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We did in class $x^2+y^2$, which was easy, and we had for homework $2x^2+2xy+3y^2$, which I did (its values (if not square) must be divisible by form primes, or of the form $x^2+5y^2$, and clearly they can't all be the latter!).

Just to test my method I took the middle coefficient down by one - and ran into trouble...

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Perhaps you ran into unexpected trouble because you picked a form of discriminant -23, which is the first time the class number is 3, so you'll need to juggle values of three quadratic forms. –  KCd Dec 25 '10 at 22:34
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Also, whereas in the other cases you looked at the primes that are represented by the given form are determined by a congruence condition modulo the discriminant (e.g. $p \equiv 1 \bmod 4$ in the case of $x^2 + y^2$, and conditions mod $20$ for the form $2 x^2 + 2 x y + 3 y^2$) there is no such congruence condition for determining the prime represented by your form of discriminant $23$. –  Matt E Dec 25 '10 at 23:02
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+1 for trying to generalize an assignment-shows thinking about the issues. –  Ross Millikan Dec 26 '10 at 6:44
    
Dear Simplequestions, Do the comments by me and KCd mean anything to you? If so, great! If not, is there some point you would like to have elaborated? –  Matt E Dec 26 '10 at 15:22
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You might find this book interesting: books.google.com/books?id=pSMlAQAAIAAJ –  Aryabhata Dec 26 '10 at 21:42

2 Answers 2

This example morally comes from class field theory. It is just easy enough that I can do it using elementary algebraic number theory but not class field theory.

Write $\theta = (1+\sqrt{-23})/2$; it obeys $\theta^2-\theta+6=0$. Let $R$ be the ring $\mathbb{Z}[\theta]$. This is the ring of integers in the field $\mathbb{Q}(\sqrt{-23})$. It turns out that the class group of $R$ is $\mathbb{Z}/3$, representatives for the three ideal classes are: $$I_0:= \langle 1 \rangle,\ I_1:=\langle 2, \theta -1 \rangle,\ I_2:=\langle 2, \theta \rangle.$$ I leave this for you to check from the standard Minkowski bounds. Note that $I_1 I_2 = \langle 4, 2(\theta-1), 2 \theta, 6 \rangle = \langle 2 \rangle$, so $I_1^{-1} = (1/2) I_2$ and vice versa. Note for future use that we are giving a $\mathbb{Z}$ basis for the latter two ideals, and that $\langle 1, \theta \rangle$ is a $\mathbb{Z}$-basis for $I_0$.

An arbitrary ideal equivalent to $I_0$ is of the form $\langle x+y\theta \rangle$, and we have $N(\langle x+y\theta \rangle) = x^2 - xy + 6 y^2$. An arbitrary ideal equivalent to $I_0$ is of the form $\langle x+y\theta \rangle$ in precisely two ways, because the unit group of $R$ is $\pm 1$. So $$\# \{ I \subseteq R : \ I \sim I_0 \ \mbox{and} \ N(I) = n \} = \frac{1}{2} \# \{(x,y) : x^2 - xy + 6y^2 =n \}. \quad (1)$$

Similarly, an ideal equivalent to $I_1$ is of the form $\alpha I_1$, for $\alpha \in I_1^{-1} = (1/2) I_2$. This $\alpha$ looks like $(1/2) (2 x + y(\theta -1))$ for some integers $x$ and $y$ and we have $N(\alpha I_1) = N(\alpha) N(I_1) = (x^2 - xy +3 y^2/2) \cdot 2 = 2x^2 - 2xy + 3 y^2$ (exercise!). Running through the same argument as before: $$\# \{ I \subseteq R : \ I \sim I_1 \ \mbox{and} \ N(I) = n \} = \frac{1}{2} \# \{(x,y) : 2 x^2 - xy + 3y^2 =n \}. \quad (2)$$ Similarly, $$\# \{ I \subseteq R : \ I \sim I_2 \ \mbox{and} \ N(I) = n \} = \frac{1}{2} \# \{(x,y) : 2 x^2 + xy + 3y^2 =n \}. \quad (3)$$

Call these numbers $a_0(n)$, $a_1(n)$ and $a_2(n)$. We want to know that $a_2(p)$ is positive for infinitely many primes.

We have unique factorization into ideals in the ring $R$. So we have the analog of the Euler product $$\sum_n \frac{a_0(n) + a_1(n) + a_2(n)}{n^s} = \prod_{p} \frac{1}{(1- p^{-s})^{a_0(p)+a_1(p)+a_2(p)}}.$$

Let $\omega$ be a primitive cube root of unity. Since the class group is $\mathbb{Z}/3$, with $[I_1]^2=[I_2]$ and $[I_1]^3=[I_0]$, we also have the identities: $$\sum_n \frac{a_0(n) + \omega a_1(n) + \omega^2 a_2(n)}{n^s} = \prod_{p} \frac{1}{(1-p^{-s})^{a_0(p)}} \frac{1}{(1-\omega p^{-s})^{a_1(p)}} \frac{1}{(1-\omega^2 p^{-s})^{a_2(p)}}.$$ $$\sum_n \frac{a_0(n) + \omega^2 a_1(n) + \omega a_2(n)}{n^s} = \prod_{p} \frac{1}{(1-p^{-s})^{a_0(p)}} \frac{1}{(1-\omega^2 p^{-s})^{ a_1(p)}} \frac{1}{(1-\omega p^{-s})^{ a_2(p)}}.$$

Define these three sums to be $L_0(s)$, $L_1(s)$ and $L_2(s)$. Using equations $(1)$, $(2)$ and $(3)$, and approximating sums by integrals, we see that $L_0(s)$ has a simple pole at $s=1$ while $L_1(s)$ and $L_2(s)$ are bounded as $s \to 1$. Using explicit numerical computation, one can check that $L_1(s)$ and $L_2(s)$ do not vanish at $s=1$.

Now the end game is exactly like the proof of Dirichlet's theorem on primes in arithmetic progressions: take logs of the above equations to deduce that $$\log \left( \frac{1}{s-1} \right) + O(1) = \sum \frac{a_0(p)+a_1(p)+a_2(p)}{p^s} + O(1)$$ $$O(1) = \sum \frac{a_0(p)+\omega a_1(p)+ \omega^2 a_2(p)}{p^s} + O(1)$$ $$O(1) = \sum \frac{a_0(p)+\omega^2 a_1(p)+ \omega a_2(p)}{p^s} + O(1)$$

We thus have $$\sum \frac{a_2(p)}{p^s} = (1/3)\log \left( \frac{1}{s-1} \right) + O(1) \ \mbox{as}\ s \to 1^{+}$$ and, in particular, $a_2(p)$ is nonzero infinitely often.


So, where is the class field theory? For $R$ the ring of integers of a number field, $H$ the class group of $R$, and $\chi: H \to \mathbb{C}^*$ a nontrivial character, define $L(\chi, s) = \sum_{I \subseteq R} \chi([I]) N(I)^{-s}$. If you wanted to generalize this proof to an arbitrary quadratic number field, you would want to know that $L(\chi, s) \neq 0$. For any particular $R$ and $\chi$, you can prove this by hand. In Dirichlet's theorem, the analogous result can be proved by looking at the zeta function of the cyclotomic field. In the general case, it can be proved by looking at the $\zeta$ function of the class field -- but only once you know this exists!

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David, in your equations (1), (2), and (3), you forgot to include "= n" on the right side each time. –  KCd Mar 29 '11 at 21:50
    
Also, you wrote a primitive cube root of unity as $\chi$ and then at the end of the answer you use $\chi$ for a character (more standard notation). Why not use a more standard notation for the cube root of unity, say $\omega$? –  KCd Mar 29 '11 at 21:51
    
Thanks, those are good suggestions. –  David Speyer Mar 30 '11 at 13:30

If you are willing to accept Dirichlet's Theorem on primes in arithmetic progression, then you can assume there are infinitely many primes of the form 4n + 3. Your form generates such numbers when y=1 and x is a multiple of 4.

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Yes, but you are not guaranteed to get infinitely many such primes this way. As I mentioned in the comments, a result of this kind is not known for any polynomial in one variable of degree greater than $1$. –  Qiaochu Yuan Mar 30 '11 at 15:05
    
@Qiaochu: My bad. Yes there will be an infinite number of numbers congruent to 3 mod 4, but I have no way of knowing infinitely many of them are indeed prime. Once again, I have proved the old adage "Haste makes waste." –  Chris Leary Mar 30 '11 at 17:12

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