An 'obvious' property of algebraic integers?

Question

I am looking at the book A Brief Guide to Algebraic Number Theory by H. P. F. Swinnerton-Dyer. I like the section on page 1 'the ring of integers' as it gives a motivation for choosing which elements we would like to regard as integers and how we get the definition in terms of monic polynomials.

He lists the 'obvious' properties which one would want the integers ${\frak{o}}_k$ of an algebraic number field $k$ to have. Property number 3 is:

${\bf{3.}} \ {\frak{o}}_{k} \otimes_{\mathbb{Z}} \mathbb{Q}= k $.

I have not come across this tensor product notation before, but I have a feeling this statement is related to the requirement that the field $k$ should be the field of fractions of ${\frak{o}}_k$. Is this the case, and if so how can the statement 3 be 'translated' into this requirement? Is it really as obvious as he claims? Why do you think he has chosen to state it in this way?

A link to the book.

Answer 1

Essentially, this means that every element of $k$ can be written as $\frac{\alpha}{n}$ where $\alpha\in\mathcal{O}_k$ and $0\neq n\in\mathbb Z$. This is stronger than the field of fractions property you've given.

Answer 2

The tensor product ${\frak o}_k\otimes_{\mathbb Z}\mathbb Q$ is not necessarily the ring of fractions - or at least not by this formulation. Instead it is simply (when viewing both ${\frak o}_k$ and $\mathbb Q$ as subsets of $k$) the set of rational linear combinations of elements of ${\frak o}_k$. In other words, the obvious property is that ${\frak o}_k$ should span the $\mathbb Q$-vector space $k$.

Answer 3

Let me try to answer this, since it has also confused me in the past.

What the tensor product with $\mathbb{Q}$ is doing is generally called an extension of scalars. You know that the ring of integers is initially a $\mathbb{Z}$-module of rank $n$, where $n = [k:\mathbb{Q}]$. So you can actually write

$$ \mathfrak{o}_k = \alpha_1\mathbb{Z} + \cdots + \alpha_n\mathbb{Z} $$

for some algebraic integers $\alpha_1, \dots, \alpha_n \in \mathfrak{o}_k$. Then when you tensor with $\mathbb{Q}$ you are basically making that $\mathbb{Z}$-module into an $n$-dimensional vector space over $\mathbb{Q}$, namely you get

\begin{align} \mathfrak{o}_k \otimes_\mathbb{Z} \mathbb{Q} &= (\alpha_1\mathbb{Z} + \cdots + \alpha_n\mathbb{Z})\otimes_\mathbb{Z} \mathbb{Q}\\ &\cong \alpha_1\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Q} + \cdots + \alpha_n\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Q}\\ &\cong \alpha_1 \mathbb{Q} + \cdots + \alpha_n \mathbb{Q}\\ &= k \end{align}

where I used the fact that tensor products commute with direct sums and also that if $M$ is an $R$-module, then $R \otimes_{R} M \cong M$.

Answer 4

See http://en.wikipedia.org/wiki/Tensor_product_of_modules for information about the tensor product. As Thomas Andrews' answer implies, saying ${\frak{o}}_{k} \otimes_{\mathbb{Z}} \mathbb{Q}= k$ is just a fancy way of saying that for any $x$ in $k$, there is a $c \in \mathbb{N}_{{>}0}$ such that $cx \in {\frak o}_k$. I would imagine the author expects the reader to know this and has stated it this way for reasons of conciseness.

As for your second question: it is not obvious to me that one would expect ${\frak o}_k$ to have this property while deciding how to define it. But it is a nice property and there is an easy argument that the definition in terms of monic polynomials delivers it: if $x$ is a root of a polynomial $f$ with integer coefficients and leading coefficient $c$ then $cx$ is the root of a monic polynomial with integer coefficients (as you can see by developing $(cx)^n = (cx)^n - c^nf(x)$ as a polynomial of degree $n-1$ in $cx$, where $n$ is the degree of $f$).

Answer 5

A property we would like to have is that the ring of integers $\mathcal{O}_K$ of an algebraic number field $K$ of degree $n$ is a free $\mathbb{Z}$-module of rank $n$. Then we can choose an integral basis $x_1,\ldots ,x_n$, which is a $\mathbb{Q}$-basis of $K$. In other words, $\mathcal{O}_K\otimes_{\mathbb{Z}}\mathbb{Q}=K$. Usually we define the ring of integers by integral elements with a monic polynomial, and the above property is a special case of the following result: if $A$ is an integrally closed domain with quotient field $K$, and $L$ a separable field extension of $K$, and $B$ the integral closure of $A$ in $L$, then $B$ is a free $A$-module, provided $A$ is a PID. In our case, $A=\mathbb{Z}$ is a PID, hence the result.