Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 2.

## Modules

Exercise

**Proof.** If $m,n$ coprime then $n$ is a unit in $Z_{m}$, so

$x⊗y=n_{−1}x⊗ny=0$

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Exercise

**Proof.**Obviously $a→A→A/a→0$ is an exact sequence, so is $a⊗M→A⊗M→(A/a)⊗M→0$. But $a⊗M≅aM$ and $A⊗M≅M$, and the first arrow is the inclusion map, so $(A/a)⊗M≅(M/aM)$.

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Exercise

**Proof.**Let $m$ be the maximal ideal of $A$ and $k=A/m$ be the residue field of $A$. Let $M_{k}$ denote $k⊗_{A}M=(M/mM)$, then by Nakayama Lemma, $M_{k}=0→M=0$. So we have $M⊗_{A}N=0⟹(M⊗_{A}N)_{k}=0⟹M_{k}⊗_{k}N_{k}=0$. But $M_{k}$ and $N_{k}$ are vector spaces over field $k$, so $M_{k}⊗_{k}N_{k}=0$ implies $M_{k}=0$ or $N_{k}=0$, hence $M=0$ or $N=0$.

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Exercise

**Proof.**$M=⨁_{i∈I}M_{i}$ is flat $⟺$ for all injective $f:N→N_{′}$, $f⊗(⨁_{i∈I}1_{M_{i}})=⨁_{i∈I}(f⊗1_{M_{i}}):N⊗M→N_{′}⊗M$ is injective. And $⨁_{i∈I}f_{i}$ is injective if and only if each $f_{i}$ is injective, so qed.

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Exercise

**Proof.**As a $A$-module, $A[x]≅⨁_{n=0}A$, so by Exercise2.4 $A[x]$ is flat (since $A$ is flat).

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Exercise

Show that $M[x]≅A[x]⊗_{A}M$.

**Proof.** By define $(∑_{i=0}a_{i}x_{i})(∑_{j=0}m_{j}x_{j})=∑_{i=0}∑_{j=0}a_{i}m_{j}x_{i+j}$, trivially the module axioms hold here.

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Exercise

**Proof.** Consider map $ϕ:A[x]→(A/p)[x]$, then $Kerϕ=p[x]$. Then $p[x]$ is prime since $(A/p)[x]$ is an integral domain.

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Exercise

1. | If $M$ and $N$ are flat $A$-modules, then so is $M⊗_{A}N$. |

2. | If $B$ is a flat $A$-algebra and $N$ is a flat $B$-module, then $N$ is flat as an $A$-module. |

**Proof.**

1. | If $U→V→W$ is an exact sequence, then so is $(U⊗M)→(V×M)→(W×M)$, hence $(U⊗M)⊗N→(V⊗M)⊗N→(W⊗M)⊗N$. but $(U⊗M)⊗N≅U⊗(M⊗N)$, qed. |

2. | Let $j:M→M_{′}$ be an injective $A$-module homomorphism. Since $B$ is flat, $(id_{B}⊗_{A}j):(B⊗_{A}M)→(B⊗_{A}M_{′})$ is injective. Consider $(id_{B}⊗_{A}j)$ as a $B$-module homomorphism, then since $N$ is flat, $id_{N}⊗_{B}(id_{B}⊗_{A}j)$ is injective. By associativity of tensor product, and $N⊗_{B}B≅N$, we have $id_{N}⊗_{A}j$ injective, hence $N$ is flat as $A$-module. $□$ |

Exercise

**Proof.**Let $f:M_{′}→M$ and $g:M→M_{′′}$ be the maps in the sequence. If $x_{1},…,x_{n}$ generate $M_{′}$ and $y_{1},…,y_{m}$ generate $M_{′′}$. For each $y_{i}$ select an element $q_{i}∈M$ such that $g(q_{i})=y_{i}$, then since $M=⋃_{i=1}(q_{i}+Kerg)=Imf+∑_{i=1}(q_{i})$, and $Imf$ is generated by $p_{i}=f(x_{i})$, so $M$ is generated by $p_{1},…,p_{n},q_{1},…,q_{m}$.

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Exercise

**Proof.**If $uˉ:M/aM→N/aN$ is surjective, then forall $y∈N$ there exists $x∈M$ such that $u(x)−y∈aN$. That means, $N=Imu+aN$. So by Nakayama Lemma, $N=Imu$, i.e. $u$ is surjective.

