Proposition:
The integral lattices in $\mathbb{R}^n$ are precisely the symmetric bilinear form modules $(S,b)$ over the ring of integers $\mathbb{Z}$, where $S$ is a free $\mathbb{Z}$ module of rank $n$ and $b:S\times S\to \mathbb{Z}$ is a positive definite symmetric bilinear form.
The proof is really taciturn. The $\Rightarrow$ part, I understood. But the other implication is really briefly explained and I have troubles understanding it:
On the other hand, let $(S,b)$ be a symmetric bilinear form module over $\mathbb{Z}$ with a positive definite symmetric bilinear form $b$. Consider the real vector space $V:=S\otimes\mathbb{R}$. Then $b$ extends to a scalar product on $V$. Let $(\varepsilon_1,...,\varepsilon_n)$ be an orthonormal basis of $V$ with respect to the scalar product $b$, and identify $V$ with $\mathbb{R}^n$ by mapping $(\varepsilon_1,...,\varepsilon_n)$ to the standard basis of $\mathbb{R}^n$. Then $b$ becomes the Euclidean scalar product of $\mathbb{R}^n$ and $S\subset \mathbb{R}^n$ is an integral lattice. $\Box$
Questions
- What does $S\otimes \mathbb{R}$ mean? The tensor product of a free $\mathbb{Z}$ module with $\mathbb{R}$? How is that a vector space?
- What does it mean for $b: S\times S \to \mathbb{Z}$ to be extended to a scalar product on $V=S\otimes\mathbb{R}$? We're appending one more component, but how does this make a scalar product? $b$ isn't non-degenerate.
- Does "identify" mean that there is an isomorphism between $V$ and $\mathbb{R}^n$ given by a transformation matrix between the basis $(\varepsilon_1,...,\varepsilon_n)$ and the standard basis of $\mathbb{R}^n$?
Indeed, $S\otimes \mathbb{R}$ is the tensor product of $S$ and $\mathbb{R}$. Technically it should be written $S\otimes_\mathbb{Z} \mathbb{R}$, since in that case $S$ and $\mathbb{R}$ are viewed as $\mathbb{Z}$-modules, but this is often omitted when there is no possible confusion. In general, if $A$ is a commutative ring, $S$ is an $A$-module, and $B$ is an $A$-algebra, then $S\otimes_A B$ is naturally a $B$-module, with the action $b\cdot (s\otimes x)=s\otimes (bx)$ for all $s\in S$ and $x,b\in B$. In your case, $A=\mathbb{Z}$ and $B=\mathbb{R}$, so $S\otimes_\mathbb{Z} \mathbb{R}$ is an $\mathbb{R}$-module, that is to say an $\mathbb{R}$-vector space.
For any $\mathbb{Z}$-bilinear form $b:S\times S\to \mathbb{Z}$, there is a unique extension of $b$ to an $\mathbb{R}$-bilinear form $$b_\mathbb{R}: (S\otimes_\mathbb{Z} \mathbb{R})\times (S\otimes_\mathbb{Z} \mathbb{R})\to \mathbb{R}.$$ It is given by the formula $b_\mathbb{R}(x\otimes a,y\otimes b)=b(x,y)\cdot ab$. We can without ambiguity use the word "extend" because there is a natural embedding $S\to S\otimes \mathbb{R}$ given by $s\mapsto s\otimes 1$. You say that in your example $b$ is not non-degenerate, but actually it is since you say it's definite positive. It's easy to show that in that case $b_\mathbb{R}$ is also definite positive, and therefore a euclidean form.
What you say is basically correct, though there is no meaningful "transformation matrix" in that case. In general, the sentence "identify $X$ with $Y$ by..." means "define an isomorphism between $X$ and $Y$ by...". Here you define an isomorphism of vector spaces between $S\otimes_\mathbb{Z} \mathbb{R}$ and $\mathbb{R}^n$, which is given by the image of a certain basis. It's not really meaningful to talk about a transformation matrix here, since to do that you need to fix a basis on both sides, and here the isomorphism precisely sends the only two mentioned bases to each other; so at best the transformation matrix is, by definition, the identity.