Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 9.

## Discrete Valuation Rings and Dedekind Domains

Exercise

**Proof.**$S_{−1}A$ is a Noetherian domain of dimension at most one, and is integrally closed. If the dimension of $S_{−1}A$ is one then it is a Dedekind domain; otherwise it’s the field of fractions of $A$.

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Suppose that $S=A−{0}$, and let $H,H_{′}$ be the ideal class groups of $A$ and $S_{−1}A$ respectively. Show that extension of ideals induces a surjective homomorphism $H→H_{′}$.

**Proof.**Let $I,I_{′}$ denote the groups of ideals of $A$ and $S_{−1}A$ respectively, $P⊆I$ and $P_{′}⊆I_{′}$ denote the subgroups of principal fractional ideals, and $K$ denote the field of fractions of $A$ (and $S_{−1}A$). For every $a∈I$, the extension $S_{−1}a∈I_{′}$; and clearly $(S_{−1}a)(S_{−1}b)=S_{−1}(ab)$. Hence the extension of ideals induces a homomorphism $I→I_{′}$. If $x_{1},…,x_{n}$ in $K$, let $a$ be the fractional ideal of $A$ generated by $x_{1},…,x_{n}$, then $S_{−1}a$ is pricisely the fractional ideal of $S_{−1}A$ generated by $x_{1},…,x_{n}$. Hence the homomorphism $I→I_{′}$ is surjective, and maps $P$ into $P_{′}$. Therefore it induces a surjective homomorphism $H→H_{′}$.

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Exercise *content* of $f$ is the ideal $c(f)=(a_{0},…,a_{n})$ in $A$. Prove the *Gauss’s Lemma* that $c(fg)=c(f)c(g)$.

**Proof.** First assume $A$ is a discrete valuation ring with the maximal ideal $m=(u)$. If $c(f)=m_{n}=(u_{n})$, then $f=u_{n}f_{1}$ such that $c(f_{1})=(1)$. So it’s sufficient to show that if $c(f)=c(g)=(1)$ then $c(fg)=(1)$. Passing to the residue field $K=A/m$, it’s equivalent to that if $f,g∈K[x]$ are non-zero, then $fg=0$, which is clearly true.

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Exercise

**Proof.** $⟸$: Obvious.

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Exercise

**Proof.** Suppose $m=(u)$. Since $⋂_{n=1}m_{n}=0$, for every non-zero $a∈A$, there exists $n≥0$ such that $a∈m_{n},a∈/m_{n+1}$. Let $v(a)=n$. Then clearly $v(xy)=v(x)+v(y)$ and $v(x+y)≥min(v(x),v(y))$. If $v(a)=n$, then $a=yu_{n}$ for some $y∈A$; and $y∈/m$ since $a∈/m_{n+1}$. Hence $y$ is a unit, therefore $(a)=m_{n}$.

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Exercise

**Proof.** Since torsion-freeness and flatness are local properties, we can passing to the localization, and assmue that $M$ is a finitely generated module over a discrete valuation ring.

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Exercise

**Proof.** Suppose $A$ is a discrete valuation ring with the maximal ideal $m$. Then $A$ is a PID, so by the structure theorem of finitely generated module over PID, $M$ is uniquely representable as a finite direct sum of modules $A/q_{i}$, for primary ideals $q_{i}$. Since the only primary ideals of $A$ is $m_{k},k=1,2,…$, it follows that $M≃⨁_{i}A/m_{k_{i}}$ uniquely.

In the case where $A$ is a Dedekind domain, for each prime ideal $p$, $M_{p}≃⨁A_{p}/(pA_{p})_{k_{i}}$ as above. By the exactness of localization, we have $A_{p}/(pA_{p})_{k_{i}}≃A/p_{k_{i}}$. Hence $M_{p}$ can be written as the finite direct sum of modules $A/p_{k_{i}}$ uniquely. Now it’s sufficient to show that there are only finite $p$ such that $M_{p}$ is not zero, and that the homomorphism $M→∏_{p}M_{p}$ is an isomorphism.

By Chapter 3, Exercise 19, since $M$ is finitely generated, we have $Supp(M)=V(Ann(M))$. Let $x_{1},…,x_{n}$ generate $M$, then $Ann(x_{k})=0$ for any $k$. Since $0$ is prime in $A$, it follows that $Ann(M)=⋂_{k=1}Ann(x_{k})=0$. Hence $A/Ann(M)$ is Artinian and hence has finite prime ideals. Therefore $Supp(M)$ is finite.

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Exercise

**Proof.** Since $a=0$, it follows that $B=A/a$ is a Noetherian ring of dimension $0$, hence an Artin ring. So there are finitely many prime ideals $p$ containing $a$, and $B=∏_{p⊇a}B_{p}$. Let $p⊇a$ be an prime ideal of $A$, then $B_{p}=(A/a)_{p}$ is a quotient ring of $A_{p}$. Therefore the maximal ideal of $B_{p}$ is principal for that $A_{p}$ is a discrete valuation ring.

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Deduce that every ideal in $A$ can be generated by at most $2$ elements.

**Proof.**Let $a$ be an non-zero ideal in $A$. Let $x∈a$ be any non-zero ideal in $a$, then $a/(x)⊂A/(x)$ is principal by the result above. Let $y∈a$ which image generates $a/(x)$, then clearly $a=(x,y)$.

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Exercise

**Proof.**By localizing at each prime ideal, then we may assume the ring is a discrete valuation ring, in which every ideal is either $0$ or a power of the maximal ideal $m$. We have $0∩a=0,0+a=a$ and $m_{u}+m_{v}=m_{min(u,v)},m_{u}∩m_{v}=m_{max(u,v)}$. Then the result is obvious.

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Exercise

**Proof.** (According to the hint) The propsition is equivalent to that the sequence of $A$-modules $Aφ i=1⨁n A/a_{i}ψ i<j⨁ A/(a_{i}+a_{j})$is exact, where $φ(x)ψ(x_{1}+a_{1},…,x_{n}+a_{n}) =(x+a_{1},…,x+a_{n})=(x_{i}−x_{j}+(a_{i}+a_{j}))_{i<j}. $

To show the sequence is exact, it is enough to show that it is exact when localized at any $p=0$. Hence it reduces to the case where $A$ is a discrete valuation ring.

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