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Exercise

**Proof.**let $m$ be a maximal ideal of $A$, and $ϕ:A_{m}→A_{n}$ an isomorphism, then $1⊗ϕ:(A/m)⊗A_{m}→(A/m)⊗A_{n}$ is an isomorphism between two $A/m$-vector spaces, hence the dim of two space are same, i.e. $m=n$.

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Exercise

**Proof.**Let $x_{1},…,x_{n}$ be a set of generators of $A_{n}$, and $y_{1},…,y_{n}∈M$ such that $ϕ(y_{i})=x_{i}$. Let $M_{′}$ be the submodule generating by $y_{1},…,y_{n}$, then clearly $M_{′}∩Kerϕ=0$, and for all $t∈M$ there exists $y∈M_{′}$ such that $f(t)=f(y)$, hence $M_{′}+Kerϕ=M$. Summarize results above we get $M≅M_{′}⊕Kerϕ$. Since $M$ is finitely generated, $Kerϕ$ must be finitely generated too.

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Exercise

**Proof.** Consider the quotient map $B⊗_{A}N→B⊗_{B}N$. Since $B⊗_{B}N≅N$, we have $h:N_{B}→N$ which maps $b⊗y$ to $by$.

Now $h∘g=id_{N}$, so $g$ is injective. Consider map $ϕ:N_{B}→N⊕Kerh$ defined by $ϕ=h⊕(id_{N_{B}}−g∘h)$. ($h(x−g(h(x)))=h(x)−h(x)=0$ so the second part of image of $ϕ$ is actually $Kerh$). We will prove $ϕ$ is an isomorphism so $N_{B}≅N⊕Kerh$.

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### Direct limits

Exercise *direct* set if for each pair $i,j$ in $I$ there exists $k∈I$ such that $i≤k$ and $j≤k$.

Let $A$ be a ring, let $I$ be a direct set and let $(M_{i})_{i∈I}$ be a family of $A$-modules indexed by $I$. For each pair $i,j$ in $I$ such that $i≤j$, let $μ_{ij}:M_{i}→M_{j}$ be an $A$-homomorphism, and suppose that the following axioms are satisfied:

1. | $μ_{ii}$ is the identity mapping of $M_{i}$ for all $i∈I$; |

2. | $μ_{ik}=μ_{jk}∘μ_{ij}$ whenever $i≤j≤k$. |

Then the modules $M_{i}$ and homomorphisms $μ_{ij}$ are said to form a *direct system* $M=(M_{i},μ_{ij})$ over the directed set $I$.

We shall construct an $A$-module $M$ called the *direct limit* of the direct system $M$. Let $C$ be the direct sum of $M_{i}$, and identify each module $M_{i}$ with its canonical image in $C$. Let $D$ be the submodule of $C$ generated by all elements of the form $x_{i}−μ_{ij}(x_{i})$ where $i≤j$ and $x_{i}∈M_{i}$. Let $M=C/D$, let $μ:C→M$ be the projection and let $μ_{i}$ be the restriction of $μ$ to $M_{i}$.

The module $M$, or MORE correctly the pair consisting of $M$ and the family of homomorphisms $μ_{i}:M_{i}→M$ is called the *direct limit* of the direct system $M$, and is written $lim M_{i}$. From the construction it is clear that $μ_{i}=μ_{j}∘μ_{ij}$ whenever $i≤j$.

**Proof.**No exercise here. For the last sentence, $μ_{i}(x)−μ_{j}(μ_{ij}(x))=μ(x−μ_{ij}(x))$, but $x−μ_{ij}(x)∈D=Kerμ$.

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Exercise

Show also that if $μ_{i}(x_{i})=0$ then there exists $j≥i$ such that $μ_{ij}(x_{i})=0$ in $M_{j}$.

**Proof.** Any element of $M$ can be written in form $∑_{i∈I}μ_{i}(x_{i})$ with only finite $x_{i}$ nonzero. But for any $i,j$ there exists $k$ such $i,j≤k$, so $μ_{i}(x_{i})+μ_{j}(x_{j})=μ_{k}(μ_{ik}(x_{i})+μ_{jk}(x_{j}))$, hence any element of $M$ can be written in form $μ_{k}(x_{k})$ for some $k$.

