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Where can I find information (at least examples) about factorization of prime ideals in biquadratic extensions of $\mathbb{Q}$. Right now I have no idea how, for example, find factorization of $(2)$ in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Another thing that I'd like to know is what are examples of factorization for all six possible combinations of $e$, $f$, $r$?

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(2) = P^4, (3) = Q^2, A/Q=GF(9), (5) = RS, A/R=A/S=GF(25), (23) splits completely. I think that is about all. –  Jack Schmidt Nov 14 '11 at 16:39
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You will never get totally inert primes in a biquadratic extension. The reason is a bit more advanced than what your course probably has covered up to this point. It comes from the fact that the Galois group of the residue class extension $GF(q^f)/GF(q)$ can be realized as a quotient group $H/K$ of two subgroups $H,K$ of the Galois group $G$ of the number field extension $G=Gal(L/K)$. We know that the Galois group of that extension of finite fields is cyclic of degree $f$, so for that value of $f$ to occur we need $H/K\simeq C_f$. Look up decomposition and inertia groups for more information. –  Jyrki Lahtonen Nov 14 '11 at 18:37
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@Jyrki: Dear Jyrki (and Alex), If one knows the splitting rule for primes in a quadratic extension and its relationship to quadratic reciprocity, one can deduce the splitting rule for primes in a biquadratic extension directly (without having to think about decomposition groups): the fact that the product of two elements with Jacobi symbol $-1$ has Jacobi symbol $1$ implies that if a prime is inert in two of the three quadratic exts. making up the biquadratic extension, then it must split in the third. Regards, –  Matt E Nov 14 '11 at 20:54
    
Thanks @MattE ! Arturo also covered this in his answer. Yours truly, OTOH, seem to have grabbed a finetuning sledgehammer :-( –  Jyrki Lahtonen Nov 14 '11 at 21:07
    
@Jyrki: Dear Jyrki, In your defense, this situation is commonly used as an example of how cyclicity of the decomposition group can be used to deduce arithmetic information, and it is a very nice sledgehammer! Best wishes, –  Matt E Nov 14 '11 at 21:14

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up vote 7 down vote accepted

Daniel Marcus's Number Fields has a lot of information on biquadratic extensions in the exercises.

Exercise 2.42 (pages 51 and 52), in five parts, determines the ring of integers of $\mathbb{Q}[\sqrt{m},\sqrt{n}]$ for $m$ and $n$ distinct squarefree integers different from $1$.

Exercise 4.6 (page 116) discusses the factorization of a prime in such a biquadratic field in terms of the factorization in each of the three quadratic subextensions.


So, if you want the solution (as opposed to just references)...

For quadratic number fields, $\mathbb{Q}(\sqrt{m})$ with $m$ square free, and $p$ a prime, we have that:

  1. If $p|m$, then $p$ ramifies.

  2. If $m$ is odd, then $2$ ramifies if $m\equiv 3\pmod{4}$, splits if $m\equiv 1\pmod{8}$, and is inert if $m\equiv 5\pmod{8}$.

  3. If $p$ is odd, and $p$ does not divide $m$, then $p$ splits if $m$ is a quadratic residue modulo $p$, and inert if $m$ is a quadratic nonresidue modulo $p$.

(Theorem 25 in Marcus, page 74).

Looking at the biquadratic extension $K=\mathbb{Q}(\sqrt{m},\sqrt{n})$, with $m$ and $n$ squarefree, the exercise suggests looking at the three quadratic subextensions, which are given by $\mathbb{Q}(\sqrt{m})$, $\mathbb{Q}(\sqrt{n})$, and $\mathbb{Q}(\sqrt{k})$, with $k=mn/\gcd(m,n)^2$. Let $p$ be a rational prime.

If $p$ is ramified in each of the quadratic subfields, then $p$ is completely ramified in $K$ and conversely: if $p$ is completely ramified in $K$, then multiplicativity of the ramification index ensures it is ramified in each quadratic subextension. Conversely, if it is not completely ramified in $K$, then the Inertia group for a prime lying over $p$ is not the whole Galois group, so $p$ does not ramify in the fixed field, which is necessarily one of the quadratic subextensions.

For $p=2$, this occurs precisely when two of $m,n,k$ are even, and the third is congruent to $3$ modulo $4$ (we cannot have all three even; if at least one of them is even, then exactly two of them are; and if all three were odd, we would need $m$, $n$, and $k$ to all be congruent to $3$ modulo $4$, which is impossible because $k\equiv k\gcd(m,n)^2 \equiv mn\equiv 1\pmod{4}$ if $m\equiv n\equiv 3\pmod{4}$); for $p$ odd, this is impossible since it requires that $p$ divide each of $m$, $n$, and $k$, but if $p$ divides $m$ and $n$, then it does not divide $k = mn/\gcd(m,n)^2$. So the only prime that can ramify completely is $p=2$, and only if exactly two of $m$, $n$, and $k$ are even and the third is congruent to $3$ modulo $4$.

