# algebraic number theory-part 1

The most important arithmetic property of the integer ring $\mathbb{Z}$ is perhaps that it is a principal ideal domain, thus is unique factorization domain, such that every non-zero non-unit element can have a decomposition into the products of irreducible elements, and this decomposition is unique up to a permutation. Yet this property is rather rare. In other words, consider the algebraic integer ring $\mathfrak{O}_K$of any algebraic number field $K$. In general it is not unique factorial. For example, there are only finitely many imaginary quadratic integer ring that are unique factorial(a conjecture of Gauss, now proved. We have mentioned this result in a post before). The easiest counterexample is $\mathbb{Z}[\sqrt{-5}]=A_{-20}=\mathfrak{O}_{\mathbb{Q}[\sqrt{-5}]}$. We have clearly, $2\times 3=(1+\sqrt{-5})\times (1-\sqrt{-5})$. Since $N(2)=4,N(3)=9,N(1+\sqrt{-5})=N(1-\sqrt{-5})=6$, we can show easily that these four numbers are all irreducible, yet examining their norms, we see that this decomposition is not unique.

How to rescue this situation? One genius idea is to consider the ideals, instead of elements, consider the decomposition of ideals into maximal ideals(note that, irreducible ideals correspond to p-maximal ideals, here a p-maximal ideal is an ideal that is maximal among the set of principal ideals), instead of decomposition of an element into irreducible elements.

Before going on, let’s reexamine the definition of unique factorization domain. One common definition is that, just as above, each non-zero non-unitary element can be decomposed into a product of irreducible elements, and this decomposition is unique up to permutation of these irreducible elements. Sometimes, this uniqueness is hard to verify. In fact, in some sense, irreducible elements are suitable to define the decomposition of an element, since, recall the definition of irreducible elements, an element $r$ is irreducible in a ring $R$ if for any expression $r=st(s,t\in R)$ we have that $s$ is a unit or $t$ is a unit. So, in this definition, there is not at all any trace of decomposition. But look at prime elements. It says that, an element $r\in R$ is prime if for any $r|st(s,t\in R)$, we have that $r|s$ or $r|t$. This definition deals with decomposition. So, there is an equivalent definition for the unique factorization domain: a ring is unique factorization domain if and only if all its non-zero non-unitary elements can be decomposed into a product of prime elements. This time, we don’t require the uniqueness of the decomposition. But this uniqueness is already guaranteed by the property of the prime elements(from these two equivalent definitions, we can see that, a ring is unique factorization domain if and only if all its irreducible elements are also prime elements). So from now on, we will always talk about the decomposition into prime elements or prime ideals, according to the situation.

So sometimes, the irreducible elements and prime elements are hard to distinguish, so are the maximal ideals and prime ideals. Yet, the good news is that for the algebraic integer ring, all non-zero prime ideals are maximal ideals, just as in the case of $\mathbb{Z}$. This involves another concept, the Dedekind domain. A domain $R$ is a Dedekind domain if it is Noetherian, all its non-zero prime ideals are maximal ideals, and it is integrally algebraically closed in its fractional field. The simples example is the integer ring, $\mathbb{Z}$. Other examples are the principal ideal domain. There is a result concerning the relation between unique factorization domains(UFD) and Dedekind domains: a Dedekind domain is UFD if and only if it is a principal ideal domain(PID). Or, equivalently according to the above comment, a UFD is Dedekind if and only if it is PID. This means that any UFD that is not PID is not Dedekind. An easy example is $R=k[x,y]$ polynomial ring over a field $k$. For example, the ideal generated by $x$, $Rx$ is a prime ideal, yet it is not maximal($Rx\subset R(x,y)$). One important result from commutative algebra is that, if $R$ is Dedekind domain, and $K$ is its fraction field, $L$ an finite extension of $K$. Then the algebraic closure $R'$ of $R$ in $L$ is again Dedekind. This last proposition gives us directly that all the algebraic integer rings are Dedekind. Dedekind domains are very suitable to talk about prime decompositions of ideals. One reason is that they are Noetherian, and non-zero prime ideals are maximal. Before going on, we have to introduce another concept: fractional ideals.

