1. Integrality

Let $\mathbb{Q}$ be the field of rational numbers. A number field is a finite extension of $\mathbb{Q}$. Algebraic number theory is a one of the fields in mathematics which studies number field, as problems in $\mathbb{Q}$ becomes a easier problem when lifted to a number field. The next post will include an example of such. The reason why we consider $\mathbb{Q}$ in number theory is because it is the quotient field of $\mathbb{Z}$, so we generalize the concept of integers by the following method.

The materials are from Prof. Elman's Lecture Notes, Lang's Algebraic Number Theory, Neukirch's Algebraic Number Theory, and Marcus's Number Fields.

Let $B$ be a commutative ring an extension of a ring $A$ (i.e. $A \subset B$), then an element $x \in B$ is called integral over $A$ if there exists a monic polynomial $f \in A[t]$ such that $f(x)= 0$.

Proposition: With the same assumption as above, the following are equivalent
        1) $x \in B$ is integral over $A$
        2) $A[x]$ is a finitely generated $A$-module.
Proof)
    ($\Rightarrow$): Any element in $A[x]$ is of the form $g(x)$ for some $g \in A[t]$, then as $f$ is monic, we can have $g = fq+r$ for some $q,r \in A[t]$. Then we have that $g(x) = r(x) \in \sum_{i=1}^{n-1} Ax^i$, hence $A[x]$ is finitely generated.
   ($\Leftarrow$): An more general statement of 1) would be the following:
M be an $A[x]$-module that is finitely generated as $A$-module, and suppose $ann_{A[x]} = 0$, then $x$ is integral over $A$.

Then as $A[x]$ is $A[x]$-module that is finitely generated, and as $1 \in A[x]$ as a subring of $B$, we may conclude that $x$ is integral over $A$.

To prove the general setting, We have, by assumption, that

$M = Ax_1 + \cdots + Ax_n$

Then as $M$ is an $A[x]$-module, multiplication by $x$ gives a $A$-endomorphism of $M$. Then we have the equation

$xm_i = \sum_{j=1}^n a_{ij}m_j$

Moving it to the otherside, we get

$\sum_{j=1}^n (\delta_{ij}x - a_{ij})m_i = 0$

Then if $\Delta$ is the determinant of the matrix defined by $(\delta_{ij} - a_{ij})$, then by Cramer's rule, $\Delta m_i = 0$ for all $i$, hence $\Delta M = 0$. Then as $\Delta \in ann_{A[x]}M$, $\Delta = 0$. The monic polynomial $det((\delta_{ij}t - a_{ij}))$ has a root $x$.

Proposition: Let $B/A$ be a commutative ring extension. Suppose $x_1, \ldots, x_n \in B$ is integral over $A$, then $A[x_1, \ldots, x_n]$ is finitely generated by $A$-module if and only if $x_1, \ldots, x_n$ are all integral over $A$. 

Proof)

    ($\Leftarrow$): $A[x_1]$ satisfies by the proposition above, so we assume that $A[x_1, \ldots, x_{n-1}]$ is finitely generated. And let $x_n$ have a monic polynomial $f \in A[t]$, then the set of generators of $A[x_1, \ldots, x_n]$ and the coefficients of $f$, which is finit, generate $A[x_1, \ldots, x_n]$.

    ($\Rightarrow$): Using the general setting above, we know that $ann_{A[x_i]} A[x_1, \ldots, x_n] = 0$ for $i=1, \ldots, n$.

If $x \in A[x_1, \ldots, x_n]$, then we have $A[x_1, \ldots, x_n, x] = A[x_1, \ldots, x_n]$, so, in particular,  $x_1, x_2$ are integral over $A$, then $x_1 + x_2, x_1x_2$ is integral over $A$.

We say that $B$ is an integral extension of $A$ if all elements in $B$ are integral over $A$. Let $C$ be an intermediate ring between $A$ and $B$, then $C$ is called a integral closure of $A$ in $B$ if 

$C = \{ b \in B ~|~ b$ integral over $A \}$

If $A=C$, then we say that $A$ is integrally closed in $B$, and a domain $A$ is integrally closed if $A$ is integrally closed in $qf~A$.

Proposition: Consider $A \subset B \subset C$ commutative rings, then $C/B$ is integral and $B/A$ is integral if and only if $C/A$ is integral.

Proof) Let $c \in C$, then there exists $f \in B[t]$ such that $f(c) = 0$, then we have $A[b_0, \ldots, b_n]$ finitely generated over $B$, and call it $B'$, then we have $A[b_0, \ldots, b_n, c]$ is finitely generated $B'$-module, and $B'$ is finitely generated $A$-module, so $A[b_, \ldots, b_n, c]$ is finitely generated. Hence the result follows. The converse is trivial.

We finish this by showing the following:

Proposition: Let $A$ be a UFD, then $A$ is integrally closed.

Proof) Suppose $a/b$ be integral over $A$, then we have

$(a/b)^n + c_{n-1}(a/b)^{n-1} + \ldots + c_0 = 0$

Then after multiplying $b^n$, we get

$a^n + (c_{n-1}b^{n-1})a^{n-1} + \cdots + c_0b^n = 0$

We may assume that $a,b$ are relatively prime, by dividing it by the greatest common divisor. Then if $p$ is a prime such that it divides $b$, then by the equation above, $p|a^n$ which is a contradiction. Hence $b=1$.