ideal classes of an integral ring

Suppose that $R$ is an integral. There are many operations that we can do on the ideals of $R$. For example, if $I,J\subset R$ two ideals, then, $I+J=$ the sum of $I,J$ being the ideal containing both $I$ and $J$. Also, $IJ=$, the product of $I,J$ being the ideal containing the product of any two elements from $I,J$. In this post, we mainly concern ourselves with the second operation, the product operation. We denote $C(R)$ to be the set of non-zero ideals of $R$. Why exclude the zero ideal? One reason is that it annulates every ideal $0I=I$, which is not so interesting. And we give this set the product structure defined above. There is a neutral element, $R$ since $RI=I$ for any ideal. Sometimes this set is too large. For example, even the simplest ring, $R=\mathbb{Z}$, we have that $C(R)=\{(n)|n\in\mathbb{Z}\}$. This set is almost the same as the ring itself. On one hand, this is good, since it reflects all the information of $R$, yet the problem is that it is too large. Studying $C(R)$ is almost the same as studying $R$ itself, so this approach will not give us much good. In fact, sometimes we concern only with some particular properties of $R$, for example, whether $R$ is a principal ring. We have seen that this kind of ring have very nice properties, especially the modules on these rings, they behave much the same way as vector spaces over a field. So perhaps we can define some equivalence relation on $C(R)$ to identify some elements to make smaller, at the same time, to reflect if $R$ is principal or not. We wish to express the idea that a ring $R$ is principal if and only if $C(R)$ contains only one equivalence class of ideals. That is, all the principal ideas are equivalent. In other words, $(i)\equiv(j)$. How to express this idea? We know that, for any $i,j\in R-0$, we have that $xi=yj$ for some $x,y\in R-0$(for example, we can take $x=j,y=i$). This is the equivalence relation we are looking for. More precisely, $I,J$ are two ideals, we say that $I\equiv J$ if there exist $x,y\in R-0$ such that $xI=yJ$. This is in deed an equivalence relation, the reflectivity, the transitivity, etc. can all be verified easily. So, we denote $Cl(R)=C(R)/\equiv$. Note that this equivalence relation is compatible with the product operation: if $I\equiv J,I'\equiv J'$, then we have that $II'\equiv JJ'$. So, now we have a much smaller set $Cl(R)$, at least it identifies all the principal ideals. What is more, we see that $Cl(R)$ contains only one element if and only if $R$ is principal. This is what we were looking for.

In general, $Cl(R)$ can still be very large. Yet for some important rings, we can show that $Cl(R)$ is finite. This is the case for $R$ a subring of some algebraic integer ring of finite type as a module over $\mathbb{Z}$, that is $R\subset \mathfrak{O}_K$ where $K$ is an algebraic number ring, that is the following result:

For an algebraic integer ring $R$, $Cl(R)$ is finite.

The proof of this important result relies on the following proposition:

For $R$ as above, there is a positive integer $c$, such that for any non-zero ideal $I\subset R$, there is an element $r\in I$ with $|I/rR|\leq c$.

Indeed, for any $I\in C(R)$, we choose $r\in I$ such that $N=|I/rR|\leq c$. Then we must have that $NI\subset rR$. Note that $rR\equiv R$. Moreover, $R\equiv NrR\subset NI$, So we have that $NrR\subset NI\subset rR$. This means that for any ideal $I$, there is a multiple of it (the multiplicator is bounded for all non-zero ideals) such that it is sandwiched between $cR$ and $R$. Note that, since $R$ is a $\mathbb{Z}$-module of finite type, this means that $R/cR$ is finite. So, this quotient must have finitely many ideals. Since any ideal $I$ has a multiple $NI/x$ that is sandwiched between $cR\subset R$, what is more, $(NI/x)/cR$ is also an ideal in $R/cR$, thus there are only finitely many equivalent ideals in $Cl(R)$.