If $μ_{i}(x_{i})=0$, i.e. $x_{i}∈Kerμ=D$, then we have

$x_{i}=j,k∈I∑ (y_{j}−μ_{jk}(y_{j}))=j∈I∑ z_{j}$

where the sum contains only finite nonzero terms, and $z_{j}$ is projection to $M_{j}$. But $x_{i}∈M_{i}$ and the equation above is in a direct sum, so all elements $z_{j}=0$ except $z_{i}=x_{i}$. Select an index $p∈I$ which $≥$ any $j,k$ appearing here, then

$μ_{ip}(x_{i})=j∈I∑ μ_{jp}(z_{j})=j,k∈I∑ (μ_{jp}(y_{j})−(μ_{kp}∘μ_{jk})(y_{j}))=0$

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Exercise

**Proof.** First prove $(M,μ_{i})$ constructed here satisfies this condition.

For any $N$ and $α_{i}:M_{i}→N$ satisfies $α_{i}=α_{j}∘μ_{ij}$, define $β:C→N$ by $β(∑_{i∈I}x_{i})=∑_{i∈I}α_{i}(x_{i})$, then for all $i≤j$ and $x_{i}∈M_{i}$ we have $β(x_{i}−μ_{ij}(x_{j}))=0$, hence $D⊆Kerβ$, so $β$ induce an $A$-homomorphism $α:M→N$ and $α_{i}=α∘μ_{i}$ for any $i∈I$. Since all elements in $M$ can be written in the form $μ_{i}(x_{i})$, the map $α$ is then unique.

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Exercise

$lim M_{i}=∑M_{i}=⋃M_{i}$

**Proof.** $∑M_{i}=⋃M_{i}$ is obviously hold since for all $i,j$ there exists some $k,M_{i}+M_{j}⊆M_{k}$.

We show $lim M_{i}≅⋃M_{i}$ by show $⋃M_{i}$ have the universal property in the previous exercise. If given $(N,α_{i})$ such that $α_{i}=α_{j}∘μ_{ij}$ for any pair $M_{i}⊆M_{j}$, then for all pair of indices $i,j$ and element $x∈M_{i}∩M_{j}$ there exists $M_{k}⊇M_{i}∪M_{j}$, so $α_{i}(x)=α_{k}(x)=α_{j}(x)$. Therefore we can define $α:⋃M_{i}→N$ that agrees each $α_{i}$ over $M_{i}$. Since any element belonging to $⋃M_{i}$ also belongs to some $M_{i}$, so the homomorphism here is unique.

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Exercise

A *homomorphism* $Φ:M→N$ is by definition a family of $A$-module homomorphisms $ϕ_{i}:M_{i}→N_{i}$ such that $ϕ_{j}∘μ_{ij}=ν_{ij}∘ϕ_{i}$ whenever $i≤j$. Show that $Φ$ defines a unique homomorphism $ϕ=lim ϕ_{i}:M→N$ such that $ϕ∘μ_{i}=ν_{i}∘ϕ_{i}$ for all $i∈I$.

**Proof.**Let $ϕ_{i}=ν_{i}∘ϕ_{i}:M_{i}→N$, then $ϕ_{j}∘μ_{ij}=ν_{j}∘ν_{ij}∘ϕ_{i}=ϕ_{i}$ whenever $i≤j$. Hence there exists a unique homomorphism $ϕ:M→N$ such that $ϕ∘μ_{i}=ϕ_{i}=ν_{i}∘ϕ_{i}$ for all $i∈I$.

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Exercise

is *exact* if the corresponding sequence of modules and module homomorphisms is exact for each $i∈I$. Show that the sequence $M→N→P$ of direct limits is then exact.

**Proof.** Let $ϕ_{i}:M_{i}→N_{i},ψ_{i}:N_{i}→P_{i}$ be the corresponding homomorphisms, $μ_{ij},ν_{ij},π_{ij}$ be the homomorphisms in system $M,N,P$, and $ϕ:M→N,ψ:N→P$ the induced homomorphisms.

First we show $ψ∘ϕ=0$. For any $x∈M$, there exists $i∈I$ and $x_{i}∈M_{i}$ such that $x=μ_{i}(x_{i})$, and then

$ψ(ϕ(x))=(ψ∘ϕ∘μ_{i})(x)=(π_{i}∘ψ_{i}∘ϕ_{i})(x)=π_{i}(0)=0$

Then if $y∈Kerψ$, again $y=ν_{i}(y_{i})$ for some $i∈I$ and $y_{i}∈N_{i}$, hence $0=ψ(ν_{i}(y_{i}))=π_{i}(ψ_{i}(y_{i}))$. So there exists $j≥i$ such that $0=π_{ij}(ψ_{i}(y_{i}))=ψ_{j}(ν_{ij}(y_{j}))$, hence $ν_{ij}(y_{j})∈Kerψ_{j}=Imϕ_{j}$, and there exists $x_{j}$ such that $ϕ_{j}(x_{j})=ν_{ij}(y_{j})$. Let $x=μ_{j}(x_{j})$, we have

