\classheader{2013-01-14}

\subsection*{Accomplishments in algebraic number theory in the 19th century}

In the middle of the 19th century, Kummer started the theory of ideals.

Two motivations of Kummer:
\begin{enumerate}
\item He hoped to prove Fermat's Last Theorem.
\item He hoped to make progress in class field theory.
\end{enumerate}

Fermat (1601-1665) is the father of modern number theory. As you should all be familiar, he stated Fermat's Last Theorem, that there are no integer solutions to
\[x^n+y^n=z^n\]
when $n\geq 3$ and $xyz\neq 0$. It was written in the margin's of Diophantus' \textit{Arithmetica}, which was written in the 3rd century.

He also wrote other things in the margins; for example, he claimed that any non-negative integer could be written as a sum of four squares, i.e. that for any $n\geq 0$, there exist some $x,y,z,u\in \Z$ such that
\[n=x^2+y^2+z^2+u^2.\]
However, it took people about 100 years to prove this.

Fermat did give a proof of Fermat's Last Theorem for $n=4$. 100 years after that, Euler gave a proof for the case $n=3$. Prior to Kummer, the cases $n=5$ and $n=7$ were also known.

Fermat's Last Theorem can be reduced to the cases $n=4$ and $n=$ a prime, because (for example) we can re-express
\[x^6+y^6=z^6\quad\implies\quad (x^2)^3+(y^2)^3=(z^2)^3.\]

\begin{theorem}[Kummer]
Assume that $n$ is an odd prime number. If the class number of $\Q(\zeta_n)$ is not divisible by $n$, then Fermat's Last Theorem is true for $n$.
\end{theorem}

The only prime numbers less than 100 which do not have this property (i.e. the class number is divisible) are $n=37,59,67$.

The class number is the order of the group of classes of fractional ideals.

Kummer could prove Fermat's Last Theorem if the ring $\Z[\zeta_n]$ had unique factorization into irreducibles (note that $\Z[\zeta_n]=\mathcal{O}_K$ where $K=\Q(\zeta_n)$), because we can re-express Fermat's Last Theorem as a multiplicative statement in this ring:
\[x^n=z^n-y^n=\prod_{i=0}^{n-1}(z-\zeta_n^iy).\]
The class number measures how badly unique factorization fails.

\subsection*{Let Us Share the Feelings of Kummer}

Kummer was very happy when he proved that
\[n\nmid\text{class number of }\Q(\zeta_n)\implies\text{FLT for }n\text{ is true}.\]
But Kummer was not happy that he could not treat the case that $n\mid \text{class number}$. 

As an example of this technique, we can prove that
\[x^3=y^2+4\text{ for }x,y\in\Z\quad\implies\quad (x,y)=(5,\pm 11)\text{ or }(2,\pm 2)\]
by using unique factorization in $\Z[i]$.

We are very happy to prove that
\[x^3=y^2+20\quad\implies\quad (x,y)=(6,\pm 14)\]
by using the fact that $3\nmid $ the class number of $\Q(\sqrt{-5})$ (which is 2); the problem can be re-expressed as
\[x^3=(y+2\sqrt{-5})(y-2\sqrt{-5}).\]

However, we are not happy that we have difficulty for $x^3=y^2+26$, because $3\mid $ the class number of $\Q(\sqrt{-26})$ (which is 6). Note that
\[3^3=(1+\sqrt{-26})(1-\sqrt{-26}),\]
so that in the world of ideals,
\[(3)=\frak{p}\frak{p}'\]
where $\frak{p}$ and $\frak{p}'$ are prime ideals such that $\frak{p}^3=(1+\sqrt{-26})$ and $(\frak{p}')^3=(1-\sqrt{-26})$.

Let $\frak{q}=(2,\sqrt{-26})$, which is a prime ideal with $\frak{q}^2=(2)$. The the class group of $\Q(\sqrt{-26})$ consists of the ideal classes
\[\{1,\;\text{class}(\frak{p}),\;\text{class}(\frak{p}^2),\;\text{class}(\frak{q}),\;\text{class}(\frak{q}\frak{p}),\;\text{class}(\frak{q}\frak{p}^2)\;\}\cong \Z/2\Z\times\Z/3\Z\cong \Z/6\Z.\]
In the class group, we have $\text{class}(\frak{p})^3=1$.

Going back to the problem $x^3=y^2+4$, we can re-express it as
\[x^3=(y+2i)(y-2i).\]
For any prime element $\pi\in\Z[i]$, if $\pi\mid y+2i$, we have $\pi\mid x^3$, and therefore $\pi\mid x$.

If $\pi\mid y+2i$ and $\pi\mid y-2i$, then $\pi\mid (y+2i)-(y-2i)=4i$, so $\pi\mid 2$, so that $(\pi)=(1+i)$.

For an $\alpha\in\Z[i]\setminus\{0\}$, define
\[\ord_\pi(\alpha)=\text{the exponent }e\text{ such that }\pi^e\mid \alpha,\;\pi^{e+1}\nmid \alpha\]
where $\pi$ is a prime element of $\Z[i]$. Then if $\pi\mid y+2i$,
\[\ord_\pi(y+2i)=3\ord_\pi(x)\text{ if }(\pi)\neq(1+i),\]
which implies that $3\mid \ord_\pi(y+2i)$. If $(\pi)=(1+i)$, then
\[\ord_\pi(y+2i)=\ord_\pi(y-2i)\]
since $y-2i=\overline{y+2i}$, so that
\[2\ord_\pi(y+2i)=3\ord_\pi(x),\]
and therefore $3\mid \ord_\pi(y+2i)$. Thus, we have shown that $3\mid \ord_\pi(y+2i)$ for all prime divisors $\pi$ of $y+2i$. Thus $y+2i=\alpha^3$ for some $\alpha\in\Z[i]$. We'll finish this next time.


