Proposition 1.17. Let , and let be ideals of and , respectively.
- i) ;
- ii) ;
- iii) If is the set of contracted ideals in and if is the set of extended ideals in , then , and is a bijective map of onto , whose inverse is .
\begin{proof}
i) Since , there is $$
a^{ec}=f^{-1}(\left\langle f(a)\right\rangle)\supseteq f^{-1}(f(a))\supseteq a
b^{ce}=(f^{-1}(b))^e=\left\langle f(f^{-1}(b))\right\rangle\subseteq\left\langle b\right\rangle =b
as $f(f^{-1}(b))\subseteq f(A)\cap b$. Now we finish the proof. ii) By i), we have $b^{cec}=(b^{ce})^c\subseteq b^c$ and $b^{cec}=(b^c)^{ec}\supseteq b^c$. It deduces that $b^c=b^{cec}$. Similarly, we have $a^e\subseteq (a^{ec})^e=a^{ece}$ and $a^{ece}=(a^e)^{ce}\subseteq a^e$. It deduces that $a^e=a^{ece}$. iii) For any $a\in C$, there exists an ideal $b\subseteq B$ such that $b^c=a$. Then by ii) $a=b^c=b^{cec}=a^{ec}$. Hence $C\subseteq \{a:a^{ec}=a\}$. On the other hand, for any $a\in \{a:a^{ec}=a\}$, $a=(a^e)^c$ is a contracted ideal. Therefore, $C=\{a:a^{ec}=a\}$. Similarly, every extended ideal $b$ satisfies $b^{ce}=b$, and every $b$ with $b=(b^{c})^e$ is an extended ideal. So $E=\{b:b^{ce}=b\}$. Define $\phi:C\to E,a\mapsto a^e$ and $\psi:E\to C,b\mapsto b^c$. Note that\psi\circ\phi(a)=a^{ec}=a,\phi\circ\psi(b)=b^{ce}=b.
Hence $\phi$ and $\psi$ are bijective, and $\phi^{-1}=\psi$. `\end{proof}` **Text-Ex. 1.2. Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an indeterminate $x$, with coefficients in $A$. Let $f=a_0+a_1 x+\cdots+a_n x^n \in A[x]$. Prove that** ^fg1mc4 - **i) $f$ is a unit in $A[x] \Leftrightarrow a_0$ is a unit in $A$ and $a_1, \ldots, a_n$ are nilpotent. (If $b_0+b_1 x+\cdots+b_m x^m$ is the inverse of $f$, prove by induction on $r$ that $a_n^{r+1} b_{m-r}=0$. Hence show that $a_n$ is nilpotent, and then use Ex. 1.)** - **ii) $f$ is nilpotent $\Leftrightarrow a_0, a_1, \ldots, a_n$ are nilpotent.** - **iii) $f$ is a zero-divisor $\Leftrightarrow$ there exists $a \neq 0$ in $A$ such that $a f=0$. (Choose a polynomial $g=b_0+b_1 x+\cdots+b_m x^m$ of least degree $m$ such that $f g=0$. Then $a_n b_m=0$, hence $a_n g=0$ (because $a_n g$ annihilates $f$ and has degree $<m)$. Now show by induction that $a_{n-r} g=0(0 \leqslant r \leqslant n)$.)** - **iv) $f$ is said to be primitive if $\left(a_0, a_1, \ldots, a_n\right)=(1)$. Prove that if $f, g \in A[x]$, then $f g$ is primitive $\Leftrightarrow f$ and $g$ are primitive.** `\begin{proof}` i) Assume that $f$ is a unit in $A[x]$, then there exists the inverse $g:=b_0+b_1x+\cdots+b_mx^m$. Sincefg=a_0b_0+\sum_{k=1}^{m+n}\left(\sum_{t=0}^{n}a_tb_{k-t}\right)x
where $b_{s}=0$ if $s\not\in\{0,\cdots,m\}$, we have $a_0b_0=1$ and $c_k:=\sum_{t=0}^n a_t b_{k-t}=0$ for all $k\in\{1,\cdots, m+n\}$. We claim that $a_n^{r+1}b_{m-r}=0$. When $k=m+n$, $c_k=a_nb_m=0$. Assume that $a_n^{r+1}b_{m-r}=0$ holds for $0\leqslant r\leqslant\ell-1\leqslant m-1$. Note that the coefficient of $x^{m+n-\ell}$ isa_nb_{m-\ell}+a_{n-1}b_{m-\ell+1}+\cdots +a_{n-\ell}b_m=0.
0=a_n^{\ell}(a_nb_{m-\ell}+a_{n-1}b_{m-\ell+1}+\cdots +a_{n-\ell}b_m)=a_n^{\ell+1}b_{m-\ell}
by induction hypothesis. Therefore, $a_n^{r+1} b_{m-r}=0$ for all $0\leqslant r\leqslant m$. Since $a_n^{m+1}b_0=0$ and $b_0\in A^\times$, $a_n^{m+1}=0$ and $a_n$ is nilpotent. Therefore, $-a_nx^n$ is nilpotent and so $f_1:=a_0+a_1x+\cdots+a_{n-1}x^{n-1}$ is a unit by Text-Ex. 1. Repeat the procedure above, we can prove that $a_1,\cdots,a_n$ are nilpotent. Conversely, assume that $a_0$ is unit and $a_1,\cdots,a_n$ are nilpotent. There exists $k_1,\cdots,k_n$ such that $a_i^{k_i}=0$. Thus $(f-a_0)^{k_1+\cdots +k_n}=0$ and so $f-a_0$ is a nilpotent element. By Text-Ex. 1, $f=a_0+(f-a_0)$ is a unit. ii) Assume that $f$ is nilpotent, then $1+f=(1+a_0)+a_1 x+\cdots+a_n x^n$ is a unit and so $1+a_0$ is unit, $a_1,\cdots,a_n$ are nilpotent. As $f$ is nilpotent, $a_0$ is nilpotent and so $a_0,\cdots,a_n$ are nilpotent. Conversely, if $a_0,\cdots,a_n$ are nilpotent, then there exist $k_0',\cdots, k_n'$ such that $a_i^{k_i'}=0$ and so $f^{k_0'+\cdots+k_n'}=0$. Hence $f$ is nilpotent. iii) If there exists $a\neq 0$ in $A$ such that $af=0$, then $f$ is a zero-divisor. Conversely, suppose $f$ is a zero-divisor and there does not exists $a\in A$ such that $af=0$. Assume that a polynomial $g$ with $fg=0$ has degree at least $m$. Note that the coefficient of $x^{n+m}$ is $a_nb_m=0$, then $a_ngf=0$. If $a_ng\neq 0$, then $g_1=a_ng$ has $\deg g_1<m$ and $g_1f=0$, which contradicts with the assumption. Thus $a_ng=0$. We claim that $a_{n-r}g=0$ for $0\leqslant r\leqslant n$. Assume that $a_{n-r}g=0$ holds for $0\leqslant r\leqslant r_0-1$. Then by induction hypothesis,fg=(a_0+a_1x+\cdots +a_{n-r_0}x^{n-r_0})(b_0+b_1x+\cdots +b_mx^m)=0
and so $a_{n-r_0}b_m=0$. If $a_{n-r_0}g\neq 0$, then $\deg a_{n-r_0}g<m$ and $a_{n-r_0}gf=0$, which is impossible. Therefore, $a_{n-r_0}g=0$. Now we have proved that $a_{n-r} g=0(0 \leqslant r \leqslant n)$. It deduces that $a_{n-r}b_0=0$ for any $0\leqslant r\leqslant n$. If $b_0=0$, then $g/x$ with $\deg g/x< m$ annihilates $f$, which is impossible. Hence, $b_0\neq 0$ in $A$ is what we desired. iv) Let $I=(a_0,a_1,\cdots,a_n)$ and let $J=(b_0,\cdots,b_m)$. Define $K=(a_0b_0,a_1b_0+a_0b_1,\cdots,a_nb_m)$, that is, the ideal generated by coefficients of $fg$. If $fg$ is primitive, then $K=(1)$ and $I,J\supseteq K=(1)$. Thus, $f$ and $g$ are primitive. If $f$ and $g$ are primitive, then $I=J=(1)$. If $K\neq (1)$, then there exists a prime element $p\in A$ such that $K\subseteq (p)$. Let $a_rx^r$ (rep. $b_sx^s$) be the first term of $f(x)$ (rep. $g(x)$) not divisible by $p$. Then the coefficient of $x^{r+s}$ is\cdots+a_{r+2} b_{s-2}+a_{r+1} b_{s-1}+a_r b_s+a_{r-1} b_{s+1}+a_{r-2} b_{s+2}+\cdots,
which is not divisible by $p$. It is impossible. Therefore, $K=(1)$ and so $fg$ is primitive. `\end{proof}` **Text-Ex. 1.4. In the ring $A[x]$, the Jacobson radical is equal to the nilradical.** `\begin{proof}` Since $\mathrm{Jac}(A[x])\supseteq\mathrm{Nil}(A[x])$, it suffices to show that any $f(x)\in \mathrm{Jac}(A[x])$ is nilpotent. Remark that $f(x)\in\mathrm{Jac}(A[x])$ yields that $1-f(x)g(x)$ is a unit for any $g(x)\in A[x]$. Assume that $f(x)=b_0+b_1x+\cdots+b_mx^m$. Take $g(x)=x$, then1-f(x)g(x)=1-b_0x-b_1x^2-\cdots -b_mx^{m+1}
is a unit. By Text-Ex. 2 i), we have $b_0,\cdots,b_m$ are nilpotent and so $f(x)$ is nilpotent by Text-Ex. 2 ii). Therefore, $f(x)\in \mathrm{Nil}(A[x])$. `\end{proof}` **Text-Ex. 1.9. Let $a$ be an ideal $\neq(1)$ in a ring $A$. Show that $\mathfrak{a}=r(\mathfrak{a}) \Leftrightarrow \mathfrak{a}$ is an intersection of prime ideals.** `\begin{proof}` If $\mathfrak{a}=r(\mathfrak{a})$, then $\mathfrak{a}=\cap_{\mathfrak a\subseteq P,P \text{ is prime}}P$ and so $\mathfrak a$ is an intersection of prime ideals. Conversely, assume that $\mathfrak a=\cap_{i\in I}P_i$ where $P_i$ are prime ideals. Then\mathfrak{a}\subseteq r(\mathfrak a)=\cap_{\mathfrak a\subseteq P,P \text{ is prime}}P\subseteq\cap_{i\in I}P_i=\mathfrak{a}
and so $\mathfrak{a}=r(\mathfrak{a})$. `\end{proof}` **Text-Ex. 1.10. Let $A$ be a ring, $\mathfrak{R}$ its nilradical. Show that the followings are equivalent:** - **i) $A$ has exactly one prime ideal;** - **ii) every element of $A$ is either a unit or nilpotent;** - **iii) $A / \Re$ is a field.** `\begin{proof}` i)->ii) If $A$ has exactly one prime ideal $P$, then $P$ is the unique maximal ideal of $A$ and $\mathfrak{R}=P$. For any $p\in \mathfrak{R}$, $p$ is nilpotent. If $a\in A\setminus \mathfrak{R}$, then $(a)+P=(1)$ and $ay+p=1$ for some $y\in A$. It deduces that $a$ is a unit by Text-Ex. 1. ii)->iii) For any $a\in A\setminus \mathfrak{R}$, $a$ is a unit. The image $\overline a$ in $A/\mathfrak R$ is also a unit. By the arbitrary of $a$, $A/\mathfrak R$ is a field. iii)->i) If $A/\mathfrak R$ is a field, then $\mathfrak R$ is a maximal ideal. Assume that $A$ has two different prime ideals $P_1$ and $P_2$, then $\mathfrak R\subseteq P_1\cap P_2$. Since either $P_1\cap P_2\subsetneq P_1$ or $P_1\cap P_2\subsetneq P_2$ holds, $\mathfrak R$ is not a maximal ideal, leading to a contradiction. Therefore, $A$ has exactly one prime ideal. `\end{proof}`