For any domain, $R$ and its fraction field $K$, a subset $I\subset K$ is called a fractional ideal if there is a non-zero element $r\in R-0$ such that $rI\subset R$ is a non-zero ideal of $R$. If $R$ is Noetherian, we have an equivalent definition, $I$ is a fractional ideal if $I$ is an $R$-module of finite type. From now on, we will restrict ourselves to the case that $R=\mathfrak{O}_K$ of some algebraic number field $K$. We set the set of fractional ideals to be $C(R)$. There is an operation on this set, multiplication, just as we did in this post. Now we want to define the inverse of a fractional ideal. Suppose that $I$ is a fractional ideal, we define $I'=\{k\in K|kI\subset R\}$. Note that this is indeed a fractional ideal(first we can assume that $I$ is an ideal of $R$ and thus generated as $I=R(e_1,...,e_n)$ since $R$ is Noetherian. So $I'=\bigcap_{1\leq i\leq n}I_i$ where $I_i=\{k\in K|ke_i\in R\}$. So, we have to show that each $I_i$ is a fractional ideal. Note that $k\in K$ lies in $I_i$ if and only if for the expression $k=\frac{s}{t}(s,t\in R)$, we have that $\frac{se_i}{t}\in R$. Thus we must have $rt\in Re_i$ for some $r\in R$. If we define $J_i=\{t\in R|\exists r\in R,rt\in Re_i\}$, then it is easy to verify that $J_i$ is an ideal of $R$, so we see that $I_i\equiv J_i\times R$, thus is a $R$-module of finite type. And this shows that $I'$ is a fractional ideal of $R$). So we have given $C(R)$ a group structure, an Abelian group structure, more precisely. As we said in the post, sometimes this group is often too large. So we can modulo the fractional principal ideal subgroup. That is, we have a natural map, $f:K^{\times}\rightarrow C(R),k\mapsto Rk=(k)$. Then the co-kernel of this map, $cok(f)=C(R)/im(f)$, is just the set $Cl(R)$ we defined in that post(of course, there we only consider the ideals, not the fractional ideals, so not all elements have inverse, and sometimes it doesn’t make a group. The only reason is that it is too small). We denote this quotient again as $Cl(R)$, this time it is a group. using the concept of fractional ideals, we can state the theorem of prime decomposition of fractional ideals:

If $I$ is a fractional ideal of $R$, then there is a unique expression $I=\prod_{P}P^{e_P}$ where the product is over all the non-zero prime ideals of $R$ and $e_P\in\mathbb{Z}$ for all $P$ and only finitely many of them is not zero.

This is really a fascinating theorem, since it points out that the law of prime decomposition works for Dedekind domains, in particular for the algebraic integer rings. So, in some sense, we don’t lose much. We have  noticed that  sometimes $C(R)$ is often too large, and we have to consider $Cl(R)$. It is easy to see that $\#Cl(R)=1$ if and only if $R$ is a PID(since $R$ is Dedekind, so this is also equivalent to that $R$ is UFD). We call $Cl(R)$ the ideal class group of $R$. So this group measures the difference between $R$ and a principal ideal domain.

To measure the difference between $R$ and $\mathbb{Z}$, we have to consider another thing, the units of the ring. We call it the unit group, and denote it by $R^{\times}$. Clearly, this group contains at least two elements.Another way to view this unit group is to reconsider the morphism $f:K^{\times}\rightarrow Cl(R)$. We see easily that $ker(f)=R^{\times}$. Concerning the unit group, there is a result which describes explicitly the structure of $R^{\times}$. Recall that we denote $\Sigma(K)$ to be the set of field morphisms from $K$ to $\mathbb{C}$, and $\Sigma_1(K)=\{\sigma\in\Sigma(K|\sigma=\overline{\sigma})\}$, and $\Sigma_2(K)$ to be the set of morphisms that are not equal to their conjugates(just pick up one from each pair), moreover we set $r_1=\#\Sigma_1(K),r_2=\#\Sigma_2(K)$. Then theorem says

We define $r=r_1+r_2-1$, and $G=\{k\in K||k|=1\}$, then we have that $R^{\times}=\mathbb{Z}^{r}\bigoplus G$.

As for the ideal class group, we have also

$Cl(R)$ is a finite Abelian group. We call $\#Cl(R)$ the class number of $R$.

Perhaps some example will help illustrate these two results.