Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 3.

## Rings and Modules of Fractions

Exercise

**Proof.** $⟹$: if $x_{1},…,x_{n}$ generate $M$, then $x_{i}/1=0/1$, i.e. exists $s_{i}∈S$ such that $s_{i}x_{i}=0$. Let $s=s_{1}s_{2}⋯s_{n}$, then $sx_{i}=0$ for all $i=1,…,n$.

$□$

Exercise

**Proof.**If $x/(1+y)∈S_{−1}a$, then $x,y∈a$. Therefore $1+x/(1+y)=(1+x+y)/(1+y)$, and $1+x+y∈S$, hence $1+x/(1+y)$ is unit, i.e. $1+S_{−1}a$ only contains units. So $S_{−1}a$ is contained in the Jacobson radical of $S_{−1}A$.

$□$

**Proof.**If $aM=M$, let $S=1+a$, then $(S_{−1}a)(S_{−1}M)=S_{−1}M$. Since $S_{−1}a$ is contained in the Jacobson radical, by Nakayama we have $S_{−1}M=0$, so by Exercise 3.1 there exists $x∈S=1+a$ such that $xM=0$.

$□$

Exercise

**Proof.** Define $f:A→U_{−1}(S_{−1}A)$ by $f(a)=(a/1)/(1/1)$, then for all $u∈ST$ we have $f(u)$ is a unit (if $x∈S$ and $y∈T$, then $f(xy)_{−1}=(1/x)/(y/1)$), so $f$ induces a map $fˉ :(ST)_{−1}→U_{−1}(S_{−1}A)$ satisfying: forall $s∈S,t∈T,fˉ (x/(st))=(x/s)/(t/1)$.

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Exercise

**Proof.** Define $g:B×S→B×T$ by $g=id_{B}×f$. By definition the action of $S$ on $B$ is $s⋅b=f(s)b$, so two elements $(b,s);(b_{′},s_{′})$ is equal in $S_{−1}B$ if and only if there exists $t∈S$ such that $f(t)(f(s)b_{′}−f(s_{′})b)=0$; this is equivalent to the equality of $(b,f(s));(b_{′},f(s_{′}))$, so $g$ induces a bijection $ϕ:S_{−1}B→T_{−1}B$. Since $f$ is ring homomorphism, it’s easy to show $ϕ$ is a group isomorphism. Finally, if $a/s∈S_{−1}A$ and $b/t∈S_{−1}B$, then

$ϕ((a/s)⋅(b/t))=ϕ((f(a)b)/(st))=(f(a)b)/f(st)=(a/s)⋅(b/f(t))=(a/s)⋅ϕ(b/t)$

$□$

Exercise

**Proof.** Let $R(A),R(A_{p})$ denote the nilradical of $A$ and $A_{p}$, then $R(A_{p})=R(A)_{p}$. So for all $p$, $R(A)_{p}=R(A_{p})=0$, hence $R(A)=0$.

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Exercise

**Proof.** Since ${1}∈Σ$, it is nonempty. By Zorn’s lemma, it’s sufficient to show every chain in $Σ$ has a upper bound. If $(S_{i})_{i∈I}$ is an ordered chain that $α<β∈I⟹S_{α}⊆S_{β}$, then consider the onion of them, denoted by $S$. Clearly $0∈/S$, and if $x∈S_{α}⊆S,y∈S_{β}⊆S$, then $xy∈S_{max(α,β)}⊆S$, so $S$ is multiplicatively closed, and $S∈Σ$ is an upperbound of $(S_{i})_{i∈I}$.

If $S$ is a maximal element of $Σ$, let $p=A−S$. We will show $p$ is an ideal. If $a∈p$, then the minimal multiplicatively closed set containing $S$ and $a$ is ${sa_{n}∣s∈S,n≥0}$. It’s a multiplicatively closed set strictly containing $S$, so by hyposthesis it must contains $0$, i.e. $∃s∈S,n>0$ such that $sa_{n}=0$. Conversely if these $s,n$ exist then clearly $a∈p$.

