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

Chapter 21.2 of Friedlander and Iwaniec's Opera de Cribro begins (essentially) as follows:

Let $$F(X, Y) = aX^2 + bXY + cY^2 + \alpha X + \beta Y + \gamma \in \mathbb{Z}[X, Y]$$ be irreducible in $\mathbb{Q}[X, Y]$, represent arbitrarily large odd integers, and satisfy $(a, b, c, \alpha, \beta, \gamma) = 1$. We want $F(X, Y)$ to depend essentially on two variables, which may be expressed by the requirement that $\frac{\partial F}{\partial X}, \frac{\partial F}{\partial Y}$ are linearly independent.

We introduce two discriminants: $$\Delta = b^2 - 4 ac,$$ $$D = a \beta^2 - b \alpha \beta + c \alpha^2 + \Delta \gamma.$$

[end of excerpt]

Wait, they do what?

F+I go on to state that any $F(X, Y)$ satisfying the above conditions represents infinitely many primes: on the order of $\frac{x}{\log x}$ primes $\leq x$ if $D = 0$ or $\Delta$ is a square, and on the order of $\frac{x}{(\log x)^{3/2}}$ otherwise.

They say only a very little bit about their proof, referring instead to this paper of Iwaniec. Obviously these discriminants are important (Iwaniec calls them the "small" and "large" discriminants respectively), but Iwaniec does not explain them in any highbrow fashion; he simply dives into the guts of some computations. Predictably, they are related to affine changes of coordinates (we want to say that $F, F'$ are equivalent if an invertible such transformation changes one into the other), but if the role of $D$ is something simple then this wasn't apparent from a quick look at the paper.

Are these two discriminants classical and familiar? Is there a simple highbrow explanation of what they say about the polynomial $F$? It seems there must be.

(See also here for a related MO question.)

Thank you!

share|improve this question
2  
If you homogenize F, you get a ternary quadratic forms. These were studied by Gauss in his Disquisitiones. There must be a lot on these objects in any book on classical invariant theory (you might want to look at the recent book by Olver with this title). –  Franz Lemmermeyer Jul 14 '12 at 19:28
    
@Franz: Qiaochu answered my question (the answer turned out to be quite simple), but this book looks quite interesting and I plan to order it. Thanks! –  Frank Thorne Jul 16 '12 at 13:16

2 Answers 2

up vote 2 down vote accepted

Here are some naive comments. If $X, Y$ are large then the contribution of the quadratic terms swamps the other terms, so first let's concentrate on the quadratic terms $$Q(X, Y) = a X^2 + b XY + c Y^2.$$

This can be written as $$Q(X, Y) = \left[ \begin{array}{cc} X & Y \end{array} \right] \left[ \begin{array}{cc} a & \frac{b}{2} \\\ \frac{b}{2} & c \end{array} \right] \left[ \begin{array}{c} X \\\ Y \end{array} \right]$$

and the matrix occurring above has determinant $ac - \frac{b^2}{4} = - \frac{\Delta}{4}$. $Q$ is positive-definite if and only if $\Delta < 0$, in which case this determinant has concrete geometric significance: it is $\pi$ times the reciprocal of the area of the ellipse $Q(X, Y) \le 1$. Consequently it describes the asymptotics of the number $q_n$ of pairs of integers $(X, Y)$ such that $Q(X, Y) \le n$ (and the same should be true of $F$). In all cases, $\Delta$ is invariant under affine change of coordinates.

$F$ itself can be written as

$$F(X, Y) = \left[ \begin{array}{ccc} X & Y & 1 \end{array} \right] \left[ \begin{array}{ccc} a & \frac{b}{2} & \frac{\alpha}{2} \\\ \frac{b}{2} & c & \frac{\beta}{2} \\\ \frac{\alpha}{2} & \frac{\beta}{2} & \gamma \end{array} \right] \left[ \begin{array}{c} X \\\ Y \\\ 1 \end{array} \right]$$

and $D$ is $4$ times the determinant of this matrix. Again this is invariant under affine change of coordinates. $D$ is a natural invariant of the homogenization of $F$ to a ternary quadratic form.


