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This is a well known example in binary quadratic forms which I have started studying. Consider binary quadratic forms of discriminant $D=-56$. By reduction it turns out that for odd primes $p$ which are congruent to one of $1,9,15,23,25,39\,(\!\!\bmod\,56)$ there are two proper equivalence classes of forms represented by the reduced forms $x^2+14y^2$ and $2x^2+7y^2$. I haven't read genus theory much but I know that these two forms are in same genus.

My question is that is it possible that a same prime $p$ congruent to one of the above residues can be represented by both the forms $x^2+14y^2$ and $2x^2+7y^2$. In other words is there a way to separate the primes of the given residues which are represented by these two forms etc. Can I say that a prime $p$ congruent to the above residues can be represented by precisely one of the above two forms but not by both??? Is it possible to come to a conclusion with only reduction theory (this is upto what I have studied till now)???

For the case $D=-20$, however the reduced forms $x^2+5y^2$ and $2x^2\pm 2xy+3y^2$ represent different primes which can easily be seen by examining the resudues mod 20 which are coprime to 20, which happen to be disjoint for the two forms. But unfortunately this fails for $D=-56$ and the forms in question, as the residues mod 56 for $x^2+14y^2$ and $2x^2+7y^2$ are identical i.e $1,9,15,23,25,39\,(\!\!\bmod\,56)$.

Reference: Advanced Algebra by Knapp (page 16).

Thanks in advance.

Aaron Hendrickson
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Alonso Babuhicrik
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1 Answers1

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The answer is yes, class of the quadratic form is uniquely determined by the prime $p$. One way to put it - it is about the factorization of the principal ideal $(p)$ in the ring $R=\mathbb{Z}[\sqrt{-14}]$ to a product of 2 prime ideals, $(p)=P_1P_2$. To $P_1$ one can then assign a quadratic form, namely $q(x,y)=|xu+yv|^2/p$, where $u,v$ is a $\mathbb Z$-basis of $P_1$. Both $P_1$ and $P_2$ represent $p$, simply because $p\in P_1$ (and also $p\in P_2$). So in the end this is about the bijection between the ideal classes and the classes of quadratic forms (in your case $P_1$ has always the same class as $P_2$ - you get from $P_1$ to $P_2$ via $x\to x$, $y\to-y$, which doesn't change the 2 quadratic forms you mention, and so it definitely doesn't change their classes).

edit - without ideals:

If a quadratic form with a negative even discriminant $D$ (in your case $D=-56$) represents a prime $p$, you can bring it to the form $px^2+2bxy+cy^2$, with $D/4=b^2-pc$ (if $q(\mathbf u)=p$ for some $\mathbf u\in\mathbb Z^2$, the vector $\mathbf u$ has relatively prime components, so we can change the basis of $\mathbb Z^2$ to make it the 1st basis vector).

Replacing $x$ by $x+ky$ (for suitable $k\in\mathbb Z$) we can achieve that $0\leq b<p$. But since $b^2\equiv D/4$ mod $p$, and since this congruence has only 2 solutions, there are only 2 such $b$'s. So $p$ is represented by at most 2 classes of quadratic forms. One can pass between these forms by $x\to x$, $y\to -y$ (to get $b\to-b$), i.e., in the end, we can pass between their classes by any transformation with determinant $-1$. And your forms ($x^2+14 y^2$ and $2x^2+7y^2$) are invariant under such a transformation.