So all things are reduced to proving this proposition. First of all, let’s have a close look at these ideals. We can first consider the case $R=\mathfrak{O}_K$ for some algebraic field $K$. For example, $R=\mathbb{Z}[i]$, the Gauss integer ring. Note that, $R$ is in fact a lattice in $\mathbb{C}=\mathbb{R}^2$. It is also the case for $R=\mathbb{Z}[\omega](\omega^3=1,\omega\neq1)$. Yet it is not the case for $R=\mathbb{Z}[\sqrt{2}]$. Can we still give it a lattice structure? Wait a minute, why do we consider a lattice structure on $R$? This is a good question. We take the case $R=\mathbb{Z}[i]$ to illustrate this usage. It is obvious that $R$ is generated by $i$. As a lattice, it is generated by $1=(1,0),i=(0,1)$. Then for any non-zero ideal $I\subset R$, and any non-zero element $r\in I$, $r,ri$ are still linearly independent. So, $I$ has sub-lattice structure with respect to $R$. Then intuitively, we have that $covol(I)/covol(R)=|R/I|=N(I)$. For any element $r\in R$, we define a norm on $R$, the norm $N(r)$ of $r$ is just the square of the modulo of $r$ as a complex number $N(r)=|r|^2$. To prove the proposition for $R$, for each $I$, we have to find a non-zero element $x\in I$ such that $|I/xR|$ is uniformly smaller than a constant $c$ independent of $I$. Note that $|I/xR|=covol(xR)/covol(I)$. This means that $covol(xR). Can we express $covol(xR)$ in terms of the norm of $x$ and the covol of $R$? Note that $covol(xR)/covol(R)=|R/xR|$, we guess that $|R/xR|=N(x)$. So, in this way, we have that $|x|^2=N(x). So here we should consider a ball $B_t=\{r\in \mathbb{R}^2=\mathbb{C}||r|\leq t\}$. Recall the Minkowski’s body theorem(cf. the post on the geometry of number), if for some $t$ we have that $vol(B_t)\geq 2^2covol(I)$, then there must be some element $r\in I-0$ such that $r\in B_t$, or equivalently, $|r|\leq t$. For the best case, we consider the $t$ such that $vol(B_t)=4covol(I)$. After some calculations, we find that $vol(B_{t'})=\pi {t'}^2=4covol(I)$. So ,for $t'$, there is a non-zero element $r$ in $I$ such that $|r|^2\leq t'^2=\frac{4}{\pi}covol(I)$. Note that we have that $N(r)=|r|^2$, so we get that $N(r)\leq \frac{4}{\pi}covol(I)$. So, finally we find an element $r\in I-0$ such that $N(r)\leq \frac{4}{\pi}covol(I)$. Compare with the above conjecture, we can take that $c /covol(R)=\frac{4}{\pi}$. Note that our original goal is to show that $|I/rR|$ is bounded. Now $|I/rR|=covol(rR)/covol(I)=N(r)covol(R)/covol(I)\leq \frac{4}{\pi}covol(R)$. At last, we showed that there exists such $r\in I-0$ with $|I/rR|\leq \frac{4}{\pi}covol(R)$ independent of $I$. Note that in this process there seems to be some coincidence: especially $N(x)$ is the square of $x$, while the volume of the ball $B_t$ is the square of $t$. All these are due to the fact that the lattice is in $\mathbb{R}^2$, a dimension $2$ vector space. This again result from the fact that $K=\mathbb{Q}[i]$ is an extension of$\mathbb{Q}$ of degree $2$. Or more pertinently, this results from the fact that $R$ is a $\mathbb{Z}$-module of rank $2$.

