HW9

Text-Ex. 5.4. Let $A$ be a subring of a ring $B$ such that $B$ is integral over $A$. Let $\mathfrak{n}$ be a maximal ideal of $B$ and let $\mathfrak{m}=\mathfrak{n}\cap A$ be the corresponding maximal ideal of $A$. Is $B_n$ necessarily integral over $A_{\mathrm{m}}$? (Consider the subring $k\left[x^2-1\right]$ of $k[x]$, where $k$ is a field, and let $\mathfrak{n}=(x-1)$. Can the element $1/(x+1)$ be integral?)

\begin{proof} The answer is no. Take $\mathfrak{n}=(x-1)$, then $k[x]/(\mathfrak{n})\simeq k$ is a field and $\mathfrak{n}$ is a maximal ideal. Note that $\mathfrak{m}=\mathfrak{n}\cap A=(x^2-1)$ and $A_{\mathfrak{m}}=(k[x^2-1]{)}_{(x^2-1)A}$. Since $x+1\notin (x-1)$, there is $1/(x+1)\in B_{\mathfrak{n}}$. If $B_{\mathfrak{n}}$ is integral over $A_{\mathfrak{m}}$, then there exists a polynomial

$$f(y)=y^k+a_1{y}^{k-1}+\cdots +a_k$$

where $a_i\in (k[x^2-1]{)}_{(x^2-1)A}$ such that $f(1/(x+1))=0$ and $a_i=f_i/g_i$ for some $f_i\in k[x^2-1]$, $g_i\notin (x^2-1)A$. It deduces that

$$f_k{h}_k(x+1{)}^k+f_{k-1}{h}_{k-1}(x+1{)}^{k-1}+\cdots +f_1{h}_1(x+1)+g_1\cdots g_k=0,$$

where $h_i=g_1\cdots g_n/g_i$. Thus, $x+1\mid g_1\cdots g_k$ and WLOG $x+1\mid g_1$. Since $g_1\in k[x^2-1]$ and $x+1\mid g_1$, one can get $x-1\mid g_1$ and so $x^2-1\mid g_1$, which is impossible. Therefore, $B_n$ is not necessarily integral over $A_{\mathrm{m}}$. \end{proof}

Text-Ex. 5.5. Let $A\subseteq B$ be rings, $B$ integral over $A$.

• i) If $x\in A$ is a unit in $B$ then it is a unit in $A$.
• ii) The Jacobson radical of $A$ is the contraction of the Jacobson radical of $B$.

\begin{proof} i) If $x\in A$ is a unit in $B$, then there exists $y\in B$ such that $xy=1$. Since $B$ is integral over $A$, there exists $a_0,\cdots ,a_n$ such that

$$y^n+a_1{y}^{n-1}+\cdots +a_n=0,$$

which deduces that

$$y=-x^{n-1}(a_1{y}^{n-1}+\cdots +a_n)=-a_1-a_{2}x-\cdots -a_n{x}^{n-1}\in A.$$

So $y\in A$ and $xy=yx=1$. Therefore, $x$ is a unit in $A$.

ii) By ^h2e8bp, $x\in \mathrm{Jac}(B)$ iff $1-xy\in B^{\times }$ for all $y\in B$. By i), if $x\in \mathrm{Jac}(B)\cap A$, then $1-xy$ is a unit of $A$ for all $y\in A$ and so $x\in \mathrm{Jac}(A)$. Hence $\mathrm{Jac}(A)\supseteq \mathrm{Jac}(B)\cap A$. Conversely, for any maximal ideal $M\subseteq B$, either $M\cap A=A$ or $M\cap A$ is a maximal ideal of $A$. Hence $\mathrm{Jac}(A)\subseteq \mathrm{Jac}(B)$ and so $\mathrm{Jac}(A)=\mathrm{Jac}(B)\cap A$. Therefore, $\mathrm{Jac}(A)$ is the contraction of $\mathrm{Jac}(B)$, where $f:A\hookrightarrow B$ is the inclusion map. \end{proof}

Text-Ex. 5.6. Let $B_1,\dots ,B_n$ be integral $A$-algebras. Show that ${\prod }_{i=1}^n{B}_i$ is an integral $A$-algebra.

\begin{proof} By definition, we say $B_i$ is an integral $A$-algebra if there exists a ring homomorphism $f_i:A\to B_i$ and $B_i$ is integral over $f_i(A)$. Define

$$f:A\to \prod _{i=1}^n{B}_i,a\mapsto (f_1(a),\cdots ,f_n(a)),$$

which is a ring homomorphism. It suffices to show ${\prod }_{i=1}^n{B}_i$ is integral over $f(A)$.

For any $b=(b_1,\cdots ,b_n)\in {\prod }_{i=1}^n{B}_i$, each $b_i$ is integral over $f_i(A)$ and so $f_i(A)[b_i]$ is a finitely generated $f_i(A)$-module. Consider subring $M={\prod }_{i=1}^n{f}_i(A)[b_i]$, which is a finitely generated $f(A)$-module and $f(A)[b]\subseteq M\subseteq {\prod }_{i=1}^n{B}_i$. By ^jy6o7s, $b$ is integral over $f(A)$. By the arbitrary of $b$, ${\prod }_{i=1}^n{B}_i$ is an integral $A$-algebra. \end{proof}

Text-Ex. 5.7. Let $A$ be a subring of a ring $B$, such that the set $B-A$ is closed under multiplication. Show that $A$ is integrally closed in $B$.

\begin{proof} It is enough to show if $x\in B$ is integral over $A$, then $x\in A$. Assume that $x\in B\setminus A$ is integral over $A$, then there exists a monic polynomial with coefficients in A that has $x$ as a root. Let $n$ be the minimal positive integer degree for such a polynomial, that is,

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

Since $x\notin A$, there is $a_n\ne 0$ and $n⩾2$. Then $x(x^{n-1}+a_1{x}^{n-2}+\cdots +a_{n-1})=-a_n\in A$, which deduces that $y=x^{n-1}+a_1{x}^{n-2}+\cdots +a_{n-1}\in A$, otherwise $x,y\in B\setminus A$ and $xy=-a_n\in B\setminus A$. Suppose $y=a^{\prime }$, then $x^{n-1}+a_1{x}^{n-2}+\cdots +a_{n-1}-a^{\prime }=0$. Thus, we have found a monic polynomial with coefficients in $A$ of degree $n-1$ that has $x$ as a root. This contradicts our choice of $n$ as the minimal degree for such a polynomial. \end{proof}