$ψ(x)=ψ(μ_{j}(x_{j}))=ν_{j}(ψ_{j}(x_{j}))=ν_{j}(nu_{ij}(y_{j}))=ν_{i}(y_{i})=y$

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### Tensor products commute with direct limits

Exercise

For each $i∈I$ we have a homomorphism $μ_{i}⊗1:M_{i}⊗N→M⊗N$, hence by Exercise 16 a homomorphism $ψ:P→M⊗N$. Show that $ψ$ is an isomorphism, so that

$lim (M_{i}⊗N)≅(lim M_{i})⊗N$

**Proof.** Let $π_{i}$ be the projection map form $M_{i}⊗N$ to $P$.

Define $ϕ_{y,i}:M_{i}→P$ by $x_{i}↦π_{i}(x_{i}⊗y)$, then clearly $ϕ_{y,i}=ϕ_{y,j}∘μ_{ij}$, so $ϕ_{y,i}$ induce a homomorphism $ϕ_{y}:M→P$. Since every $ϕ_{y,i}$ is $A$-linear over $y$, by the construction of $ϕ_{y}$ it’s easy to show so is $ϕ_{y}(x)$. So we have a homomorphism $ϕ:M⊗N→P$ defined by $ϕ(x⊗y)=ϕ_{y}(x)$. We will show that $ϕ$ is the inverse of $ψ$. We have:

$ϕ(μ_{i}(x_{i})⊗y)ψ(π_{i}(x_{i}⊗y)) =ϕ_{y}(μ_{i}(x_{i}))=ϕ_{y,i}(x_{i})=π_{i}(x_{i}⊗y)=(μ_{i}⊗1)(x_{i}⊗y)=μ_{i}(x)⊗y $

Since all $x∈M$ can be written in form $μ_{i}(x_{i})$ and all $p∈P$ can be written in form $π_{i}(x_{i}⊗y)$, it’s clearly $ϕ∘ψ=id_{P},ψ∘ϕ=id_{M⊗N}$. So

$lim (M_{i}⊗N)≅(lim M_{i})⊗N$$□$

Exercise *direct limit* of the system $(A_{i},α_{ij})$.

If $A=0$ prove that $A_{i}=0$ for some $i∈I$.

**Proof.** Let $α_{i}:A_{i}→A$ be the mappings. In $A$ every element is some $α_{i}(a_{i})$ where $i∈I$ and $a_{i}∈A_{i}$. For any $α_{i}(a_{i})$ and $α_{j}(a_{j})$, let $k$ be an index $≥i,j$, we define $α_{i}(a_{i})⋅α_{j}(a_{j})=α_{k}(α_{ik}(a_{i})α_{jk}(a_{j}))$. If there are two indices $k_{1},k_{2}≥i,j$, find an index $p≥k_{1},k_{2}$ we can show that the definition does not depend on the choice of $k$. The ring axioms are easy to verify, with the identity element be any $α_{i}(1)$ (they are all equal).

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Exercise

If each $A_{i}$ is an integral domain, then $lim A_{i}$ is an integral domain.

**Proof.** If $x∈R_{i}$, then clearly $α_{ij}(x_{i})∈R_{j}$. So $(R_{i},αˉ_{ij})$ is a direct system where $αˉ_{ij}$ is the restriction of $α_{ij}$.

Let $A$ denote the direct limit of $A_{i}$. An element $μ_{i}(x_{i})∈A$ is nilpotent iff. $∃n>0$, $μ_{i}(x_{i})=0$ iff. $∃n>0$ and $j≥i$ such that $μ_{ij}(x_{i})_{n}=0$, i.e. exists $j≥i$ such that $μ_{ij}(x_{i})$ is nilpotent in $A_{j}$. That is, an element $x∈A$ is nilpotent if and only if it can be written in form $μ_{j}(x_{j})$ where $x_{j}∈R_{j}$. So the nilradical of $A$ is $lim R_{i}$, the proposition holds.

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Exercise *tensor product* of the family $(B_{λ})_{λ∈Λ}$.