How about the general case? We can generalize the above argument without any difficulty once we can identity $R$ with a lattice in an Euclidean space of a dimension the same as the rank of $R$ as a $\mathbb{Z}$-module. How to do this? An interesting problem. Note that from the above remarks we see that sometimes the direct identification doesn’t work($R=\mathbb{Z}[\sqrt{2}]$ is such a case). So, we have to consider other methods. Here is one. Suppose that $K$ the quotient field of $R$. So $K$ is an algebraic field. Then we know that the degree of the extension of $K$ over $\mathbb{Q}$ is the same as the rank of $R$ as a $\mathbb{Z}$-module, that is to say $[K:\mathbb{Q}]=rank_{\mathbb{Z}}(R)$. Then we consider the set of field morphisms from $K$ to $\mathbb{C}$, that is $\Sigma(K)$. Note that if $\sigma\in \Sigma(K)$, then $\overline{\sigma}(z)=\overline{\sigma(z)}(z\in K)$ also lies in $\Sigma(K)$. We separate two cases, the first is that $\overline{\sigma}=\sigma$, the other is that the conjugate of $\sigma$ is not equal to itself. So, we define $\Sigma_1(K)=\{\sigma\in \Sigma(K)|\sigma=\overline{\sigma}\}$. For the rest elements in $\Sigma(K)$, since they are in pairs, so we can choose one element from each pair($\Sigma(K)$ is a finite set, so this choice is practical), and they form the set $\Sigma_2(K)$. We set $r_1=\#\Sigma_1(K), r_2=\#\Sigma_2(K), n=[K:\mathbb{Q}]$. So we have that $r_1+2r_2=n$. Note that for $\sigma\in \Sigma_1(K)$, we have that $\sigma(z)\in\mathbb{R}$. For convenience, we denote the elements in these two sets as $\Sigma_1(K)=\{\sigma_1,...,\sigma_{r_1}\}, \Sigma_2(K)=\{\sigma_{r_1+1},...,\sigma_{r_1+r_2}\}$. And we define a map $f:R\rightarrow V=\mathbb{R}^{r_1}\times \mathbb{C}^{r_2}, r\mapsto(\sigma_1(r),...,\sigma_{r_1+r_2}(r))$. Note that $dim(V)=r_1+2r_2=n$. So it remains to show that $f(R)$ is a lattice in $V$. This is not hard to see. We have to find a basis $(e_1,...,e_n)$ for $R$, and then show that the ‘lattice’ generated by $f(e_1),...,f(e_n)$ has positive co-volume. Note that for each $r\in R$, $f(r)$ is a vector whose components are the conjugates of $r$. If we identify $\mathbb{C}=\mathbb{R}\bigoplus \mathbb{R}i$, and we set $P=(f(e_1)^{\dagger},...,f(e_n)^{\dagger})$ the matrix of the image of the basis for $R$. Then we have that $|det(P)|=2^{-r_2}|disc_{K/\mathbb{Q}}(e_1,...,e_n)|^{1/2}$. So, we have that $f(R)$ has positive co-volume. Then we have to show that $f(R)$ is discrete in $V$. We give $V$ the norm $||v||=\sup_{1\leq i\leq r_1+r_2}|v_i|$(where $v_1,...,v_{r_1}\in\mathbb{R}$ while $v_{r_1+1},...,v_{r_1+r_2}\in\mathbb{C}$). So, we have to show that for any $t>0$, the ball $D_t=\{v\in V| ||v||\leq t\}$ contains only finitely many elements of $f(R)$. But for any $r\in R$, note that $||f(r)||\leq t$ is the same as $|\sigma_i(r)|\leq t(\forall i)$. Are there finitely many algebraic integers whose conjugates are bounded by a number $t$? How to attack this problem? Note that we have a very strong restriction: $r$ is an algebraic integer. So, we can consider its minimal polynomial. We see immediately that the minimal polynomials of all such algebraic integers have coefficients bounded from above by some constant depending only on $t$. Clearly, we have only finitely many such polynomials, and thus $f(R)$ is a lattice.Besides, $covol(f(R))=|det(P)|=2^{-r_2}|disc(R)|^{1/2}$ This done, the rest proceeds much the same way as above. Hence in the following we just copy the results:

For any non-zero ideal $I\subset R$, $I$ has the same rank as $R$ viewed as $\mathbb{Z}$-modules, and thus the quotient field of $I$ is also $K$, the same as that of $R$. So, we can see that $|R/I|=\sqrt{\frac{|disc(I)|}{|disc(R)|}}$. Note that, the smaller an ideal, in some sense, the larger its discriminant. We can think it this way: the smaller an ideal, the smaller the lattice in $V$, and thus the larger its co-volume. And we have seen that this implies that the discriminant of the ideal is larger. For any ideal $I$ in $R$, we define $N(I)=|R/I|$ the norm of $I$. This definition is compatible with the norm $N=N_{K/\mathbb{Q}}$ in that $N(rI)=|N(r)|N(I)(r\in R)$. With this, we have the following result due to Minkowski

Suppose that $I\subset R$ is a non-zero ideal, then there exists a non-zero element $r\in I$ such that $|N(r)|\leq C(r_2,n)N(I)|disc(R)|^{1/2}$ where $C(r,n)=(\frac{4}{\pi})^r\frac{n!}{n^n}$.

Note that with this and $|N(r)|=N(rR)$, we have that $|N(r)|/N(I)=N(rR)/N(I)=\frac{|R/rR|}{|R/I|}=|I/rR|$. So, we have that $|I/rR|\leq C(r_2,n)|disc(R)|^{1/2}$, proving the theorem in the general case, and showing that $Cl(R)$ is a finite set for any algebraic integer ring.