algebraic integer ring

The basic object in number theory is perhaps the algebraic number field and the corresponding algebraic integer ring.

First things first, what is an algebraic number over the $\mathbb{Q}$? Or more generally, over a certain field $R$? Algebraic means polynomials, so this algebraic number must satisfy a polynomial with coefficients in $R$. To define an algebraic integer, we have to know what is the algebraic ring of $R$. It is not so obvious, and perhaps we should define what is the algebraic integer ring for the most basic field, the rational number field. In fact, it is first we know that there is the integer numbers, then we can define what is the algebraic integer ring over a number field.

By a number field, we mean a field which contains the rational number field. By an algebraic number field, we mean an finite algebraic extension of the rational number field. Suppose that $K\subset L$ two algebraic number fields, we define the degree of the extension of $L$ over $K$ to be the dimension of $L$ viewed as a vector space over $K$, $[L:K]=dim_K(L)$. In this post, we always assume that the degree of extension to be finite.

Note that all algebraic number fields are contained in the complex number field $\mathbb{C}$. This is an important observation, since later we will talk maps from a field to another that we don’t know exactly, and in this case, we always assume that this field is mapped to the complex number field, although we can show that in fact this kind of maps can’t get to every element in $\mathbb{C}$.

Having defined algebraic number fields, then the natural things to do are of course the homomorphisms. Note that a homomorphism between two fields is very special, it is either injective, either zero. This looks like the morphisms between simple groups, simples modules, simple algebras, etc..  What should we first consider as homomorphisms? One such between two given fields? Perhaps it is a good idea. But it is better to first consider only homomorphisms starting from a given field to the complex number field.

This motivates the following definition: suppose that $K$ is an algebraic number field, then $\Sigma(K)$ to be all the homomorphisms of fields from $K$ to $\mathbb{C}$. Note that, if $\sigma\in\Sigma(K)$, then it send $1$ to $1$, and thus sends the rational numbers to itself. So this gives that $\sigma_{\mathbb{Q}}=id_{\mathbb{Q}}$. Noting that $K$ is a finite extension of $\mathbb{Q}$, so this motivates again the following consideration,

$\Sigma(L/K)=\{\sigma\in\Sigma(L), \sigma_K=id_K\}$ where $K\subset L$ are two algebraic number fields.

How can we determine a homomorphism of fields? First of all we can view this homomorphism as a homomorphism of $\mathbb{Q}$-vector spaces. So easily we get that a such homomorphism is determined by the action of this morphism on a basis of vector spaces for $K$. So this inspires us to find what a basis of a field looks like. First, we consider the field $\mathbb{Q}(\sqrt{2})$. An obvious basis for this $\mathbb{Q}$-vector space is $\{1,\sqrt{2}\}$. And if $\sigma$ is a homomorphism from $\mathbb{Q}(\sqrt{2})$ to the complex number field, then it is determined by $\sigma(\sqrt{2})$. This is a very general idea, that is, if $x$ is algebraic over a field $K$, then for any $\sigma\in\Sigma(K(x)/K)$, it is determined by $\sigma(x)$. So there is a one-to-one correspondence between $\Sigma(K(x)/K)$ and the set of possible values of $\sigma(x)$. So we must find this set of possible values. Note that if a polynomial $P(T)\in K[T]$ annulate $x$, then we must have that $P(\sigma(x))=\sigma(P(x))=\sigma(0)=0$. So, the possible values of $\sigma(x)$ are those satisfying the minimal polynomial for $x$. So, there are not many possible values, and we get that(if we set the minimal polynomial of $x$ over $K$ to be $P_{x,K}(T)$):

$\#\Sigma(L/K)=deg(P_{x,K})$.

