Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Elementary algebra shows that the product of two numbers in the form $x^2 + ny^2$ again has the same form, since if $p = (a^2 + nb^2)$ and $q = (c^2 + nd^2)$, $$pq = (a^2 + nb^2)(c^2 + nd^2) = (ac \pm nbd)^2 + n(ad \mp bc)^2$$ My question is: Assuming that a number $z$ can be factored into primes of the form $x^2 + ny^2$, does every representation of $z$ in this form arise from repeated applications of this formula to the prime factors?

share|improve this question
    
Cross-posted: math.stackexchange.com/questions/229201/… –  Andres Caicedo Nov 4 '12 at 22:05
add comment

2 Answers

up vote 8 down vote accepted

The answer is yes. To see this, consider the ring $R=\mathbb{Z}[\sqrt{-n}]$. If $z=p_1\dots p_k$ is the decomposition of $z$ into rational primes, then by assumption each $p_j$ decomposes in $R$ as $p_j=q_j\bar q_j$. We need to show that any decomposition $z=r\bar r$ in $R$ can be gotten as follows: for each $j$ let $r_j$ be either $q_j$ or $\bar q_j$, and then put $q=wr_1\dots r_k$, where $w$ is a unit in $R$. In $R$ the ideal $(z)$ decomposes into prime ideals as $(z)=(q_1)(\bar q_1)\dots (q_k)(\bar q_k)$, hence it suffices to show that in $R$ the ideal $(z)$ and its divisors decompose uniquely into prime ideals.

If $n$ is square-free and congruent to $1$ or $2$ mod $4$, then $R$ is the full ring of integers in $\mathbb{Q}(\sqrt{-n})$, hence it is a Dedekind domain. So in this case we are done.

If $n$ is square-free and congruent to $3$ mod $4$, then $R$ is a quadratic order of conductor $2$ in $\mathbb{Q}(\sqrt{-n})$, hence unique factorization holds in $R$ for ideals prime to $2$. Clearly, each $p_j$ above is odd, hence $(z)$ is coprime to $2$, and we are done.

If $n$ is not square-free, then $R$ is a quadratic order of some conductor $f\mid 2n$ in $\mathbb{Q}(\sqrt{-n})$, hence unique factorization holds in $R$ for ideals prime to $f$. Clearly, each $p_j$ above is at least $n$, hence $(z)$ is coprime to $2n$, and we are done.

For the quoted result on quadratic orders see Exercise 7.26 in Cox: Primes of the form $x^2+ny^2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.