Now if $a,b∈p$, then there exists $s,t∈S$ and $n,m>0$ so that $sa_{n}=tb_{m}=0$. Consider $st(a−b)_{n+m}$, by binomial expanding it’s a sum of some $sta_{i}b_{j}$, where $i≥n$ or $j≥m$, so equals $0$; hence $a−b∈p$. Together with $0∈p$ we have $p$ is an additive subgroup of $A$. If $a∈A,x∈p$, then $sx_{n}=0$ for some $s∈S,n>0$, so $s(ax)_{n}=0$ and $ax∈p$, and so $p$ is an ideal of $A$. If $x,y∈/p$, that is, $x,y∈S$, then $xy∈/p$, and $p$ is an prime ideal. If $p_{′}⊆p$ is some prime ideal, then $A−p_{′}$ is a multiplicatively closed set containing $S$, so equals to $S$, an $p_{′}=p$. This means $p$ is minimal.

$□$

Exercise *saturated* if

$xy∈S⟺x∈Sandy∈S$

Show that

1. | $S$ is saturated $⟺A−S$ is a union of prime ideals. |

2. | If $S$ is any multiplicatively closed subset of $A$, there is a unique smallest saturated multiplicatively closed subset $Sˉ$ containing $S$, and that $Sˉ$ is the complement in $A$ of the union of the prime ideals which do not meet $S$. ($Sˉ$ is called the |

If $S=1+a$, where $a$ is an ideal of $A$, find $Sˉ$.

**Proof.**

1. | $⟸$: If $A−S=⋃_{i∈I}p_{i}$, then for any $x,y∈A$: $xy∈S⟺∀i∈I,xy∈/p_{i}⟺∀i∈I,x∈/p_{i}andy∈/p_{i}⟺x∈Sandy∈S$so $S$ is saturated. $⟹$: If $S$ is saturated and $a∈/S$, then $(a)$ is disjoint from $S$. Let $Σ$ be all ideals containing $a$ and disjoint from $S$, then $(a)∈Σ$, and all increasing chains are closed in $Σ$, so by Zorn’s Lemma, there is a maximal element $p∈Σ$. If $x,y∈/p$, then $(x)+p,(y)+p$ strictly contain $p$, so not in $Σ$, i.e. they meet with $S$. Let $u∈((x)+p)∩S,v∈((y)+p)∩S$, then $uv∈((xy)+p)∩S$, so $xy∈/p$, hence $p$ is prime. Every element in $A−S$ is contained in a prime ideal disjoint from $S$, so $A−S$ is union of prime ideals. |

2. | Let $Sˉ$ be the complement of the union of all prime ideals disjoint from $S$: $Sˉ=A−∪{p∈Spec(A)∣p∩S=∅}$, then by (1), $Sˉ$ is saturated. If $T$ is any saturated multiplicatively closed subset containing $S$, then $T$ is the complement of union of some prime ideals disjoint from $S$, hence $T⊇Sˉ$. |

$□$

Exercise

1. | $ϕ$ is bijective. |

2. | For each $t∈T$, $t/1$ is a unit in $S_{−1}A$. |

3. | For each $t∈T$ there exists $x∈A$ such that $xt∈S$. |

4. | $T$ is contained in the saturation of $S$. |

5. | Every prime ideal which meets $T$ also meets $S$. |

**Proof.**

i)$⟹$ii): | If $ϕ$ is bijective, then since $ϕ(t/1)=t/1$ is a unit in $T_{−1}A$, it must be a unit in $S_{−1}A$. |

ii)$⟹$i): | Consider $ψ:T_{−1}A→S_{−1}A$ defined by $ψ(x/t)=(x/1)⋅(t/1)_{−1}$, clearly $ϕ$ and $ψ$ are inverses, so $ϕ$ is bijective. |

ii)$⟹$iii): | $t/1∈S_{−1}A$ is a unit $⟹$ there exists $s,s_{′}∈S,x_{′}∈A$ such that $stx_{′}=ss_{′}$ ($(t/1)(x_{′}/s_{′})=1$). Let $x=sx_{′}$, then $xt∈S$. |

iii)$⟹$iv): | Let $Sˉ$ denote the saturation of $S$, then $xt∈S⊆Sˉ⟹t∈Sˉ$, so $T⊆Sˉ$. |

iv)$⟹$v): | If $p∩T=∅$, then $p∩Sˉ=∅$. Since $Sˉ$ is disjoint from any prime ideal which doesn’t meet $S$, so there must be $p∩S=∅$. |

v)$⟹$ii): | Since every prime ideal of $S_{−1}A$ is $S_{−1}p$ for some $p$ not meet $S$, so: $t/1$ is unit in $S_{−1}A⟺$ Any prime ideal in $S_{−1}A$ doesn’t contain $t/1⟺$ for any $p$ disjoint from $S$, we have $t∈/p$. By v), every prime ideal disjoint from $S$ also disjoint from $T$, so for all $t∈T$, $t/1$ is a unit in $S_{−1}A$. $□$ |

Exercise

**Proof.** If $xy∈S_{0}$, then $x$ and $y$ must not be zero-divisors, so $x∈S_{0}$ and $y∈S_{0}$. Conversely $x∈S_{0}$ and $y∈S_{0}$ implies $xy∈S_{0}$, so $S_{0}$ is a saturated multiplicatively closed subset.

