MATH 3240 Final Review

number-theory

MATH 3240

Review Session Notes

Wilson’s Theorem

He won’t ask us to prove the main law of quadratic reciprocity, nor the supplementary law for $2$. This seems like a hint that another supplementary law will be on there!

We will get a list of squares $\mathrm{mod}p$ (up to $13$).

Theorem Problem 7

For primes $p>3$,

• $\left(\frac{3}{p}\right)=1⟺p\equiv 1,11\mathrm{mod}12$.
• $\left(\frac{6}{p}\right)=1⟺p\equiv 1,5,19,23\mathrm{mod}24$.

Recall the main law of quadratic reciprocity $\left(\frac{q}{p}\right)=(-1{)}^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right)$.

So $\left(\frac{3}{p}\right)=(-1{)}^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p}{3}\right)=(-1{)}^{\frac{p-1}{2}}\left(\frac{p}{3}\right)$. It is easy to decide when each term is $1$ or $-1$.

$\left(\frac{3}{p}\right)=1⟺(-1{)}^{\frac{p-1}{2}}=1,\left(\frac{p}{3}\right)=1$ OR $(-1{)}^{\frac{p-1}{2}}=-1,\left(\frac{p}{3}\right)=-1$.

In the first case, we have $p\equiv 1\mathrm{mod}4$ and $p\equiv 1\mathrm{mod}3$, thus $p\equiv 1\mathrm{mod}12$.

In the second case, we have $p\equiv 3\mathrm{mod}4$ and $p\equiv 2\mathrm{mod}3$, thus $p\equiv 11\mathrm{mod}12$.

Now for $\left(\frac{6}{p}\right)$

$\left(\frac{6}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)$

He won’t ask us to do work with Bezout’s Identity in $\mathbb{Z}[\sqrt{2}]$

Know how to do division algorithm in all the fields we used ($\mathbb{Z}$, $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{d}]$, $\mathbb{F}[x],$ etc)

Understand the proofs on the review sheets, and the ones on the exam should be no issue.

Same with numerical problems.

Know how to determine if there exists a solution to a quadratic congruence.

Division Algorithm in

$\mathbb{Z}$: For $a,b\in \mathbb{Z}$ with $b\ne 0$ $\mathbb{Z}[i]$: For $\alpha ,\beta \in \mathbb{Z}[i]$, with $N(\beta )\ne 0$, there exists $\gamma ,\rho \in \mathbb{Z}[i]$ such that

Know the main law of quadratic reciprocity as well as the two supplementary laws we covered (-1 and 2).z

Look up in every $\mathbb{Z}[\sqrt{d}]$ that $N(\alpha )=\pm p$ for prime number $p$ implies that $\alpha$ is prime in $\mathbb{Z}[\sqrt{d}]$.