Let me also say some naive things about the two conditions appearing in the OP. By the quadratic formula, the quadratic polynomial $ax^2 + bx + c$ has roots $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Consequently, $\Delta$ is a square if and only if this polynomial has rational roots, hence if and only if $Q$ splits into a product of factors $$Q(X, Y) = (dX + eY)(fX + gY)$$

over $\mathbb{Z}$ (by Gauss' lemma). As Will Jagy says, these quadratic forms have very different behavior as far as representing primes compared to quadratic forms that do not factor this way.

If $\Delta$ is not a square, then it is in particular not zero. Now, if $D = 0$, then the $3 \times 3$ matrix above has nontrivial nullspace. I believe this is equivalent to being able to write $F$ as the product of two linear polynomials (after a suitable quadratic extension) as opposed to the product of two linear polynomials plus a constant.

share|improve this answer
    
"$D$ is 4 times the determinant of this matrix": aagh, somehow missed that. There's the simple explanation! +1, sir. –  Frank Thorne Jul 16 '12 at 13:14

I made two corrections in your formula for $D.$ It is now consistent with the 1974 Acta Arithmetica paper.

Some examples. If $\alpha, \beta, \gamma$ are all $0,$ we have a quadratic form, which is required primitive. For positive forms, the proportion of primes represented is a constant times the full count of primes, this is Cebotarev density. See Theorem 9.12 on page 188 of David A. Cox, Primes of the Form $x^2 + n y^2.$

In comparison, $X^2 + Y^2 + 1$ represents a constant times $\frac{x}{( \log x )^{3/2}}$ primes up to $x.$ I was not aware of this. Note that this case was proved by IWANIEC 1972 where the pdf can be downloaded. He does this special case and improves on estimates of Motohashi.

My understanding is that, for nondegenerate indefinite quadratic forms, that is $aX^2 + b XY + c Y^2$ with $\Delta = b^2 - 4 a c$ nonnegative but not zero or a square, the number of primes $p$ represented and the number of primes $q$ such that $-q$ is represented both obey a Cebotarev-like law. Franz would know details. I am taking primes as positive only. For example, $X^2 - 3 Y^2$ represents all (positive) primes $p \equiv 1 \pmod {12},$ and all $-q$ for positive primes $q \equiv 11 \pmod {12}.$

If, instead, I take a degenerate indefinite form as $X^2 - Y^2,$ I can represent all multiples of $4$ and all odd numbers. So the number of primes up to some positive bound $x$ is just the usual $\frac{x}{\log x}$ from the Prime Number Theorem, without any constant multiplier.

Finally, what happens with $X^2 + Y^2 + 2 X + 1$ or $X^2 - Y^2 + 2 X + 1?$ These can be rewritten with $(X+1)^2 = X^2 + 2 X + 1.$

So, overall, they are saying that two distinct cases give PNT times a constant, either degenerate (representing an entire arithmetic progression containing more than one prime) or a genuine nondegenerate quadratic form, which does not represent any arithmetic progression but may represent all primes in an arithmetic progression (such as $X^2 + Y^2$ and $4n+1,$ which behavior requires very small class number) but in any case follows a Cebotarev-like rule for primes.

As an example without arithmetic progressions, $x^2 + xy + 6 y^2$ and $2x^2 \pm xy + 3 y^2.$ Both represent only primes $23$ and $p$ that satisfy $(p|23) = 1.$ The first represents those for which $z^3 - z + 1$ has a root $\pmod p,$ accounting for $1/6$ of all primes in the long run, the latter pair of forms (both) represent the others, accounting for $1/3$ of all primes. Oh, $23$ itself is represented by the first one.

The other case is like $X^2 + Y^2 + 1,$ representing fewer primes but still infinite.

So, they have managed to put together a number of different cases with two constants.

share|improve this answer
    
nice comments, thank you! –  Frank Thorne Jul 16 '12 at 13:19

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.