A little digression before going on. An efficient way of viewing the field extension is perhaps the point of view of algebras. If $K\subset L$ are two fields, then $L$ is in fact an algebra over $K$(which is, moreover, divisible and commutative). And suppose further that $L\subset M$, $M$ is a field, then in the vector space point of view, we can choose a $K$-basis for $L$($L=K(e_1,...,e_l)$) and a $L$-basis for $M$($M=L(f_1,...,f_m)$). So how do view $M$ as a $K$-vector space? In fact this is trivial since $K$ is contained in $L$. Then what is the $K$-basis for $M$? The answer is $\{e_if_j\}$.

Return to our morphism stuff. In this way, we can show that

$\#\Sigma(L/K)=[L:K]$.

This result is not hard to prove. We can show it by using recurrence on $[L:K]$.

An inverse problem is that, given a element $\sigma \in\Sigma(K)$, can we extend it to an element of $\Sigma(L)$? This is not difficult, either. We can extend $\sigma\in\Sigma(K)$ to $K(x)$($x\in L-K$) in using the same method as above. So, we have that,

For any $\sigma\in\Sigma(K)$, we can extend it to $L$.

Now let’s have a closer look at $\Sigma(L/K)$. Its elements fix $K$. The question is that can these morphisms determine $K$? Suppose that $x\in L-K$, then according to the above argument, we can find a morphism $\sigma\in \Sigma(K(x)/K)$ such that $\sigma(x)\neq x$(the roots of the minimal polynomial of $x$ are all distinct, which is due to the fact that the characteristic of these number fields is $0$). Then we can extend this $\sigma$ to $L$, thus in this way, we have shown that

$K=\bigcap_{\sigma\in\Sigma(L/K)}Fix(\sigma)$ where $Fix(\sigma)=\{x\in L|\sigma(x)=x\}$.

Sometimes we also denote $Fix(\sigma)=L^{\sigma}$. Using the same idea, we can get the primitive element theorem:

If $K\subset L$ are two fields, then there exists $x\in L$ such that $L=K(x)$.

We can extend $id_K$ to $L$, and we consider the set $H=K-\bigcup_{\sigma\in\Sigma(L/K)}L^{\sigma}$. An element is in this set means that it is not fixed by any $\sigma\in \Sigma(L/K)$. If $K(x)\neq L$, then we can extend $id_{K(x)}$ to $L$ to get some morphism $\sigma$, but this time $\sigma(x)=x$, a contradiction. It remains to show that $H$ is not empty. Viewed as $\mathbb{Q}$-vector spaces, $H$ is just the complement of an union of a finite number of hyperplanes, which is not empty since $\mathbb{Q}$ is infinite. Thus we proved the primitive element theorem.

There is a particular type of morphism in from $L$ to itself. Suppose that $x\in L$, then the morphism $f_x:L\rightarrow L, y\mapsto xy$. This morphism is $L$-linear, that is $f_x(ay)=xay=axy=af_x(y)(a\in L)$, so for any subfield $K\subset L$, $f_x$ is also $K$-linear. This is one of the most important morphisms over a field. From this morphism we can define many important maps on $L$.

Suppose that $K\subset L$ two algebraic number fields, and $x\in L$. Then $L$ is a $K$-vector space, and thus $f_x$ is in fact a $K$-transformation on $L$, that is $f_x\in End_K(L)$. So we can define the trace, the determinant, the characteristic polynomial of $f_x$, as follows,

For $x\in L$, we define $tr_{L/K}(x)=trace(f_x)$, $N_{L/K}(x)=det(f_x)$, and $\chi_{x,L/K}(T)=det(T*id_L-f_x)$.

Note that the norm of $x$, $N_{L/K}(x)$ is just the determinant of $f_x$. What is more, $tr_{L/K}(x),N_{L/K}(x)\in K, \chi_{x,L/K}(T)\in K[T]$.

