## One variable polynomials over a field

**Lemma:** If $f(x)\in F[x]$, and $r\in F$, then $f(r)=0$ if and only if $(x-r)$ divides $f$.

**Proposition:** A polynomial of degree $n$ over a field $F$ has at most $n$ roots in $F$.

## Finite subgroups of fields

**Proposition:** Let $U\in F^{\times}$ be a finite subgroup of a field $F$. Then $U$ is cyclic.

**Corollary:** $\Zn{p}^{\times}$ is cyclic. Note: the number theorists call a generator of $\Zn{p}^{\times}$ a *primitive root* modulo $p$. There are $\phi(p-1)$ primitive roots modulo $p$.

## More discussions of quadratic rings

**Proposition:** (Fermat) A prime number is the sum of two squares if and only if it is $2$ or is congruent to $1$ mod $4$.

**Lemma:** The congruence $x^2\equiv -1\pmod{p}$ has a solution modulo a prime $p$ if and only if $p=2$ or $p\equiv 1\pmod{4}$.

**Proof:** If $p=2$, $1$ is a solution. If $p$ is odd, and $x^2=-1$ has a solution, then $(\Zn{p})^{*}$ has an element of order $4$, so $4\divides (p-1)$. Notice that $(\Zn{p})^{*}$ has only two elements of order dividing $2$, because of $x^2\equiv 1\pmod{p}$ then $p\divides (x^2-1)$, so $p\divides (x+1)(x-1)$, so either $x\equiv 1\pmod{p}$ or $x\equiv -1\pmod{p}$. If $4\divides (p-1)$ then let $H$ be the Sylow $2$-subgroup of $(\Zn{p})^{*}$. If $H$ were not cyclic, then there would be too many elements of order $2$ in $H$. So $H$ must be cyclic and therefore there is an element of order $4$.

Now suppose that $p\equiv 1\pmod{4}$. Let $u$ be a solution to $x^2+1\equiv 0\pmod{p}$. Consider the ideal $I=(p,u+i)\subset \Z[i]$. This is a maximal ideal. If $\pi=a+bi$ is a generator of this ideal, then $p=x\pi$. If $x$ were a unit, then $u+i$ would have to be a multiple of $p$, which it visibly isn’t. Therefore $N(\pi)$ must be $p$.

But $N(\pi)=a^2+b^2$, so we’ve found our representation.

**Proposition:** The ring $\Z[\sqrt{-5}]$ is not a Euclidean ring. In fact, the ideal $(3,1+\sqrt{-5})$ is not principal. It is a proper ideal, because the quotient of $\Z[\sqrt{-5}]$ by this ideal is $\Zn{3}$. If $\pi$ were a generator of this ideal, then $3=x\pi$ means that either $N(\pi)=3$ or $N(\pi)=9$. Also $(1+5i)=y\pi$ means that $N(\pi)$ divides $6$. Since $\pi$ is not a unit, $N(\pi)=3$. But the equation $x^2+5y^2=3$ has no integer solutions, so there is no element of norm 3 in this ring.

**Proposition:** $\Z[\sqrt{2}]$ is Euclidean with respect to the norm. It has an infinite unit group.

**Proposition:** Let $\rho=e^{2\pi/3}$. Then $\Z[\rho]$ is Euclidean (and it has six units).

## Back to the Gaussian integers

The irreducibles in $\Z[i]$ are:

- $(1+i)$
- $p\in\Z$ with $p\equiv 3\pmod{4}$
- $a\pm bi$ where $a^2+b^2=p$ for $p\in Z$ and $p\equiv 1\pmod{4}$.

A positive integer is a sum of two squares if and only if it factors \(n=2^k p_1^{e_1}\cdots p_k^{e_k}q_{1}^{f_1}\cdots q_{r}^{f_{r}}\) where the $p_{i}\equiv 1\pmod{4}$ and the $q_{i}\equiv 3\pmod{4}$ and all the $f_{i}$ are even.

The proof follows from the question of when is $n=N(x)$ for some $x\in \Z[i]$.