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The class number of $\mathbb{Q}(\sqrt{-23})$ is $3$, and the form $$2x^2 + xy + 3y^2 = z^3$$ is one of the two reduced non-principal forms with discriminant $-23$.

There are the obvious non-primitive solutions $(a(2a^2+ab+3b^2),b(2a^2+ab+3b^2), (2a^2+ab+3b^2))$.

I'm pretty sure there aren't any primitive solutions, but can't seem to find an easy argument. Are there?

In general, is it possible for a non-principal form to represent an $h$-th power primitively (where $h$ is the class number of the associated quadratic field)?


[EDIT]

I think I've solved the first question and have posted an answer below.

Since the proof is very technical I don't see how it can generalize to the greater question above ($h$-th powers).

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please don't rely on the title for content; the body of the message should be self-contained, as a service to your readers. –  Arturo Magidin Jan 25 '11 at 19:31
    
You may want to post that as an answer, rather than an edit to the question. It is perfectly acceptable to post answers to your own questions if you manage to figure them out. Then you can add just the closing paragraph of your addition to the question, referencing your work below. –  Arturo Magidin Jan 25 '11 at 21:19

1 Answer 1

If $4|y$, then $2|z$, which in turn, by reducing mod $4$, $2|x$, contradicting primitivity. Multiply by $2$ and factor the equation over $\mathbb{Q}(\sqrt{-23})$: $$(\frac{4x+y+\sqrt{-23}y}{2})(\frac{4x+y-\sqrt{-23}y}{2}) = 2z^3$$

Note that both fractions are integral. The gcd of the two factors divides $\sqrt{-23}y$ and $4x+y$. If $23|4x+y$ then $23|z$, and reducing the original equation modulo $23^2$ we see that $23|y$, hence also $23|x$, contradicting primitivity. So the gcd divides $y$ and $4x+y$, and by the above argument, it then must be either $1$ or $2$ according to wether $y$ is odd or even, respectively.

First, assume that $y$ is odd. So the gcd is $1$. Thus, for some ideal $I$: $$(2x+y\frac{1+\sqrt{-23}}{2})=(2,\frac{1+\sqrt{-23}}{2})I^3$$

Which implies that $(2,\frac{1+\sqrt{-23}}{2})$ is principal, leading to a contradiction. Now assume that $y$ is even, and that $x$ is therefore odd and $z$ is even. Put $y=2u$, $z=2v$, $u$ odd, so that: $$(2x+u+\sqrt{-23}u)(2x+u-\sqrt{-23}u)=16v^3$$

Both factors are divisble by $2$, so that: $$(x+u\frac{1+\sqrt{-23}}{2})(x+u\frac{1-\sqrt{-23}}{2})=4v^3$$

As before, the gcd is 1, and since $x+u\frac{1+\sqrt{-23}}{2}=x+u+u\frac{-1+\sqrt{-23}}{2} \in (2, \frac{-1+\sqrt{-23}}{2})$ we must have for some ideal $I$: $$(x+u\frac{1+\sqrt{-23}}{2}) = (2, \frac{-1+\sqrt{-23}}{2})^2I^3$$

Contradicting that $(2, \frac{-1+\sqrt{-23}}{2})^2$ is non-principal (the ideal above $2$ appears squared since the product of factors is divisible by $4$). We are done!


It actually seems that the above can be generalised, but I have to use a major theorem, which seems like it might be a bit of an overkill. Without further ado:

Let $aX^2+bXY+cY^2$ be a primitive non-principal quadratic form of discriminant $\Delta=b^2-4ac$, and $h$ be the class number of the associated quadratic field. Assume there is a solution $x,y,z$: $$ax^2+bxy+cy^2=z^h$$

Recalling the Chebotarev Density Theorem (OVERKILL), there is an equivalent form with $a$ an odd prime that doesn't divide $\Delta$, and since the invertible change of variables preserves primitivity, we reduce to this case.

Multiplying by $a$ and factoring over $\mathbb{Q}(\sqrt{\Delta})$: $$(ax+\frac{b+\sqrt{\Delta}}{2}y)(ax+\frac{b-\sqrt{\Delta}}{2}y)=az^h$$

The gcd of the factors divides $(2ax+by,\sqrt{\Delta}y)$. Say $\Delta |2ax+by$, then since $$(2ax+by)^2-\Delta y^2=4az^h$$ we must have $\Delta |z$, so $\Delta |y$, and finally $\Delta |x$ (unless $\Delta=\pm 2$, which is impossible), contradicting primitivity.

Hence the gcd divides $(2a,y)$.

1) gcd$=1$: $$(ax+\frac{b+\sqrt{\Delta}}{2}y) = (a,\frac{b+\sqrt{\Delta}}{2})I^h$$ contradicting that the form is non-principal.

2) gcd$=2$ or $2a$: then $y$ is even, so $z$ is too, and since $a$ is odd, $x$ must also be even, contradicting primitivity.

3) gcd$=a$: thus $a|z$. If $a^2|y$, then reducing modulo $a$ we see that $a|x$. So $a||y$. Dividing this gcd out of the two factors, we must have: $$(x+\frac{b+\sqrt{\Delta}}{2}\frac{y}{a}) = (a,\frac{b-\sqrt{\Delta}}{2})^{h-1}I^h$$ for some ideal $I$. Note the particular ideal above $a$, which appears since its conjugate cannot appear as $x$ isn't divisble by $a$. Since $(a,\frac{b-\sqrt{\Delta}}{2})^{h-1} \sim (a,\frac{b+\sqrt{\Delta}}{2})$, this contradicts that the form is non-principal.

We are done!

There must be a way to avoid applying a major theorem such as Chebotarev... Please tell me if you have an idea how :D

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how did you learn to prove using Ideals like that? –  quanta Jan 25 '11 at 21:47
    
other than a number theory course at uni, Milne's Algebraic Number Theory notes has a few calculations like this, and it made an impression. –  simplequestions Jan 26 '11 at 7:02

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