The idea of defining these quantities is to control $x$ in using the informations from $K$. It is obvious that $tr_{L/K}(x)$ is additive, while $N_{L/K}(x)$ is multiplicative. What is more, $\chi_{x,K(x)/K}(T)$ is exactly the minimal polynomial of $x$ over $K$, and thus, we have that $\chi_{x,L/K}(T)=\chi_{x,K(x)/K}(T)^{[L:K(x)]}$. One last point, for $K\subset K(x)$, there are exactly $[K(x):K]$ many elements in $\Sigma(K(x)/K)$, each $\sigma\in\Sigma(K(x)/K)$ corresponds to an conjugate of $x$. According to their definition, these conjugates of $x$ must satisfy the same minimal polynomial as $x$, so counting the cardinality, we have that $\chi_{x,K(x)/K}(T)=\prod_{\sigma\in\Sigma(K(x)/K)}(T-\sigma(x))$. Using the same idea, we have that $tr_{L/K}(x)=\sum_{\sigma\in \Sigma(L/K)}\sigma(x)$, and $N_{L/K}(x)=\prod_{\sigma\in\Sigma(L/K)}\sigma(x)$.

There is another quantity concerning the whole $L$ over $K$. Suppose $x_1,x_2,...,x_k\in L$ are $k$ elements, then we define the discriminant of these elements to be the discriminant of the matrix whose entries are $tr_{L/K}(x_ix_j)(1\leq i,j\leq k)$. That is, $disc_{L/K}(x_1,x_2,...,x_k)=det((tr_{L/K}(x_ix_j)))$. The motivation for this quantity is that, if these elements for a basis of $L$ over $K$, then any number $x\in L$ can be written as $x=\sum_i a_ix_i$. Now we want to know what these coefficients $a_i\in K$ are. What do we usually do in an Euclidean space? One natural way is to consider make scalar product of both sides with this basis. But, here we haven’t defined any scalar product. But recall what is a scalar product. A scalar product is, first of all, a bilinear form on the vector space to the ground field. Note we have seen something linear, yes, that is the trace function on $L$, at that time we say that it is additive, which is, in some sense equivalent to linearity. How do we define a bilinear form with the help of the trace function. Note that $L\times L\rightarrow L,(x,y)\mapsto xy$ is already bilinear, so we can take the trace of the image, thus we can define $\phi: L\times L\rightarrow K, (x,y)\mapsto tr_{L/K}(xy)$. So for the basis, we naturally have a matrix $(tr_{L/K})(x_ix_j)$. So to find these $a_i$, we have to consider the linear system $tr_{L/K}(xx_i)=\sum_j tr_{L/K}(a_jx_jx_i)=\sum_j a_jtr_{L/K}(x_jx_i)$. Is this system invertible? Clearly this depends on the invertibility of the basis matrix $(tr_{L/K}(x_ix_j))$. How to determine the invertibility of this matrix? A little remark perhaps helps. Note that $tr_{L/K}(x_ix_j)=\sum_k{\sigma}\sigma(x_i)\sigma(x_j)$. If we numerate these $\sigma$ as $\sigma_l(1\leq l\leq n)$($n=[L:K]$), then we see that this matrix is in fact the product of two matrices, or more precisely the product of a matrix with its transpose. This matrix is $M=(\sigma_l(x_i))$, a row of $M$ looks like $(\sigma_l(x_1),\sigma_l(x_2),...,\sigma_l(x_k))$, and $(tr_{L/K}(x_ix_j))=M^{\dagger}M$. So, if these $x_i$ are $K$-linearly independent, then we must have that $rank(M)=k$, which implies that $M^{\dagger}M$ is invertible, so $disc_{L/K}(x_1,...,x_k)\neq 0$. The converse is also true. Another character of discriminant is the relation with the discriminant of a polynomial. If $x\in L$, then for the minimal polynomial of $x$, $P_{x,K}(T)$, its discriminant is just $disc(P_{x,K})=\prod_{i where these $y_i$ are just the roots of this polynomial, and thus the conjugates of $x$. So if $(x_1,x_2,...,x_k)=(1,x,x^2,...,x^{k-1})(k=[K(x):K])$, then we must have that $M=(\sigma_i(x^j))=((\sigma_i(x))^j)$ which is Vandermonde matrix, thus it is clear that $disc_{L/K}(1,x,x^2,...,x^{k-1})=disc(P_{x,L/K})$.

