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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.

user50229
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  • See also: Why do we use this definition of “algebraic integer”? – Xodarap Oct 08 '13 at 16:52
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    I feel obligated to mention, just for academic interest, Dino Lorenzini's take on this. He says that the obvious properties $\mathcal{O}_K$ should have is that $\mathcal{O}_K\cap \mathbb{Q}=\mathbb{Z}$ (for obvious reasons), $\text{Frac}(\mathcal{O}_K)=K$ (to mimic the case of $\mathbb{Z}\subseteq\mathbb{Q}$, and that if $K/\mathbb{Q}$ is Galois, then $\sigma(\mathcal{O}_K)=\mathcal{O}_K$ for all $\sigma\in\text{Gal}(K/\mathbb{Q})$ (otherwise any "naturality" in our choice of $\mathcal{O}_K$ is lost--why not pick one of its Galois conjugates?). You can prove that, as a fun exercise – Alex Youcis Oct 08 '13 at 19:22
  • that already in the case of a Galois extension $K/\mathbb{Q}$ the above three properties describe precisely the standard definition of $\mathcal{O}_K$ as the integral closure of $\mathbb{Z}$ in $K$. Also, it's worth mentioning that there are also strong analogies to the study of curves, deep, deep analogies. Just as a simple observation, in the theory of curves, one often obtains non-singular curves by taking finite extensions of $k[x]$ and looking at it's integral closure. For example, if $k[x,y]/(f(x,y))$ is a non-singular plane curve (monic in $y$ and irreducible) then – Alex Youcis Oct 08 '13 at 19:25
  • the closure of $k[x]$ in $\text{Frac}(k[x,y]/(f(x,y)))$ is precisely $k[x,y]/(f(x,y))$. So, taking integral closures is analogous to producing smooth curves. As an even further analogy, note that if you have a singular curve $k[x,y]/(f(x,y))$ then $k[x,y]/(f(x,y))$ will actually be an order in the integral closure of $k[x]$ in $\text{Frac}(k[x,y]/(f(x,y)))$. This tells us why we think of orders in number rings as being "almost non-singular"--they just have a few singularities (non-invertible fractional ideals, or points whose localization is not a DVR). – Alex Youcis Oct 08 '13 at 19:28
  • There are deep reasons for why this is the "right" definition of the ring of integers. Sorry for so many comments! – Alex Youcis Oct 08 '13 at 19:29
  • @Alex No problem, thanks for your interesting comments. Are you referring to the study of elliptic curves or something more general? – user50229 Oct 08 '13 at 19:33
  • @user50229 I am referring to the general fact that if you squint hard enough, basic algebraic number theory and basic theory of curves over an algebraically closed start to look eerily similar--this is no mistake. – Alex Youcis Oct 08 '13 at 19:36
  • @Alex It sounds like I would have to look into this a lot further! I suppose something like Swinnerton-Dyer's introduction is the closest we can get to an 'elementary' level motivation for the definition of algebraic integers. It's just that like you mentioned, I think it makes sense to assume Frac(${\mathfrak{O}}_K) = K$ rather than the point 3 above. – user50229 Oct 08 '13 at 19:42
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    @user50229 Depending on your level, if you have some mathematical maturity and are comfortable with algebra, then I highly, highly reccomend Dino Lorenzini's book An Invitation to Arithmetic Geometry (don't let the title scare you, it's not nearly as advanced as it sounds). This is the best introduction to number theory I have yet seen, and it does so while introducing you to the theory of plane curves, and drawing the amazing, non-coincidental parallels between them . – Alex Youcis Oct 08 '13 at 19:45

5 Answers5

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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.

Thomas Andrews
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    Yes I have seen this in other textbooks. Why is it reasonable to assume this stronger property? I would have thought it was reasonable to assume every element of $k$ can be written as the ratio of two algebraic integers, but why assume it can be written as $\frac{\alpha}{n}$ as you say? – user50229 Oct 08 '13 at 15:09
  • I'd have to see the chapter to understand the context. I'm not sure what the motivation would be for wanting/expecting this property. – Thomas Andrews Oct 08 '13 at 15:14
  • I have added a link to the book on Google Books if it's any help. – user50229 Oct 08 '13 at 15:19
  • It is reasonable basically because Dedekind found it works well. He was probably led to it because you commonly calculate the inverse to an algebraic number $\alpha$ as the product of its (other) conjugates divided by its norm. That is, you commonly present all denominators as real rational numbers. – Colin McLarty Oct 08 '13 at 16:24
  • Yeah, I was thinking along those lines - you can always rationalize the denominator with the numerator an algebraic integer. But that seems unsatisfying, on some level, particularly in the context of the book, where it is trying to argue without apriori experimentation, what the "right" definition of algebraic integer is. – Thomas Andrews Oct 08 '13 at 17:55
  • @Colin I like this point, maybe coming from the perspective of algebraic numbers in general, you can see that the denominator can be rationalised and therefore you can assume the form $\frac{\alpha}{n}$ without loss of generalisation. – user50229 Oct 08 '13 at 20:16
  • But by what intuition do you know you can always get a numerator that $\alpha$ is an algebraic integer? @user50229 – Thomas Andrews Oct 08 '13 at 20:19
  • @Thomas I suppose you have to use the other two points that (1) the conjugates of an integer are integers and (2) the integers are a ring? – user50229 Oct 08 '13 at 20:23
  • But that doesn't give an intuition for why this property. That, rather, just proves that if $k$ is the field of fractions for $\mathcal O_k$ then $\mathcal O_k\otimes \mathbb Q\cong k$. – Thomas Andrews Oct 08 '13 at 20:26
  • I think a more intuitive rule is that if $k_1\subset k_2$ then $\mathcal O_{k_2} \cap k_1 = \mathcal O_{k_1}$. That is, whether a number is an integer is independent of which field it is in. In particular, then, the product of the conjugates of an algebraic integer, being a rational number and an algebraic integer, must be in $\mathbb Z$. – Thomas Andrews Oct 08 '13 at 21:29
  • @Thomas That is his property 4 on page 1. – user50229 Oct 09 '13 at 10:17
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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 $k$.

Hagen von Eitzen
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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 $R \otimes_{R} M \cong M$.

Adrián Barquero
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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 $f$).

Rob Arthan
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  • Thanks for the link. Yes I suppose my question could also be rephrased why can we assume this property a priori, because as you say it can be easily shown once you have the monic polynomial definition, but I can't see why it would be obvious beforehand. – user50229 Oct 08 '13 at 15:32
  • The property that would be obviously desirable to me would be the weaker property that $k$ is the field of fractions. – user50229 Oct 08 '13 at 15:33
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    I agree that it would be pedagogically better to say that we want $k$ to the field of fractions and then observe that the definition using monic polynomials delivers that, because, via an easy argument, it delivers the stronger property. – Rob Arthan Oct 08 '13 at 15:39
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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.

Dietrich Burde
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