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Definition If $A$ is a commutative ring, then $A_{P}$ is its localisation at the prime ideal $P$. In particular, if $A$ is an integral domain, then $A_{0}$ is the fraction field of $A$.
Let $A\subseteq B$ be commutative rings. $b\in B$ is integral over $A$ if $p(b)=0$ for some monic polynomial $p\in A[X]$, or, equivalently, if the subring $A[b ]\subseteq B$ is a finitely generated $A$-module. The set of elements in $B$ which are integral over $A$ is called the integral closure of $A$ in $B$. If $B$ happens to be the quotient field of $A$, then the integral closure of $A$ in $B$ is also called the normalisation of $A$. If every element of $B$ is integral over $A$ then $B$ is called an integral extension of $A$. If no element of $B-A$ is integral over $A$ then $A$ is said to be integrally closed in $B$. In this case, there are no integral extensions intermediate between $A$ and $B$ except $A$ itself. $A$ is said to be integrally closed or normal when $A$ is integrally closed in its quotient field.
Example $\sqrt{2}\in\mathbb R$ is integral over $\mathbb Z$. Further, $\mathbb Z [\sqrt{2}]$ is an integral extension of $\mathbb Z$. The set of all complex numbers which are integral over $\mathbb Z$ is called the algebraic integers. We now show that $\mathbb Z$ is integrally closed, that is, a rational number is integral over $\mathbb Z$ iff it is an integer.
Proposition Let $A\subseteq B$ be commutative rings. $b\in B$ is integral over $A$ iff $b\in C$ for some subring $C$, $A\subseteq C\subseteq B$, that is a finitely generated $A$-module.
Proposition If $A\subset B$ are commutative rings, then the integral closure of $A$ in $B$ is itself a ring.
Proof Let $b_1,b_2\in B$ be integral over $A$. Then $A[b_1]$ is a finitely generated $A$-module. Since $b_2$ is integral over $A$, it is clearly integral over $A[b_1]$. Hence $A[b_1][b_2]=A[b_1,b_2]$ is a finitely generated $A[b_1]$-module. But this implies that $A[b_1,b_2]$ is a finitely generated $A$-module. Hence $b_1\pm b_2, b_1b_2 \in A[b_1,b_2]$ and so these elements are integral over $A$ as well.
Proposition Let $A$ be a UFD (i.e. factorial). Then A is integrally closed.
Proof Suppose $k=\frac{a}{b}\in A_{0}$ is integral over $A$ and is in lowest terms. Then $\exists\alpha_i\in A$ with $\alpha_n=1$ and $\sum_{i=0}^{n} \alpha_i k^i=0$. But then\[a^n=-b\left(\alpha_{n-1}a^{n-1}+\cdots+\alpha_0b^{n-1}\right)\]and by UF $b\mid a$. This implies $b$ is invertible in $A$, i.e. $k\in A$. So $A$ is integrally closed.
Definition Let $A$ be a UFD, and $f\in A[X]$. Then the content of $f$, written $c(f)$, is the gcd of the coefficients of $f$.
Remark Notice that $A$ is a UFD iff A is a domain and $A/A^\times$ is, multiplicatively, a free monoid. This is because any element of $A$ can be written uniquely as $a=u\prod_i p_i^{e_i}$.
Lemma (Gauss' Lemma) Let $A$ be a UFD, and suppose $f,g\in A[X]$. Then $c(fg)=c(f)c(g)$.
Corollary $A$ is a UFD iff $A[X]$ is.
Proof Let $\phi: A[X] \rightarrow A \times K[X]/K^{\times}$ be defined by $\phi(f) = (c(f), f)$ where $c(f)$ is the content of $f$. By the lemma, $\phi$ is a monoid homomorphism. It is clearly surjective. To show injectivity, say $(c(f),f)=(c(g),g)$ in $A \times K[X]/K^\times$. Then
Proof If $f\in A[X]$ then define the content $c(f)$ of $f$ to be the gcd of
the coefficients of $f$. Then
Lemma (Generalisation of Gauss' Lemma) Let $A$ be a commutative ring, and $B$ be an extension of $A$, and monic polynomials $p(x),q(x)\in B[X]$ are
such that $p(x)q(x)$ has coefficients integral over $A$. Then $p(x)$ and $q(x)$ have coefficients integral over $A$. That is, a factorisation in an extension produces a factorisation in the integral closure.
Corollary Let $A$ be an integrally closed integral domain, and monic polynomials $p,q\in K[X]$ such that $pq\in A[X]$. Then $p,q\in A[X]$.
Lemma If $A$ is integrally closed and $p\in K[X]$ ($K=A_0$) then $p$ is integral over $A[X]$ iff its coefficients are in $A$.
Proof Since $p$ is integral over $A[X]$ $\exists\ \alpha_i\in A[X]$ with\[p^n+\alpha_{n-1}p^{n-1}+\cdots+\alpha_0=0\]Now we know $p=\sum_{i=0}^k p_i x^i$ with $p_i\in K$ and $\alpha_i=\sum_j {\alpha_i}_j x^j$ with ${\alpha_i}_j\in A$. We now show inductively that each coefficient $p_i$ is integral over $A$ (and therefore in $A$). The $x^0$ component of the previous equation gives us\[p_0^n+{\alpha_{n-1}}_0p_0^{n-1}+\cdots+{\alpha_0}_0=0\]which gives us immediately that $p_0$ is integral over $A$, and so $p_0\in A$. Now suppose that $p_0,\ldots,p_i$ are all in $A$. Then the $x^{i+1}$ component of the equation would never involve coefficients higher than $p_{i+1}$
Theorem Let $A$ be an integral domain. Then $A$ is integrally closed iff $A[X]$ is.
Proof Clearly if $A[X]$ is integrally closed then $A$ is as well. We prove the converse. Suppose $p\in A[X]_0$ is integral over $A[X]$. Then trivially $p$ is
integral over $K[X]$, which is a UFD and therefore integrally closed. Hence $p\in K[X]$. Since $p$ is integral over $A[X]$, the coefficients of $p$ must be integral over $A$ and therefore, by the integrality of $A$, $p\in A[X]$. Hence $A[X]$ is integrally closed.
Corollary Let $A$ be an integral domain. Then $A$ is integrally closed iff for some $n$ $A[X_1,\ldots,X_n]$ is. |
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