Number Theory: Lecture 7

In which we meet the law of quadratic reciprocity.

  • Proposition 23 (Gauss‘s Lemma): Let p be an odd prime, and let a be coprime to p.  Then \genfrac{(}{)}{}{}{a}{p} = (-1)^{\nu}, where \nu = \#\{k \in \{1, 2, \dotsc, \frac{p-1}{2} \} : \langle ka \rangle < 0 \}.  We proved this by showing that \langle a \rangle, \langle 2a \rangle, …, \langle (\frac{p-1}{2})a \rangle correspond to \pm 1, \pm 2, …, \pm \frac{p-1}{2} in some order, where each of these has a definite sign, and then multiplying them all together to show that a^{\frac{p-1}{2}} \equiv (-1)^{\nu} \pmod{p}.  Then we finished off using Euler’s criterion.
  • Corollary 24: Let p be an odd prime.  Then \genfrac{(}{)}{}{}{2}{p} = (-1)^{\frac{p^2 - 1}{8}}.  We used Gauss’s lemma to prove that 2 is a quadratic residue modulo p if p \equiv \pm 1 \pmod{8} and is a quadratic non-residue modulo p if p \equiv \pm3 \pmod{8}.
  • Theorem 25 (Law of quadratic reciprocity): Let p and q be odd primes.  Then \genfrac{(}{)}{}{}{q}{p} = (-1)^{\frac{p-1}{2} \frac{q-1}{2}} \genfrac{(}{)}{}{}{p}{q}.  We proved this using Gauss’s lemma, thinking about counting lattice points in regions within a rectangle.
  • Exercise: Use the law of quadratic reciprocity to show that if p and q are odd primes with p \equiv \pm q \pmod{4a} then \genfrac{(}{)}{}{}{a}{p} = \genfrac{(}{)}{}{}{a}{q}.

Understanding today’s lecture

Can you use Gauss’s lemma to prove any other results with the flavour of Corollary 24 (the one about \genfrac{(}{)}{}{}{2}{p})?

Can you find the values of the Legendre symbols \genfrac{(}{)}{}{}{19}{73} and \genfrac{(}{)}{}{}{34}{97}?  We’ll do these as examples at the start of the next lecture.

Can you use the exercise to give another proof of quadratic reciprocity (first prove the result in the exercise directly from Gauss’s lemma, then deduce quadratic reciprocity from the result in the exercise).

Further reading

Davenport (The Higher Arithmetic), Baker (A concise introduction to the theory of numbers), and Jones and Jones (Elementary number theory) all deal with these topics, as do many other books.  There are several ways to prove the law of quadratic reciprocity; Davenport and Baker (for example) choose different approaches.  The exercise above is to deduce a result from the law of quadratic reciprocity.  I encourage you to try to prove the result directly from Gauss’s lemma, and then to deduce the law of quadratic reciprocity from the result.  This is the approach that Davenport describes, and that Gareth Taylor has summarised in these online notes.  Baker and Jones & Jones both give the lattice-point proof of quadratic reciprocity that we saw in lectures.

The law of quadratic reciprocity is really central to number theory.  The Wikipedia page about quadratic reciprocity has a bit of discussion about the history of the ideas involved, and about ways in which mathematicians moved on to study higher reciprocity laws, with more suggestions for further reading.

Preparation for Lecture 8

Next time, following a couple of numerical examples, we’ll think about how to generalise the Legendre symbol to situations when the number on the bottom isn’t prime (we’ll end up defining the Jacobi symbol).  Which of the results we’ve proved for the Legendre symbol will turn out to hold for the Jacobi symbol too?


One Response to “Number Theory: Lecture 7”

  1. Number Theory: Lecture 8 | Theorem of the week Says:

    […] Expositions of interesting mathematical results « Number Theory: Lecture 7 […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: