HW7

(1) Suppose $A$ is a Noetherian ring, $B=A/I$ is a quotient ring of $A$. Show that:

• (i) $B$ is a Noetherian $A$-module.
• (ii) $B$ is a Noetherian ring.

\begin{proof} i) Since $A$ is a Noetherian ring, any submodule of the $A$-module $A$ is finitely generated. Define $\pi :A\to A/I$ as the canonical map. For any submodule $M$ of $B=A/I$, there exists $C\subseteq A$ such that $\pi (C)=M$ and $C$ is finitely generated. Assume that $C=⟨a_1,\cdots ,a_n⟩$, then $M=⟨\pi (a_1),\cdots ,\pi (a_n)⟩$ and so $M$ is finitely generated. Therefore, $B$ is a Noetherian $A$-module.

ii) It suffices to show any submodule of $B$-module $B$ is finitely generated. Assume that $N$ is a submodule of $B$-module of $B$, then for any $a\in A$ and $\overline{b}=b+I\in B$, we have that $a(b+I)=ab+I=\overline{a}\overline{b}\in B$ and so $N$ is a submodule of $A$-module $B$. By i), we know $N$ is finitely generated and so $B$ is a Noetherian ring. \end{proof}

(2) Let $k$ be a field, then $k[x]$ is an Euclidean domain.

\begin{proof} Define $N:k[x]\setminus \{0\}\to \mathbb{N},f(x)\to \mathrm{deg}f$. Then for any $f,g\in k[x]$, if $N(f)<N(g)$, then $f=0\cdot g+f$ with $d(f)<d(g)$. Now we assume that $N(f)⩾N(g)$. It deduces that $f(x)$ can be written as $f(x)=h(x)g(x)+r(x)$ where $\mathrm{deg}hg=\mathrm{deg}f$ and $\mathrm{deg}r<\mathrm{deg}g$. Hence, the quotient is $h$ and the remainder is $r$, where $N(r)<N(g)$. Therefore, $k[x]$ is an Euclidean domain. \end{proof}

(3) Find an example, where $A$ is a Noetherian ring, $B\subset A$ is a subring, but $B$ is NOT a Noetherian ring.

\begin{proof} Let $A=\mathbb{R}[x_1,x_2]$, then $A$ is a Noetherian ring. Let $B$ be the subring generated by $\{x{y}^i:i⩾0\}$. Then $B$ is not a Noetherian ring, because $B$ is not finitely generated.

Alternating proof. Let $B=k[x_1,x_2,\cdots ,x_n,\cdots ]$ and $A$ be the fractional field of $B$. \end{proof}

(4) Let $M$ be a Noetherian $A$-module and $u:M\to M$ a module homomorphism. If $u$ is surjective, then $u$ is an isomorphism. (Hint: consider the submodules $\ker (u^n)$.)

\begin{proof} Since $u$ is a module homomorphism, $u^n:M\to M$ is also a module homomorphism for all $n\in {\mathbb{N}}_{+}$. It follows that $\ker (u^n)$ is a submodule of $M$ for all $n\in {\mathbb{N}}_{+}$. Since

$$\ker (u)\subseteq \ker (u^2)\subseteq \cdots \subseteq \ker (u^n)\subseteq \cdots$$

and $M$ is Noetherian, there exists $k$ such that $\ker (u^k)=\ker (u^{k+1})$. For any non-zero $m\in N$, there is $m^{\prime }$ such that $u^k(m^{\prime })=m$ as $u$ is surjective. If $u(m)=0$, then $m^{\prime }\in \ker (u^{k+1})\setminus \ker (u^k)$, which is impossible. Therefore, $u(m)\ne 0$ for any non-zero $m\in N$ and so $u$ is an isomorphism. \end{proof}