Lattice with conditions

Question

In Book's Algebraic number theory, I. Stewart page 142:

Theorem 7.2: If $p$ is prime of the form 4k+1 then $p$ is sum of two squares.

Proof: The multiplicative group $G$ of the field $\mathbb Z/(p)$ is cyclic and has order $p-1=4k$. It therefore contains an element $u$ of order $4$. Then $u^2\equiv -1 \ (mod\ p)$ since $-1$ is the only element of order $2$ in $G$.

Let $L\subset \mathbb Z^2$ be the lattice in $\mathbb R^2$ consisting of all pairs $(a,b)$ with $a,b\in\mathbb Z$ such that

$$b\equiv ua\ (mod \ p)$$.

This is a subgroup of $\mathbb Z^2$ of index $p$ ...

They could help in the following steps:

a) How prove is a Lattice?

b) Why is a subgroup of index $p$?

Thank you all.

Answer 1

A lattice is a finitely generated free abelian group in this context. The set of all given $(a,b)$ is the free abelian group generated by $(1,u)$ and $(0,p)$, so it is a lattice. $\mathbb{Z}\oplus \mathbb{Z}$ has a decomposition as the direct sum of the subgroup generated by $(1,u)$ and the subgroup generated by $(0,1)$. Thus for any element $x$ of $\mathbb{Z}\oplus \mathbb{Z}$ we have that $px$ is in the subgroup we are considering, so the subgroup has index at most $p$. Thus the only possibilities are index 1 or index $p$, and since there are elements not contained in the subgroup it must be of index $p$.

Answer 2

Since you are refering To I.Stewart and Tall's Book, I think he had another proof in mind, using the following Result from the Book:

Theorem: $H \leq G$ is a subgroup of a free abelian group of rank $n$:

If $|G/H|<\infty$, then $H$ is of rank $n$ and $|G/H| = |\det(B_G^H)|$ (with $B_G^H$ the change of basis matrix from $G$ to $H$).

Once you see that the lattice $L:= \left\{x\begin{pmatrix}1\\u \end{pmatrix} + y \begin{pmatrix}0\\p \end{pmatrix}\Big|\ x,y \in \mathbb{Z} \right\}$, generates the desired subgroup. The index is $|\mathbb{Z}^2/L|$ and you can see that it is finite since $p\begin{pmatrix}1\\u \end{pmatrix} + y \begin{pmatrix}0\\p \end{pmatrix} \in L$.

So we can apply the Theorem and here, $B_G^H = \begin{pmatrix}1 & u \\0 & p\end{pmatrix}$ and thus $\det(B_G^H) = p= |\mathbb{Z}^2/L|$.