**Proof.** For $J={λ_{1},…,λ_{n}}$ and $J_{′}=J∪{λ_{n+1},…λ_{n+m}}$ we have a canonical map $β_{JJ_{′}}:B_{J}→B_{J_{′}}$ defined by $b_{1}⊗⋯⊗b_{n}↦b_{1}⊗⋯⊗b_{n}⊗1⊗⋯⊗1$. Clearly $β_{JJ_{′}}$ are $A$-algebra homomorphism.

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### Flatness and Tor

Exercise

1. | $M$ is flat; |

2. | $Tor_{n}(M,N)=0$ for all $n>0$ and all $A$-modules $N$; |

3. | $Tor_{1}(M,N)=0$ for all $A$-modules $N$. |

**Proof.** i) $⟹$ ii): $M$ is flat means $⊗M$ is exact functor. If $P→N$ is a projective resolution of $N$, then $P⊗M$ is exact, and $Tor_{n}(M,N)=H_{i}(P⊗M)=0$.

ii) $⟹$ iii): Trivial.

iii) $⟹$ i): Consider a exact sequence $0→N_{′}→N→N_{′′}→0$, then there is a long exact sequence:

$⋯→Tor_{1}(M,N_{′})→→N_{′}⊗M→N⊗M→N_{′′}⊗M→0$

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Exercise

**Proof.** For all $A$-module $M$ there exists a long exact sequence:

$⋯→Tor_{2}(M,N_{′′})→Tor_{1}(M,N_{′})→Tor_{1}(M,N)→Tor_{1}(M,N_{′′})→⋯→0$

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Exercise

**Proof.** By (2.19), $N$ is flat $⟺⊗N$ preserves injective map of finitely generated $A$-module, so $⟺Tor_{1}(M,N)=0$ for all finitely generated $M$.

Now if $M$ is generated by $x_{1},…,x_{n}$, and let $M_{i}$ be the submodule generated by $x_{1},…,x_{i}$. Then there is exact sequence $0→M_{i−1}→M_{i}→M_{i}/M_{i−1}$. So by the last exercise, it’s sufficient to show $Tor_{1}(M_{i}/M_{i−1},N)=0$. But $M_{i}/M_{i−1}≃A/a$ for some ideal $a$, so it’s sufficient to show $Tor_{1}(A/a,N)=0$ for all $a$.

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Exercise

1. | $A$ is absoutely flat. |

2. | Every principal ideal is idempotent. |

3. | Every finitely generated ideal is a direct summand of A. |

**Proof.** i) $⟹$ ii): Let $x∈A$, then $A/(x)$ is flat. Let $f:(x)⊗A→A$: $f(x,a)=ax$, tensor with $A/(x)$ gives an injective map $α:(x)⊗A/(x)→A/(x)$. By the commutative diagram, $(x)⊗A→A→A/(x)$ (the zero map) equals $(x)⊗A→(x)⊗A/(x)→A/(x)$. Since $α$ is injective, $Im((x)⊗A→(x)⊗A/(x))=0$. So $(x)⊗A/(x)=0$ and $(x)=(x_{2})$.

ii) $⟹$ iii): For any $x∈A$, there exists $a$ such that $x=ax_{2}$, so $e=ax$ is idempotent and $(x)⊆(ex)⊆(e)⊆(x)$, i.e. $(e)=(x)$. For any two idempotent $e,f$, we have $(e,f)=(e+f−ef)$ cause $e=e(e+f−ef)$. So any finitely generated ideal of $A$ is idempotent principal ideal $(e)$, and $A=(e)⊕(1−e)$.

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Exercise

If $A$ is absolutely flat, every non-unit in $A$ is a zero-divisor.

**Proof.** In a Boolean ring every element is idempotent, so Exercise 27 ii) holds, and it’s absolutely flat.

In the ring of Exercise 1.7 (every $x$ satisfy $x_{n}=x$ for some $n=n(x)>1$), we have $(x)=(x_{n})⊂(x_{2})⊂(x)$, so $(x_{2})=(x)$ and Exercise 27 ii) holds.

If $A$ is absolutely flat and $a⊆A$, then for any $(b/a)⊆(A/a)$, we have $(b/a)_{2}=b_{2}/a=b/a$. So $A/a$ is also absolutely flat.

If $A$ is a absolutely flat local ring with maximal ideal $m=0$, then there exists $0=x∈m$, so $A=(x)⊕A_{′}$ for some $A_{′}=0$. But then $0⊕A_{′}$ is a ideal not contained in $m$ (otherwise $A=m$), contradiction. So $m=0$ and $A$ is a field.

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