$□$

*the total ring of fractions*of $A$. Prove that

1. | $S_{0}$ is the largest multiplicatively closed $A$ for which the homomorphism $A→S_{0}A$ is injective. |

2. | Every element in $S_{0}A$ is either a zero-divisor or a unit. |

3. | Every ring in which every non-unit is a zero-divisor is equal to its total ring of fractions (i.e. $A→S_{0}A$ is bijective). |

**Proof.**

1. | $x∈$ the kernel of $A→S_{−1}A$ if and only if there exists $s∈S$ such that $sx=0$, i.e. $Ker(A→S_{−1}A)=Ann(S)$. Hence $A→S_{−1}A$ is injective if and only if $S$ only contains non-zero-divisors, so $S_{0}$ is the largest of them. |

2. | If $x∈S_{0},s∈S_{0}$, then $(x/s)=(s/x)_{−1}∈S_{0}A$ is a unit. On the other hand if $x∈/S_{0},s∈S_{0}$, then $x$ is a zero-divisor in $A$, hence there is a non-zero element $y$ so that $xy=0$. Since $A→S_{0}A$ is injective, $y/1=0∈S_{0}A$. Then $(x/s)(y/1)=0$ and $x/s$ is a zero-divisor in $S_{0}A$. |

3. | If $A$ satisfies the condition, then $S$ only contains unit. Then for any $x∈A,s∈S_{0}$, $(x/s)=(xs_{−1}/1)∈S_{0}A$. So $A→S_{0}A$ is surjective, and by i) it is injective, hence bijective. $□$ |

Exercise

1. | If $A$ is absolutely flat, and $S$ is any multiplicatively closed subset of $A$, then $S_{−1}A$ is absolutely flat. |

2. | $A$ is absolutely flat $⟺$ $A_{m}$ is a field for each maximal ideal $m$. |

**Proof.**

1. | For any $(x/s)∈S_{−1}A$, there exists $a$ such that $x=ax_{2}$; so $(x/s)=as(x/s)_{2}$ and $(x/s)=((x/s)_{2})$. |

2. | $⟹$: $A_{m}$ is a absolutely flat local ring, hence a field. $⟸$: For any $A$-module $M$, $M_{m}$ is a vector space of $A_{m}$, hence flat. Since flatness is a local property, $M$ is flat. |

$□$

Exercise

1. | $A/R$ is absolutely flat ($R$ being the nilradical of $A$). |

2. | Every prime ideal of $A$ is maximal. |

3. | $Spec(A)$ is a $T_{1}$-space (i.e. every subset consisting of a single point is closed). |

4. | $Spec(A)$ is Hausdorff. |

If these conditions are satisfied, show that $Spec(A)$ is compact and totally disconnected (i.e. the only connected subsets of $Spec(A)$ are those consisting of a single point).

**Proof.** Cause ii) iii) iv) holds for $A$ if and only if it holds for $A/R$, without loss of generality we may assume $R=0$.

i)$⟺$ ii): If $p$ is a prime ideal and contained in the maximal ideal $m$, then $p_{m}$ is a prime ideal of $A_{m}$, which is a field. Hence $p=m$ is maximal. Conversely if ii) holds, then $A_{m}$ has only $m_{e}$ as it’s prime ideal. But it’s nilradical is $R_{m}=0$, so $m_{e}=0$ and $A_{m}$ is a field.

ii)$⟹$iv): Let $p,q$ be distinct prime (hence maximal) ideals of $A$, and let $f∈p−q$. Consider $X_{f}={a∈Spec(A)∣f∈a}$, it’s a open subset of $Spec(A)$, and $p∈X_{f},q∈/X_{f}$, so it’s sufficient to show $X_{f}$ is close:

引理

**Proof.** Every prime ideal which does not contains $f$ correspends to a prime ideal of $X_{f}$, and every prime ideal which contains $I$ correspends to a prime ideal of $A/I$. So it’s sufficient to show $ϕ:A/I→A_{f}$ maps prime ideal to prime ideal (the preimage of a prime ideal is automaticly prime).