user8268
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    The book by Knapp mentions that $p$ can be represented by either $x^2+14y^2$ or $2x^2+7y^2$ since the number of proper equivalence classes representing $p$ is 1 or 2. This is what is not clear to me. However thanks for your answer. I'll have to check the theory for correspondence between ideals and classes. That comes much later in the chapter. I was assuming that this has a more elementary solution with only reduction theory that I was not getting somehow. – Alonso Babuhicrik Jul 01 '21 at 18:49
  • @AlonsoBabuhicrik I see - I added a bit to my answer for getting it without ideals (I hope there is enough detail). – user8268 Jul 01 '21 at 20:16
  • Let me rephrase for my convenience. Suppose $p$ can be represented by both $(1,0,14)$ and $(2,0,7)$. Since these two froms are NOT properly equivalent (under $SL_2(\mathbb{Z})$ ), being in reduced form (i.e $|b|\leq a\leq c$) one of them will be properly equivalent to $(p,b,\frac{b^2-D}{4p})$ and the other will be properly equivalent to $(p,-b,\frac{b^2-D}{4p})$ for some $b$ such that $b^2\equiv D(mod, 4p)$. I don't see how this gives any contradiction. I am assuming that you wanted to say that given one such $p$ it can only be represented by one of the two forms but not both. – Alonso Babuhicrik Jul 01 '21 at 20:44
  • By proper equivalence I mean under action of $SL_2(\mathbb{Z})$. – Alonso Babuhicrik Jul 01 '21 at 20:48
  • @AlonsoBabuhicrik The 2 forms, one with $b$ and the other with $-b$, are improperly equivalent (via a matrix with $\det=-1$). But $(1,0,14)$ and $(2,0,7)$ are not: they are invariant wrt. the improper trafo $x\to x, y\to -y$, and so any improper (det=-1) trafo acts on them as a proper one, and they are not properly equivalent. – user8268 Jul 01 '21 at 21:34
  • Sorry, I don't understand the phrase "acts on them as a proper one". From the comment I can only say that (1,0,14) and (2,0,7) are improperly equivalent. If $A\in SL_2(\mathbb{Z})$ takes (1,0,14) to the form with $b$ (say) and $B\in SL_2(\mathbb{Z})$ takes the form (2,0,7) to the form with $-b$ and $diag(1,-1)$ takes the form with $b$ to the one with $-b$, then $Adiag(1,-1)B^{-1}$ (det=-1) takes (1,0,14) to (2,0,7) and $det(Adiag(1,-1)B^{-1})=-1$, so they are improperly equivalent. This is not a contradiction to me. Sorry, I may not be aware of some terminology that's not in Knapp. – Alonso Babuhicrik Jul 02 '21 at 04:52
  • From your answer you have mentioned one can pass between the two classes with any matrix of det=-1. And your way of contradiction was to apply $diag(1,-1)$ to the form (1,0,14) and see that its unchanged and hence not in the other class. Would you please expand on your former statement: how to pass between two classes with any det=-1 matrix? – Alonso Babuhicrik Jul 02 '21 at 05:52
  • @AlonsoBabuhicrik By a matrix I always meant a matrix with integer coeffs. Any matrix with det=-1 is of the form $A\operatorname{diag}(1,-1)$ with $\det A=1$. Since $\operatorname{diag}(1,-1)$ doesn't change your quadratic forms (1,0,14) and (2,0,7), they cannot be improperly equivalent. – user8268 Jul 02 '21 at 06:10
  • So if $F_1$ is a form in the class of $B_1=(p,b,\ast)$ and $A$ is such that $det(A)=-1$ then $F_1\cdot A$ will be in the class of $B_2=(p,-b,\ast)$ since we have $F_1\cdot A=F_1\cdot (CC^{-1}A)=B_1\cdot (C^{-1}A)=B_2\cdot (diag(1,-1)C^{-1}A)$, where $C\in SL_2(Z)$ takes $F_1$ to $B_1$ and $diag(1,-1)C^{-1}A\in SL_2(Z)$. Hence $(1,0,14)\cdot diag(1,-1)$ should land in the class of $B_2$ but $(1,0,14)\cdot diag(1,-1)=(1,0,14)$, contradiction. I hope this is what you meant to say in essence. – Alonso Babuhicrik Jul 02 '21 at 06:13
  • @AlonsoBabuhicrik - yes (and sorry for not being clear enough) – user8268 Jul 02 '21 at 07:47