Next we want to say something about the integer ring of $K$, an algebraic number field. We set $\overline{\mathbb{Z}}$ to be the set of algebraic integers (over $\mathbb{Z}$). And we define $\mathfrak{O}_K=K\bigcap\overline{\mathbb{Z}}$. This is the so-called algebraic integer ring of $K$. Then the most important problem is how to decide if an element in $K$ lies in $\mathfrak{O}_k$. One tends to believe that there is a simple description of $\mathfrak{O}_K$, for example $\mathfrak{O}_K=\mathbb{Z}(x)$ for some $x\in K$.  This indeeds works for the quadratic fields over $\mathbb{Q}$, just like the post ‘quadratic integer ring’, where all algebraic integer ring over a quadratic field looks like $\mathbb{Z}(x)$. But in general, this is not the case, as we shall see an example.

We have said that the quantities like trace, norm, determinant gives some information for an element in $K$ is using the information only in $\mathbb{Q}$. Now for an algebraic integer, can we say something about it using only the information from $\mathbb{Z}$? Note that the trace and the norm use only partial information of the characteristic polynomial, so perhaps a good candidate for this task is the characteristic polynomial or minimal polynomial(in $K=\mathbb{Q}(x)$, they coincide). Note that, if $x\in K$ is an algebraic integer, then all its conjugates are algebraic integers, too. So, its minimal polynomial is just $P_{x,\mathbb{Q}}(T)=\prod_{\sigma\in\Sigma(\mathbb{Q}(x)/\mathbb{Q})}(T-\sigma(x))\in \overline{\mathbb{Z}}[T]\bigcap\mathbb{Q}[T]$. Noting that an algebraic integer is a rational number, then it must be an integer. So, we have that $P_{x,\mathbb{Q}}\in\mathbb{Z}[T]$. Conversely, we know that the characteristic polynomial of $x$ in $\mathbb{Q}(x)$ over $\mathbb{Q}$ annulates $x$, so if this polynomial has integer coefficients, then $x$ must be an algebraic integer.  If $K\neq \mathbb{Q}(x)$, then note that the characteristic polynomial is just a power of the minimal polynomial, thus we get that

For an element $x\in K$, we have that $x\in\mathfrak{O}_k$ if and only if $P_{x,\mathbb{Q}}(T)=\chi_{x,\mathbb{Q}(x)/\mathbb{Q}}(T)\in\mathbb{Z}[T]$, if and only if $\chi_{x,K/\mathbb{Q}}(T)\in\mathbb{Z}[T]$.

Now let’s say something more about this algebraic integer ring $\mathfrak{O}_K$. According to the primitive element theorem, we can have that $K=\mathbb{Q}(x)$. Since $x$ satisfies some polynomial, $a_nx^n+a_{n-1}x^{n-1}+...+a_0=0(a_i\in\mathbb{Z}, a_n\neq0)$, then multiplying both sides by $a_n^{n-1}$, we see that $a_nx$ is in fact an algebraic integer of $K$, but still we have that $K=\mathbb{Q}(a_nx)$. So we can suppose directly that $K=\mathbb{Q}(x)(x\in\mathfrak{O}_K)$. So, we must have that $\mathbb{Z}(x)\subset \mathfrak{O}_K$. Can we control this algebraic integer ring from superior? The answer is yes. In deed, for any $z\in K$, we can write $z=\sum_ia_ix^i(a_i\in\mathbb{Q})$. We must show that the denominators of these $a_i$ for any algebraic integer in $K$ can’t be too large. This is not easy to see. But perhaps we should first get these $a_i$ out of this expression. How? Very simple, using the matrix $(tr_{K/\mathbb{Q}}(x^ix^j))$. Note that, the entries of $(tr_{K/\mathbb{Q}}(x^ix^j))$ have a largest denominator, say $m$. Then the vector $(tr_{K/\mathbb{Q}}(zx^i))$ has integer components since $zx^i$ are all algebraic integers. Thus we get that the denominators of these $a_i$ for any $z\in K$ can not exceed $m$, thus showing that $\mathfrak{O}_K\subset \frac{1}{m}\mathbb{Z}(x)$. Now that $\mathfrak{O}_K$ is sandwiched between two lattices of $\mathbb{R}^{deg(P_{x,\mathbb{Q}})}$, thus it must has a $\mathbb{Z}$-basis with $[K:\mathbb{Q}]$ elements. That is to say, $\mathfrak{O}_K=\bigoplus_{1\leq i\leq [K:\mathbb{Q}]}\mathbb{Z}x_i$.