$□$

iv)$⟹$iii): Trivial.

$□$

Exercise *torsion element* of $M$ if $Ann(x)=0$, that is if $x$ is killed by some non-zero element of $A$. Show that the torsion elements of $M$ form a submodule of $M$.

**Proof.**If $a,b=0∈A;x,y∈M;ax=by=0$, then $ab(x+y)=0,a(bx)=0$, and $ab=0$ since $A$ is an integral domain, so the torsion elements are closed under addition and scalar-multiplication, and $Ann(0)=0$ so it contains $0$, hence a submodule of $M$.

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*torsion submodule*of $M$ and is denoted by $T(M)$. If $T(M)=0$, the module $M$ is said to be

*torsion-free*. Show that

1. | If $M$ is any $A$-module, then $M/T(M)$ is torsion-free. |

2. | If $f:M→N$ is a module homomorphism, then $f(T(M))⊆T(N)$. |

3. | If $0→M_{′}→M→M_{′′}$ is an exact sequence, then the sequence $0→T(M_{′})→T(M)→T(M_{′′})$ is exact. |

4. | If $M$ is any $A$-module, then $T(M)$ is the kernel of the mapping $x↦1⊗x$ of $M$ into $K⊗_{A}M$, where $K$ is the field of fractions of $A$. |

**Proof.**

1. | If $xˉ$ is a torsion element in $M/T(M)$, then there exists $a=0∈A$ such that $ax∈T(M)$, so there exists another $a_{′}=0∈A$ such that $(a_{′}a)x=a_{′}(ax)=0$, hence $x∈T(M)$ and $xˉ=0∈M/T(M)$. |

2. | If $a=0∈A,x∈T(M)$ such that $ax=0$, then $af(x)=f(ax)=0∈N$, so $f(x)∈T(N)$. |

3. | If $0 M_{′}f Mg M_{′′}$ is exact, let $fˉ :T(M_{′})→T(M)$ and $gˉ :T(M)→T(M_{′′})$ denote the induced map, then clearly $gˉ ∘fˉ =0$, and $fˉ $ is injective. If $x∈Kergˉ =Kerg∩T(M)=f(M_{′})∩T(M)$, then there exists $x_{′}∈M_{′}$ such that $f(x_{′})=x$. Since $x∈T(M)$, there exists $a=0∈A$ such that $f(ax_{′})=af(x_{′})=ax=0$. Since $f$ is injective, $ax_{′}=0$ and $x_{′}∈T(M_{′})$, that is, $Kergˉ ⊆Imfˉ $. |

4. | Let $S=A−{0}$, then $K=S_{−1}A$ and $K⊗_{A}M≅S_{−1}M$ by $(a/s)⊗m↦(am/s)$, hence $Ker(x↦1⊗x)=Ker(x↦(x/1))$. Then $(x/1)=0∈S_{−1}M⟺∃a=0∈A,ax=0⟺x∈T(M)$, so $T(M)=Ker(x↦1⊗x)$. $□$ |

Exercise

1. | $M$ is torsion-free. |

2. | $M_{p}$ is torsion-free for all prime ideal $p$. |

3. | $M_{m}$ is torsion-free for all maximal ideal $m$. |

**Proof.** Let $x∈TM$, then exists $a=0∈A$ such that $ax=0$, so for all $s∈S$, $a(x/s)=(ax)/s=0∈S_{−1}M$, and $S_{−1}(TM)⊆T(S_{−1}M)$. Conversely, if $(a/t)(x/s)=0$, then there exists $r∈S$ such that $rax=0$. Since $A$ is integral, $ra=0$, and $x∈TM$, so $T(S_{−1}M)⊆S_{−1}(TM)$.

$□$

Exercise

**Proof.** Consider $A/a$-module $M/aM$, then any maximal ideal in $A/a$ is in the form $m/a$ for some maximal ideal $m⊇a$. For any maximal ideal $m⊇a$, we have $0=M_{m}/(aM)_{m}≅(M/aM)_{m}≅A_{m}⊗_{A}M/aM$. For any $x∈a,s∈/m,mˉ∈M/aM$, we have $(x/s)⊗mˉ=(1/s)⊗(xmˉ)=0∈A_{m}⊗_{A}M/aM$, so it is isomorphic to $(A/a)_{m}⊗_{A/a}M/aM=0$.

