HW10

Text-Ex 5.8.

• i) Let $A$ be a subring of an integral domain $B$, and let $C$ be the integral closure of $A$ in $B$. Let $f,g$ be monic polynomials in $B[x]$ such that $fg\in C[x]$. Then $f,g$ are in $C[x]$. (Take a field containing $B$ in which the polynomials $f,g$ split into linear factors: say $f=\Pi \left(x-{\xi }_i\right),g=\Pi \left(x-{\eta }_i\right)$. Each ${\xi }_i$ and each ${\eta }_j$ is a root of $fg$, hence is integral over $C$. Hence the coefficients of $f$ and $g$ are integral over $C$.)

\begin{proof} i) Assume that ${\xi }_1,\cdots ,{\xi }_n$ are roots of $f$, and assume ${\eta }_1,\cdots ,{\eta }_m$ are roots of $g$. Since $B$ is an integral domain, roots of $fg$ are $\{{\xi }_i:1⩽i⩽n\}\cup \{{\eta }_j:1⩽j⩽m\}$. Then by $fg\in C[x]$, we know each ${\xi }_i$ and each ${\eta }_j$ is integral over $C$. By Vita’s theorem, the coefficients of $f$ and $g$ are integral over $C$. Since $f,g\in B[x]$, the coefficients of $f,g$ are in $B$ and they are integral over $C$. As $C$ is the integral closure of $A$ in $B$, we know $f,g\in C[x]$. \end{proof}

Text-Ex. 5.9. Let $A$ be a subring of a ring $B$ and let $C$ be the integral closure of $A$ in $B$. Prove that $C[x]$ is the integral closure of $A[x]$ in $B[x]$. (If $f\in B[x]$ is integral over $A[x]$, then

$$f^m+g_1{f}^{m-1}+\cdots +g_m=0\left(g_i\in A[x]\right)$$

Let $r$ be an integer larger than $m$ and the degrees of $g_1,\dots ,g_m$, and let $f_1=$ $f-x^r$, so that

$${\left(f_1+x^r\right)}^m+g_1{\left(f_1+x^r\right)}^{m-1}+\cdots +g_m=0$$

or say

$$f_1^m+h_1{f}_1^{m-1}+\cdots +h_m=0$$

where $h_m={\left(x^r\right)}^m+g_1{\left(x^r\right)}^{m-1}+\cdots +g_m\in A[x]$. Now apply Exercise 8 to the polynomials $-f_1$ and $f_1^{m-1}+h_1{f}_1^{m-2}+\cdots +h_{m-1}$.)

\begin{proof} Assume that $f\in B[x]$ is integral over $A[x]$, and we aim to show $f(x)\in C[x]$. Since $f$ is integral over $A[x]$, there exists $g_1,\cdots ,g_m\in A[x]$ such that

$$f^m+g_1{f}^{m-1}+\cdots +g_m=0\left(g_i\in A[x]\right).$$

Let $r$ be an integer larger than $m$ and the degrees of $g_1,\dots ,g_m$ and let $f_1=f-x^r$. Then

$${\left(f_1+x^r\right)}^m+g_1{\left(f_1+x^r\right)}^{m-1}+\cdots +g_m=0$$

yields

$$f_1(f_1^{m-1}+h_1{f}_1^{m-2}+\cdots +h_{m-1})=-h_m=-({\left(x^r\right)}^m+g_1{\left(x^r\right)}^{m-1}+\cdots +g_m)\in A[x],$$

where $f_1$ and $f_1^{m-1}+h_1{f}_1^{m-2}+\cdots +h_{m-1}$ are monic. By Exercise 8, one can check $f_1\in C[x]$ and so for $f$. Now we finish the proof. \end{proof}

Proposition 5.18.

Let $B$ be an integral domain, $K$ its field of fractions. $B$ is a valuation ring of $K$ if, for each $x\ne 0$, either $x\in B$ or $x^{-1}\in B$ (or both).

• i) $B$ is a local ring.
• ii) If $B^{\prime }$ is a ring such that $B\subseteq B^{\prime }\subseteq K$, then $B^{\prime }$ is a valuation ring of $K$.
• iii) $B$ is integrally closed (in $K$).

\begin{proof} i) Define $m=\{x\in K:x\in B,x^{-1}\notin B\}\cup \{0\}$. For $x\in B\setminus m$, we have $x,x^{-1}\in B$ and so $x$ is a unit of $B$.

For any $y\in B$ and $x\in m$, we aim to show $xy\in m$. It is easy to check $xy\in B$. If $x^{-1}{y}^{-1}\in B$, then $(x^{-1}{y}^{-1})y=x^{-1}\in B$, which is impossible. In addition, for any nonzero $x,y\in m$, there is $x+y\in B$. If $x+y\ne 0$, it remains to show $(x+y{)}^{-1}\notin B$. Otherwise, $x+y$ is a unit of $B$. WLOG we assume $x/y\in B$, then $(x/y+1)y=x+y$ is a unit and so $y$ is a unit, contradicting with $y\in m$. Now we proved that $m$ is an ideal.

Therefore, $m$ is the unique maximal ideal and so $B$ is a local ring.

ii) It suffices to show, for each $t\ne 0$ in $K^{\prime }=\mathrm{Frac}{B}^{\prime }$, either $t\in B^{\prime }$ or $t^{-1}\in B^{\prime }$. Note that $B^{\prime }\subseteq K$ yields $\mathrm{Frac}{B}^{\prime }\subseteq \mathrm{Frac}K=K$, so $K=K^{\prime }$.

For any nonzero $t\in K=K^{\prime }$, either $t\in B\subseteq B^{\prime }$ or $t^{-1}\in B\subseteq B^{\prime }$. Therefore, $B^{\prime }$ is a valuation ring of $K$.

iii) It suffices to show for any $x\in K$ such that $x$ is a root of a monic polynomial $f(x)\in B[x]$, there is $x\in B$. Since $B$ is a valuation ring of $K$, either $x\in B$ or $x^{-1}\in B$. If $x\in B$, we have done. Now we assume that $x^{-1}\in B$ and

$$f(x)=x^n+a_1{x}^{n-1}+\cdots +a_n=0,$$

where $a_1,\cdots ,a_n\in B$. As $x^{-n+1}\in B$, there is

$$x+a_1+a_2{x}^{-1}+\cdots +a_n{x}^{-n+1}=0$$

and so $x=-a_1-a_2{x}^{-1}-\cdots -a_n{x}^{-n+1}\in B$. Now we finish the proof. \end{proof}