so the field $L = \mathbb{Q}(\sqrt{-3},\sqrt{5})$ has discriminant $-15^2 = 225$ and so is an everywhere unramified extension of K, and it is abelian.

Q1. What concept are they using to say the discriminant of $L$ is $225$ as stated? I mean why is it square of $-15$? Do $L'=\mathbb{Q}(\sqrt{3},\sqrt{-5}), L'' = \mathbb{Q}(\sqrt{-1},\sqrt{15}), L''' = \mathbb{Q}(\sqrt{-15},\sqrt{1})$ also have the discriminant $-15^2 = 225$?

Q2. What does it mean by the bold part and how is it said so?

Q3. Am I right in thinking the discriminant of $L'''' = \mathbb{Q}(\sqrt{3},\sqrt{13}) = (4\cdot39)^2$? Doesn't it matter that $3\equiv3(mod\ 4)$ and $13\equiv1(mod\ 4)$? So for my question above, is the discriminant of $K = \mathbb{Q}(\sqrt{5},\sqrt{-1}) = -20^2 = 400?$

Q4. What about cubic root combination number fields?

From one of the topics on StackExchange Answer #2:

The intermediate fields of $L = \mathbb{Q}(\sqrt{-5},\sqrt{-1})$ are $\mathbb{Q}(\sqrt{-5}), \mathbb{Q}(\sqrt{-1}),\mathbb{Q}(\sqrt{5})$. Is the third intermediate field always the product of the $m_i$? This intuitively makes some sense but I don't fully understand why is it so? My instructor didn't give any explanation for it when I asked her. She asked me to refer the textbook. In the last paragraph of the answer it says ...$p$ has to ramify in both. I don't get it. Why would p have to ramify in both? I think that's a silly question and if I put a bit more thought I might get it but I nevertheless asked if somebody is willingly to answer all these.

Thank you so much in advance for explaining me.

Answer 1

First, to avoid confusion, we must recall that the discriminant of a number field $K$ is the discriminant of any $Z$-basis of its ring $R_K$ of integers. Because of this arithmetic flavor, the computation of disc($K$) in general is not easy, unless we know in advance a description of $R_K$ (which is rarely the case). Coming back to your questions:

Q1. For quadratic fields $K, R_K$ is known, hence disc($K$) is also. You can find in any text book on algebraic number theory that, if $K = Q(\sqrt d)$, where $d \in Z$ is square free, a $Z$-basis of $R_K$ is {$1,\frac 12(1+\sqrt d)$} or {$1, \sqrt d$} according as $d $ is congruent or not to $1 mod 4$. So the disc. is $d$ in the first case, $4d$ in the second. I let you compute and check the result announced by Wiki. for $d=-15$. The question now is how to compute the disc $Q(\sqrt d, \sqrt d')$, or more generally to compute the disc. of the composite $MN$ of two number fields $M, N$ of respective degrees $m, n$. A sufficient result is known, see e.g. Marcus' "Number Fields", pp. 33-34: let $\delta$ = gcd (disc($M$), disc ($N$)) and suppose that the degree of $M.N$ over $Q$ is $mn$; then $R_{MN}$ is contained in $\frac 1\delta R_M .R_N$ . In particular, if $\delta=1$, then $R_{MN}=R_M .R_N$, and you can compute disc $MN$ from those of $M,N$. In the particular case of biquadratic fields, the condition on the degree $mn$ is void. For your biquadratic example or that of Wiki., this should also work (I'm too lazy to check). But beware : you must compute separately the cases $L'$ and $L"$, because these fields are not equal; as for your L''', it is quadratic, not biquadratic.

Q2. I cannot recall the definition of a "ramified prime ideal $P$" in an extension of number fields $L/K$, you must look for yourself in any textbook. The extension $L/K$ is said to be "unramified everywhere" if any prime ideal $P$ of $R_K$ is unramified in $L$. There is a link between ramification and the relative (ideal) discriminant $D(L/K)$ of $L/K$ : $P$ ramifies iff it contains $D(L/K)$. But here, knowing the biquadratic field $L$ and its quadratic subfields, you can proceed directly, happily enough. Take a prime ideal $P$ of $K$ and the prime number $p$ "under it", i.e. $P \cap Z = pZ$. As above, $p$ is ramified in $K$ iff $p$ divides disc($K$). In the Wiki. example, if lucky, you can check that the "candidates" $p$ to ramification are not common divisors of the disc. of $K$ and of another quadratic subfield of $L$ and you are done. In Samuel's booklet "Algebraic Number Theory",you can look at exercise 1, chapter 6, to get an idea of the road map.

Q3. Just apply the recipe in Q1.

Q4. The combination of cubic fields is much more complicated. Even if you can apply the sufficient result above, the ring of integers of a cubic field is awful, even for a "pure cubic" one, i.e. of the form $Q(\sqrt [3] d)$. See e.g. Marcus, p.38.

Pffft... I'm exhausted.