$□$

Exercise

**Proof.**Let $e_{1},…,e_{n}$ be the canonical basis of $F$, and $x_{1},…,x_{n}$ be a set of generators of $F$. Define $ϕ:F→F$ by $ϕ(e_{i})=x_{i}$, then it is surjective, and we have to prove it’s a isomorphism. By Propsition 3.9, injective is a local property, so it’s sufficient to assume $A$ is a local ring. Let $N=Kerϕ$ and $k=A/m$ is the residue field of $A$. Since $0→N→Fϕ F→0$ is exact, tensoring with $k$ we have $k⊗N→k⊗F1⊗ϕ k⊗F→0$ exact. Since $k⊗F=k_{n}$ is a vector space, and $1⊗ϕ$ is a surjective automorphism over it, there must be $1⊗ϕ$ injective, and $k⊗N=Ker(1⊗ϕ)=0$, i.e. $(A/m)⊗N=N/(mN)=0,N=mN$. Also by Chapter 2, Exercise 12, $N$ is finitely generated; since $m$ is the unique maximal ideal, by Nakayama’s Lemma we have $N=0$.

$□$

**Proof.**If $x_{1},…,x_{m}$ generate $F$ for some $m<n$, extend it by random non-zero elements $y_{1},…,y_{n−m}$. By the results above, $x_{1},…,x_{m},y_{1},…,y_{m}$ is a basis of $F$; but $y_{1}=∑_{i=1}a_{i}x_{i}$ for some coefficient $a_{i}∈A$, contradiction.

$□$

Exercise

1. | $a_{ec}=a$ for all ideals $a$ of A. |

2. | $Spec(B)→Spec(A)$ is surjective. |

3. | For every maximal $m$ of $A$ we have $m_{e}=(1)$. |

4. | If $M$ is any non-zero $A$-module, then $M_{B}=0$. |

5. | For every $A$-module $M$, the mapping $x↦1⊗x$ of $M$ into $M_{B}$ is injective. |

**Proof.**

i) $⟹$ ii): By (3.16), $p$ is contraction if and only if $p_{ec}=p$.

ii) $⟹$ iii): Trivial.

iii) $⟹$ iv): Since $B$ is flat, $M_{′}⊆M$ then $M_{B}=0$ implies $M_{B}=0$. So it’s sufficient to consider only $M=Ax$ for some $x∈M$, and in this case we have $M=A/a$ for some ideal $a$ of $A$. Then $M_{B}=(A/a)⊗B=B/(a_{e})$, and $a_{e}⊆m_{e}=(1)$ for some maximal ideal $m$ of $A$. Hence $M_{B}=0$.

iv) $⟹$ v): Let $M_{′}=Ker(x↦1⊗x)$, then $0→M_{′}→M→M_{B}$ is exact. By flatness of $B$ we have $0→M_{B}→M_{B}→(M_{B})_{B}$ exact. By Exercise 2.13, $M_{B}→(M_{B})_{B}$ is injective, hence $M_{B}=0$, so $M_{′}=0$.

$□$

Exercise

**Proof.** If $ϕ:M→N$ is a injective $A$-module homomorphism, we want to prove $1⊗ϕ:B⊗_{A}M→B⊗_{A}N$ is injective. By Exercise 16 v), $B⊗_{A}M→C⊗_{B}(B⊗_{A}M)$ is injective. But $C⊗_{B}(B⊗_{A}M)=(C⊗_{B}B)⊗_{A}M=C⊗_{A}M$. By the following commutative diagram:

[TikZ 编译错误]

$□$

Exercise

**Proof.** Cause flatness is local we have $B_{p}$ is flat over $A_{p}$. Also $B_{q}$ is a localization of $B_{p}$, so flat over $B_{p}$. So $B_{q}$ is flat over $A_{p}$.

$□$

Exercise

1. | $M=0⟺Supp(M)=∅$. |

2. | $V(a)=Supp(A/a)$. |

3. | If $0→M_{′}→M→M_{′′}→0$ is an exact sequence, then $Supp(M)=Supp(M_{′})∪Supp(M_{′′})$. |

4. | If $M=∑M_{i}$, then $Supp(M)=⋃Supp(M_{i})$. |

5. | If $M$ is finitely generated, then $Supp(M)=V(Ann(M))$ (and is therefore a closed subset of $Spec(A)$. |

