Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 7.

## Noetherian Rings

Exercise

**Proof.** Since $Σ$ is nonempty and satisfies conditions of Zorn’s Lemma, it must have maximal elements.

Let $a$ be one of the maximal elements of $Σ$. Suppose $x∈/a$, then $a+(x)$ is finitely generated. If it’s generated by $a_{1}+u_{1},…,a_{m}+u_{m}$ where $a_{k}∈a,u_{k}∈(x)$, let $a_{0}=(a_{1},…,a_{m})⊂a$ be the finitely generated ideal contained in $a$, then $a+(x)=a_{0}+(x)$. Particularly, $a=(a+(x))∩a=a_{0}+(x)∩a=a_{0}+x⋅(a:x).$Since $a$ is not finite generated, neither is $(a:x)$. But $(a:x)⊇a$, so $(a:x)=a$ by the maximality of $a$ in $Σ$.

$□$

Hence a ring in which every prime ideal is finitely generated is Noetherian (I. S. Cohen).

Exercise

**Proof.** By Chapter 1, Exercise 5, $f$ is nilpotent implies that each $a_{n}$ is nilpotent.

$□$

Exercise

1. | $a$ is primary; |

2. | for every multiplicatively closed subset $S$ of $A$ we have $(S_{−1}a)_{c}=(a:x)$ for some $x∈S$. |

3. | the sequence $(a:x_{n})$ is stationary, for every $x∈A$. |

**Proof.**

i) $⟹$ ii): | Suppose $a$ is $p$-primary. If $S$ meets $p$, then $S_{−1}a=S_{−1}A$ and contracts to $(1)$. In this case, $S∩a=∅$. Select $x∈S∩a$, then $(S_{−1}a)_{c}=(1)=(a:x)$. If $S$ is disjoint from $p$, then $(S_{−1}a)_{c}=a$. Choose $x∈S$ arbitrary, then $x∈/p$, hence $(S_{−1}a)_{c}=a=(a:x)$. |

ii) $⟹$ iii): | Let $S={x_{n}:n≥0}$, then $(S_{−1}a)_{c}=(a:x_{n})$ for some $n≥0$. Clearly $(a:x_{t})⊆(S_{−1}a)_{c}$ for every $t≥0$, so $(a:x_{n})=(a:x_{n+1})=…$. |

iii) $⟹$ i): | (Similar to (7.12)) Suppose $xy∈a$ and $y∈/a$. If $(a:x_{n})=(a:x_{n+1})=…$, then $(x_{n})∩(y)⊆a$, for if $a∈(y)$, then $ax∈a$; and if $a=bx_{n}$ then $bx_{n+1}∈a$, hence $b∈(a:x_{n+1})=(a:x_{n})$, so $a∈a$. Since $a$ is irreducible and $(y)⊆a$, we must have $(x_{n})⊆a$ (Consider $a=((x_{n})+a)∩((y)+a)$). Hence $a$ is primary. $□$ |

Exercise

1. | The ring of rational functions of $z$ having no pole on the circle $∣z∣=1$. |

2. | The ring of power series in $z$ with a positive radius of convergence. |

3. | The ring of power series in $z$ with an infinite radius of convergence. |

4. | The ring of polynomials in $z$ whose first $k$ derivatives vanish at origin ($k$ being a fixed integer). |

5. | The ring of polynomials in $z,w$ all of whose partial derivatives with respect to $w$ vanish for $z=0$. |

In all cases the coefficients are complex numbers.

**Proof.** We denote the ring described as $A$.

1. | Let $S⊂C[z]$ be the set of all polynomials, which having no zeros on the circle $∣z∣=1$. Then $A=S_{−1}C[z]$, hence is Noetherian (cause $C[z]$ is Noetherian). |

2. | If $f=∑_{n=0}a_{n}z_{n}∈A$, and $a_{0}=0$, then there exists $r>0$ such that $f(z)=0$ for every $∣z∣<r$. Hence $f_{−1}∈C[[x]]$ has a positive radius of convergence (at least $r$). Therefore the only ideals of $A$ is $(z_{n}),n≥0$ ($A$ is a valuation ring). Hence $A$ is Noetherian. |

3. | Let $a_{k}={f∈A:f(k)=f(k+1)=⋯=0}$. Clearly $a_{k}$ are ideals of $A$, $a_{k}⊂a_{k+1}$, and $x−ksin(πx) ∈a_{k+1}−a_{k}.$Hence the increasing sequence $a_{0}⊂a_{1}⊂…$ is strict, so $A$ is not Noetherian. |