But the problems is that in general we can not write this as $\mathfrak{O}_K=\bigoplus_i\mathbb{Z}x^i$ for some element $x$. Here is such an example. We take $K=\mathbb{Q}(\sqrt{m},\sqrt{n})$ where $n,m$ two distinct square-free integer such that $n=m=1(\text{mod} 8)$. We set $a=\frac{1+\sqrt{n}}{2},b=\frac{1+\sqrt{m}}{2}$. We first show that $\mathfrak{O}_K=\mathbb{Z}+\mathbb{Z}a+\mathbb{Z}b+\mathbb{Z}ab$. Note that, since $[K:\mathbb{Q}(\sqrt{m})][\mathbb{Q}(\sqrt{m}):\mathbb{Q}]=[K:\mathbb{Q}]$, which shows that $[K:\mathbb{Q}]=4$. And since $1,a,b,ab$ are $\mathbb{Q}$-independent, so they form a basis for $K$ over $\mathbb{Q}$. Note any $\sigma\in\Sigma(K/\mathbb{Q})$, it is determined by $\sigma(a),\sigma(b)$. But the only other conjugate of $\sqrt{n}$ is its inverse $-\sqrt{n}$, so we have that there are only four elements in $\Sigma(K/\mathbb{Q})$. And we can show that $disc_{K/\mathbb{Q}}(1,a,b,ab)=m^2n^2$. It is clear that $a,b,ab\in\mathfrak{O}_K$. So for an element $z=c+d\sqrt{n}+e\sqrt{m}+f\sqrt{mn}\in\mathfrak{Q}_K$, we have that $tr_{K/\mathbb{Q}}(z)\in\mathbb{Z}$, that is $4c\in\mathbb{Z}$. Similarly, we have that $4d,4e,4f\in\mathbb{Z}$. So we can rewrite $z=\frac{1}{4}(c+d\sqrt{n}+e\sqrt{m}+f\sqrt{nm})=(c-d-e+f)/4+(d-f)a/2+(e-f)b/2+fab$. In using the trace matrix corresponding to the basis $1,a,b,ab$, we have that these coefficients only have possible denominators like $n^2m^2$, but $n,m$ are odd integers, so we must have that these coefficients must also be integers, in this way we proved that $\mathfrak{O}_K=\mathbb{Z}[1,a,b,ab]$. Note that $a^2-a=\frac{n-1}{4}$, an even integer, the same for $b$, this means that the quotient ring $\mathfrak{O}_K/2\mathfrak{O}_K$ is isomorphic to the ring $R=\mathbb{Z}/2\mathbb{Z}[X,Y]/(X^-X,Y^2-Y)$. So if $\mathfrak{O}_K=\mathbb{Z}[x]$, then we must have that $R=\mathbb{Z}[x]/2\mathbb{Z}[x]$. But consider the ring homomorphism from $R$ to $F_2=\mathbb{Z}/2\mathbb{Z}$. Counting the cardinality, we can see that this is impossible. So $\mathfrak{O}_K=\mathbb{Z}[x](x\in K)$.