6. | If $M,N$ are finitely generated, then $Supp(M⊗_{A}N)=Supp(M)∩Supp(N)$. |

7. | If $M$ is finitely generated and $a$ is an ideal of $A$, then $Supp(M/aM)=V(a+Ann(M))$. |

8. | If $f:A→B$ is a ring homomorphism and $M$ is finitely generated $A$-module, then $Supp(B⊗_{A}M)=f_{∗−1}(Supp(M))$. |

**Proof.**

1. | $M=0⟺$ for all prime ideal $p$, $M_{p}=0⟺Supp(M)=∅$. |

2. | $(A/a)_{p}=A_{p}/a_{p}$, So $(A/a)_{p}=0⟺a_{p}=(1)⟺a⊆p$. |

3. | For every prime ideal $p$ we have exact sequence $0→M_{p}→M_{p}→M_{p}→0$, so $M_{p}=0⟺M_{p}=0orM_{p}=0$. |

4. | If $M=∑M_{i}$, then $M_{p}=∑M_{ip}$, so $M_{p}=0⟺$ at least one $M_{ip}=0$. |

5. | If $M=∑Ax_{i}$, then $Supp(M)=⋃Supp(Ax_{i})$, and each $Ax_{i}$ is isomorphic to some $A/a_{i}$. By ii), $Supp(Ax_{i})=V(a_{i})$, so $Supp(M)=⋃V(a_{i})=V(⋂a_{i})=V(Ann(M))$. |

6. | If $(M⊗_{A}N)_{p}=M_{p}⊗_{A_{p}}N_{p}=0$, then either $M_{p}$ or $N_{p}=0$. So $Supp(M⊗_{A}N)=Supp(M)∩Supp(N)$. |

7. | If $M$ is finitely generated then also is $M/aM$. So $Supp(M/aM)=V(Ann(M/aM))$. Now if $x∈Ann(M/aM)$, and $y_{1},…,y_{m}$ generate $M$, then $xy_{i}=∑_{j}a_{ij}y_{j}$ for some $a_{ij}∈a$. By linear algebra we know $det(xI−(a_{ij}))∈Ann(M)$, that is, $x_{m}∈a+Ann(M)$. so $x∈r(a+Ann(M))$ and hence $a+Ann(M)⊆Ann(M/aM)⊆r(a+Ann(M))$. By $V(r(b))=V(b)$ we know $V(Ann(M/aM))=V(a+Ann(M))$. |

8. | If $M=∑Ax_{i}$ then $B⊗_{A}M=∑B⊗_{A}Ax_{i}$. So $Supp(M)=⋃Supp(Ax_{i})$ and $Supp(B⊗_{A}M)=⋃Supp(B⊗_{A}Ax_{i})$. Cause $f_{∗−1}(U∪V)=f_{∗−1}(U)∪f_{∗−1}(V)$, it’s sufficient to show when $M=Ax≃A/a$ for some ideal $a$ of $A$. If $M=A/a$ then $B⊗_{A}M≃B/a_{e}$, and $Supp(B⊗_{A}M)=V(a_{e})=f_{∗−1}(V(a))=f_{∗−1}(Supp(M))$. $□$ |

Exercise

1. | Every prime ideal of $A$ is a contracted ideal $⟺f_{∗}$ is surjective. |

2. | Every prime ideal of $B$ is an extend ideal $⟹f_{∗}$ is injective. |

Is the converse of ii) true?

**Proof.**

1. | If $p$ is contracted ideal then $p_{ec}=p$. By (3.16) $p$ is contraction of some prime ideal, i.e. $f_{∗}$ is surjective. The converse is trivial. |

2. | Let $p,q$ be distinct prime ideal of $B$, then $p=p_{ce},q=q_{ce}$, so $p_{c}=q_{c}$. |

$□$

Exercise

1. | Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$, and $ϕ:A→S_{−1}A$ the canonical homomorphism. Show that $ϕ_{∗}:Spec(S_{−1}A)→Spec(A)$ is a homeomorphism of $Spec(S_{−1}A)$ onto its image in $X=Spec(A)$. Let this image be denote by $S_{−1}X$. In particular, if $f∈A$, the image of $Spec(A_{f})$ in $X$ is the basic open set $X_{f}$. |