If $p$ splits completely in each of the quadratic subfields, then $p$ splits completely in $K$, and conversely (multiplicativity of the inertia and ramification degrees, and looking at the decomposition group). For $p=2$, this requires each each of $m$, $n$, and $k$ to $1$ modulo $8$; this occurs if at least two of $m$, $n$, and $k$ are congruent to $1$ modulo $8$. If $p$ is odd, we need each of $m$, $n$, and $k$ to be quadratic residues modulo $p$, which occurs if at least two of $m$ and $n$ are quadratic residues modulo $p$.

Now note that $p$ cannot be inert in $K$: this would require $p$ to be inert in each of the quadratic subextensions. For $p=2$, we would need $m\equiv n\equiv k\equiv 5\pmod{8}$; but if $m\equiv n\equiv 5 \pmod{8}$, then $mn\equiv 1\pmod{8}$. Since every square is congruent to $0$, $1$, or $4$ modulo $8$, $\gcd(m,n)^2 \equiv 1\pmod{8}$, so $1\equiv mn= k\gcd(m,n)^2 \equiv k\pmod{8}$. Hence $2$ cannot be inert in $K$.

For $p$ odd, then we would need $\gcd(p,mn)=1$, and all of $m$, $n$ and $k$ to be quadratic nonresidues modulo $p$. But if $m$ and $n$ are quadratic nonresidues, then $mn$ is a quadratic residue, hence so is $k$ (since $mn=k\gcd(m,n)^2$). So no odd prime can be inert in $K$.

Thus, the only other possibilities besides completely ramified and completely split are that $p$ splits as $\mathfrak{P}{Q}$, as $\mathfrak{P}^2\mathfrak{Q}^2$, or as $\mathfrak{P}^2$.

Consider first odd $p$.

If $p$ does not ramify in any of the quadratic subextensions but does not split completely in all of them, then $\gcd(p,mn)=1$ and exactly one of $m$, $n$, and $k$ is a quadratic residue modulo $p$. Then $p$ must split as $\mathfrak{PQ}$ in $K$.

Then note that $p$ cannot ramify in exactly one of the subextensions (it cannot divide exactly one of $m$, $n$, and $mn/\gcd(m,n)^2$). If it ramifies in two of the subextensions, and is inert in the third, then $p$ must split as $\mathfrak{P}^2$ in $K$

And if $p$ ramifies in two of the subextensions and splits in the third, then it must split as $\mathfrak{P}^2\mathfrak{Q}^2$ in $K$.

Now consider $p=2$, and that $2$ does not totally ramify in $K$. If $2$ ramifies in exactly two of the subextensions (say, when exactly two of $m$, $n$, and $k$ are even and the third is not congruent to $3$ modulo $4$; e.g., $m=6$, $n=14$), then the behaviour of $(2)$ in $K$ will depend on how it splits in the third quadratic subextension. If it is inert in the third quadratic subextension, then $(2)=\mathfrak{P}^2$ in $K$ (look at the multiplicativity of $f$ and $e$); if it splits in the third, then you'll get $(2)=\mathfrak{P}^2\mathfrak{Q}^2$.

If $m$, $n$, and $k$ are all odd, and $2$ ramifies in exactly one of the subextensions, and splits in at least one, then $(2)=\mathfrak{P}^2\mathfrak{Q}^2$; for this to occur, we would need say $m\equiv 3\pmod{4}$, and either $n$ or $k$ congruent to $1$ modulo $8$. If it ramifies in exactly one and is inert in the other two, you'll get $(2)=\mathfrak{P}^2$. And if it ramifies in none of them, but does not split in all of them, then it splits in one and is inert in the other two, and you'll get $(2)=\mathfrak{PQ}$ in $K$.

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I know this textbook but exercise 4.6 is just another way to ask my questions, so here I ask how to solve 4.6. –  Alex Nov 14 '11 at 15:16
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@Alex: You said you wanted a reference; the exercise does more than simply ask the question, it walks you through the considerations to make to answer the question. If what you actually want is a solution, then using the [reference-request] tag is misleading: you are asking for a solution. –  Arturo Magidin Nov 14 '11 at 17:09
    
Thank you, but could you, please, explain why "If $p$ is ramified in each of the quadratic subfields, then $p$ ramifies in $K$ and conversely." –  Alex Nov 15 '11 at 4:24
    
@Alex: That should be "totally ramified in $K$". If $p$ is totally ramified in $K$, then by multiplicativity of the ramification index it has to be ramified in each of the quadratic subextensions. If it is ramified in $K$ but not completely ramified, then think about the fixed field of the inertia group to see it cannot be ramified in each of the quadratic subextensions. –  Arturo Magidin Nov 15 '11 at 4:46

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