Skip to main content Link Search Menu Expand Document (external link) Copy Copied
Abstract Algebra - Graduate Course

##

Proposition: A polynomial $f(x)\in F[x]$ has a root $r\in F$ if and only if $(x-r)$ divides $f$.

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

Proposition: A finite subgroup of the multiplicative group of a field is cyclic.

Proof: Let $U$ be such a subgroup. By the fundamental, theorem of abelian groups, $U$ is the product of its Sylow $p$-subgroups. Let $U(p)$ be such a subgroup. If $U(p)$ were not cyclic, then $U(p)$ and hence $U$ would have more than $p$ elements that are solutions to the equation $x^p=1$. But $x^p-1$ has at most $p$ roots. Since $U(p)$ is cyclic for each $p$ dividing the order of $U$, $U$ itself is cyclic.

Corollary: The group of units $(\Zn{p})^{\times}$ is cyclic.

Generators of this group are called primitive roots mod $p$.

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 $x\in \Z[i]$.

Algorithm for Fermat’s theorem

Suppose $p\equiv 1\pmod{4}$. To write $p=a^2+b^2$, find a solution $u$ to the congruence $u^2\equiv -1\pmod{p}$. Then use the Gaussian Euclidean algorithm to find a generator $\pi$ for the ideal $(p,u+i)$ in $\Z[i]$. This generator divides $p$ so its norm is a divisor of $p^2$. If its norm were $p^{2}$, then $\pi$ would be an associate of $p$ and this would mean $p$ divides $u+i$, which it visibly does not. If its norm were $1$, then the ideal $(p,u+i)$ would be all of $\Z[i]$ and so we would have $px+(u+i)y=1$ in $\Z[i]$. But in that case, multiplying by $(u-i)$ would be $px(u-i)+(u^2+1)y=(u-i)$ and since $p$ divides the left side we’d have $p$ dividing $u-i$, which is not true. So therefore $N(\pi)=p$ and so if $\pi=a+bi$ we have $a^2+b^2=p$.