2. | Let $f:A→B$ be a ring homomorphism. Let $X=Spec(A)$ and $Y=Spec(B)$, and let $f_{∗}:Y→X$ be the mapping associated with $f$. Identifying $Spec(S_{−1}A)$ with its canonical image $S_{−1}X$ in $X$, and $Spec(S_{−1}B)(=Spec(f(S)_{−1}B))$ with its canonical iamge $S_{−1}Y$ in $Y$, show that $S_{−1}f_{∗}:Spec(S_{−1}B)→Spec(S_{−1}A)$ is the restriction of $f_{∗}$ to $S_{−1}Y$, and that $S_{−1}Y=f_{∗−1}(S_{−1}X)$. |

3. | Let $a$ be an ideal of $A$ and let $b=a_{e}$ be its extension in $B$. Let $fˉ :A/a→B/b$ be the homomorphism induced by $f$. If $Spec(A/a)$ is identified with its canonical image $V(a)$ in $X$, and $Spec(B/b)$ with its image $V(b)$ in $Y$, show that $fˉ _{∗}$ is the restriction of $f_{∗}$ to $V(b)$. |

4. | Let $p$ be a prime ideal of $A$. Take $S=A−p$ in ii) and then reduce mod $S_{−1}p$ as in iii). Deduce that the subspace $f_{∗−1}(p)$ of $Y$ is naturally homeomorphic to $Spec(B_{p}/pB_{p})=Spec(k(p)⊗_{A}B)$, where $k(p)$ is the residue field of the local ring $A_{p}$. |

$Spec(k(p)⊗_{A}B)$ is called the fiber of $f_{∗}$ over $p$.

**Proof.**

1. | If $p$ in $A$ is the contraction of some prime ideal in $S_{−1}A$ then it must be $S_{−1}p$. So $ϕ_{∗}$ is injective. Consider $f=x/s∈S_{−1}A$ and $q$ a prime ideal of $S_{−1}A$, then $f∈q⟺x∈q_{c}$. So $ϕ_{∗−1}$ is also continuous, hence $ϕ_{∗}$ is a homomorphism between $Spec(S_{−1}A)$ and it’s image in $X$. If $f∈A$, then a prime ideal $p$ of $A$ is the contraction of some prime ideal of $A_{f}⟺f∈p$, i.e. $ϕ_{∗}(Spec(A_{f}))=X_{f}$. |

2. | To show $S_{−1}f_{∗}$ is a restriction of $f_{∗}$, it’s equivalent to say that the right diagram below commutes. Cause the left diagram commutes, so is the induced maps. [TikZ 编译错误] If $f_{∗}(p)∈S_{−1}X$, i.e. $p_{c}$ is disjoint from $S$, then $p$ is disjoint from $f(S)$, so $p∈S_{−1}Y$. Hence $f_{∗−1}(S_{−1}X)=S_{−1}Y$. |

3. | Like ii), the map $A→A/afˉ B/b$ is the same as $Af B→B/b$, so the induced maps are the same too, i.e. $fˉ _{∗}$ is the restriction of $f_{∗}$. Also we have $f_{∗−1}(V(a))=V(b)$. |

4. | ${p}$ is identified with $V(p)$ in $A_{p}$ (since it’s maximal in $A_{p}$), and $f_{∗−1}(Spec(A_{p}))$ is identified with $Spec(B_{p})$, hence $f_{∗−1}(p)=V(pB_{p})$ in $Spec(B_{p})$, so it is $Spec(B_{p}/pB_{p})$. And $B_{p}/pB_{p}=A_{p}/p⊗_{A_{p}}B_{p}=(k(p)⊗_{A}B)_{p}$. But image of $p$ in $k(p)⊗_{A}B$ is zero, so $B_{p}/pB_{p}=k(p)⊗_{A}B$. $□$ |

Exercise

**Proof.**The intersection of all the open neighborhoods of $p$ is $⋃_{f∈/p}X_{f}$, i.e. all prime ideals contained in $p$. This is equivalent to be a contraction ideal of $A_{p}$.

$□$

Exercise

1. | If $U=X_{f}$, show that the ring $A(U)=A_{f}$ depends only on $U$ and not on $f$. |

2. | Let $U_{′}=X_{g}$ be another basic open set such that $U_{′}⊆U$. Show that there is an equation of the form $g_{n}=uf$ for some integer $n>0$ and some $u∈A$, and use this to define a homomorphism $ρ:A(U)→A(U_{′})$(i.e. $A_{f}→A_{g}$) by mapping $a/f_{m}$ to $au_{m}/g_{mn}$. Show that $ρ$ depends only on $U$ and $U_{′}$. This homomorphism is called the restriction homomorphism. |

3. | If $U=U_{′}$, then $ρ$ is the identity map. |

4. | If $U⊇U_{′}⊇U_{′′}$ are basic open sets in $X$, show that the diagram [TikZ 编译错误] (in which the arrowsare restriction homomorphism) is commutative. |

5. | Let $x=p$ be a point of $X$. Show that $x∈Ulim A(U)≃A_{p}$ |

The assignment of the ring $A(U)$ to each basic open set $U$ of $X$, and the restriction homomorphism $ρ$, satisfying the condition iii) and iv) above, constitutes a presheaf of rings on the basis of open sets $(X_{f})_{f∈A}$. v) says that the stalk of this presheaf at $x∈X$ is the correspending local ring $A_{p}$.

**Proof.** If $X_{g}⊆X_{f}$ then $r((g))⊆r((f))$. So there exists $n,u$ such that $g_{n}=uf$. So we can define a map $ρ_{fg}:A_{f}→A_{g}$ by $a/g_{t}↦au_{t}/g_{nt}$. If there are another pair of $n_{′},u_{′}$ s.t. $g_{n_{′}}=u_{′}f$, then $au_{t}/g_{nt}=au_{′t}/g_{n_{′}t}$ because $au_{t}g_{n_{′}t}=au_{t}u_{′t}f_{t}=au_{′t}g_{nt}$. So $ρ_{fg}$ only depends on $f$ and $g$.

If there are $X_{h}⊆X_{g}⊆X_{f}$, and $g_{n}=uf,h_{m}=vg$, then $h_{nm}=uv_{n}f$. So $(ρ_{gh}∘ρ_{fg})(a/f_{t})=ρ_{gh}(au_{t}/g_{nt})=au_{t}v_{nt}/h_{nmt}=ρ_{fh}(a/f_{t})$. So $ρ_{fh}=ρ_{gh}∘ρ_{fg}$.

1. | If $U=X_{f}=X_{g}$, then $ρ_{fg}ρ_{gf}=ρgg=id_{X_{g}}$ And $ρ_{gf}ρ_{fg}=ρ_{ff}=id_{X_{f}}$. So we can identity $X_{f}$ with $X_{g}$. |

2. | If $X_{g}⊂X_{f}$ then $ρ_{fg}$ is a map from $A(U)=A_{f}$ to $A(U_{′})=A_{g}$. Also if $U=X_{f_{′}},U_{′}=X_{g_{′}}$, then $ρ_{f_{′}g_{′}}=ρ_{gg_{′}}∘ρ_{fg}∘ρ_{f_{′}f}$. By the definition in i), $ρ(gg_{′})$ and $ρ(f_{′}f)$ are identity map, so this map only depends on $U,U_{′}$. |

3. | Trivial. |

4. | Trivial if we select $U=X_{f},U_{′}=X_{g},U_{′′}=X_{h}$. |

5. | Prove by the universal property of direct limit and fractions. If there is another ring $B$ and a family of maps $A_{f}→B$ whenever $f∈/p$, and these maps are conpatible; that is, a map $ϕ:A→B$ with all $ϕ(f)$ invertible for $f∈/p$, then it induces a unique map on $lim _{x∈U}A(U)$, also induces a unique map on $A_{p}$. Hence they have the same universal property, so must be isomorphic. $□$ |

Exercise

**Proof.** First, we can choose a finitely covering of $(U_{i})_{i∈I}$, so without loss of generality (cause $s$ is unique) we assume that $I={1,…,m}$.

By replace $f_{i}$ by $f_{i}$ (Cause $A_{f}≃A_{f_{n}}$), we can assume $s_{i}=a_{i}/f_{i}$. If the image $s_{i}$ and $s_{j}$ are same in $A_{f_{i}f_{j}}$, i.e. $(a_{i}f_{j})/(f_{i}f_{j})=(a_{j}f_{i})/(f_{i}f_{j})∈A_{f_{i}f_{j}}$, then there exists $m$ such that $(f_{i}f_{j})_{m}a_{i}f_{j}=(f_{i}f_{j})_{m}a_{j}f_{i}$. Choose a large $m$ such that this equation is satisfied for every pair of $i,j$, and rewrite it as

$f_{j}f_{i}a_{i}=f_{i}f_{j}a_{j}$

Since $f_{i}$ generate $A$, so do $f_{i}$. So we have $1=∑_{i}b_{i}f_{i}$ for some $b_{i}∈A$. Let $a=∑b_{i}f_{i}a_{i}$, then

$f_{j}a=i∑$