4. | Every elements of $A$ is in the form $a+z_{k+1}f$, where $a∈C$ and $f∈C[x]$. For every ideal $a$ of $A$, assume $a∩C=0$. Let $f∈a$ be a monic polynomial of the minimal degree. For any $g∈a$, if $g=g_{0}z_{t}+…$ and $t=gg>gf+k$, then $g−g_{0}z_{t−degf}f∈a$. Proceeding in this way, we can get $g_{′}∈a$ such that $gg_{′}≤gf+k$ and $g−g_{′}∈(f)$. For every $i=1,…,k$, choose a polynomial $f_{i}$ in $a$ with degree $gf+i$ (if it exists). Then clearly $g∈(f,f_{1},…,f_{k})$. Hence $a=(f,f_{1},…,f_{k})$ is finitely generated, and $A$ is Noetherian. |

5. | One can show that $zw_{k}∈/(zw,…,zw_{k−1})$ in $A$. Hence $(zw,zw_{2},…)⊂A$is not finitely generated, and $A$ is not Noetherian. $□$ |

Exercise

**Proof.**We know that (Chapter 5, Exercise 12) $B$ is integral over $B_{G}$. Since $A⊆B_{G}⊆B$, $A$ is Noetherian and $B$ is finite generated over $A$, it follows that $B_{G}$ is finite generated over $A$.

$□$

Exercise

**Proof.** If $charK=0$, then $Z⊂Q⊆K$.

Since $K$ is finitely generated over $Z$, it’s finitely generated over $Q$. Hence by (7.9)^{6}

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Exercise

**Proof.**Let $a⊂k[t_{1},…,t_{n}]$ be the ideal generated by $f_{α}(α∈I)$. Let $Σ$ be the set of ideals generated by a finite subset of ${f_{α}}_{α∈I}$. Since $k[t_{1},…,t_{n}]$ is Noetherian, $Σ$ has a maximal element, and clearly it must be $a$. Hence $a$ can be generated by $f_{α}(α∈I_{0})$ for a finite subset $I_{0}⊆I$. Thus $X$ is given by the equations $f_{α}(t_{1},…,t_{n})=0$ for $α∈I_{0}$.

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Exercise

**Proof.**Yes. $A=A[x]/(x)$ is a homomorphic image of $A[x]$.

$□$

Exercise

1. | for each maximal ideal $m$ of $A$, the local ring $A_{m}$ is Noetherian; |

2. | for each $x=0$ in $A$, the set of maximal ideals which contain $x$ is finite. |

Show that $A$ is Noetherian.

**Proof.** Let $a=0$ be a ideal of $A$, $m_{1},…,m_{r}$ be the maximal ideals which contain $a$. Choose $x_{0}=0$ in $a$ arbitrary and let $m_{1},…,m_{r+s}$ be the maximal ideals which contain $x_{0}$. Since $m_{r+1},…,m_{r+s}$ do not contain $a$, there exists $x_{j}∈a$ such that $x_{j}∈/m_{r+j}(1≤j≤s)$. Since each $A_{m_{i}}(1≤i≤r)$ is Noetherian, the extension of $a$ in $A_{m_{i}}$ is finitely generated. Hence there exist $x_{s+1},…,x_{t}$ in $a$ whose images in $A_{m_{i}}$ generate $A_{m_{i}}a$ for $i=1,…,r$. Let $a_{0}=(x_{0},…,x_{t})⊆a$. Then for every maximal ideal $m$ of $A$:

1. | If $m=m_{i}$ for some $1≤i≤r$: the images of $x_{s+1},…,x_{t}$ generate $A_{m}a$, hence $A_{m}a=A_{m}a_{0}$. |

2. | If $m=m_{r+i}$ for some $1≤i≤s$: then $x_{i}∈/m$, hence $a,a_{0}⊆m$; and $A_{m}a=A_{m}a_{0}=A_{m}$. |

3. | Otherwise, $x_{0}∈/m$. so $A_{m}a=A_{m}a_{0}=A_{m}$. |

$□$

Exercise

**Proof.** Suppose $N⊆M[x]$ is a $A[x]$-submodule. Define $N_{0}⊆M$ to be the set of all leading coefficients of polynomials in $N$; then $N_{0}$ is a submodule of $M$, so finitely generated, say by $v_{1},…,v_{m}$. For each $1≤k≤m$, choose $f_{k}∈N$ with leading coefficient $v_{k}$. Let $r_{k}$ denote the degree of $f_{k}$, and $r=max_{k=1}r_{k}$. Let $N_{′}$ be the $A[x]$-submodule generated by $f_{1},…,f_{m}$.

For any $f∈N$, if $f=ux_{t}+…$ with $m≥r$, write $u=∑_{k=1}p_{i}u_{i}$, then $f_{′}=f−∑_{k=1}p_{k}x_{t−r_{k}}f_{k}$ satisfies $f_{′}∈N$ and $gf_{′}<gf$. Proceeding in this way we get $g∈N,h∈N_{′}$ with $f=g+h$ and $gg<r$.

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Exercise

**Proof.** No (in general). If $A$ is a Boolean ring, then $A_{p}=F_{2}$ (the following proof), but $A$ need not to be Noetherian.

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Exercise

**Proof.** If $B$ is faithfully flat, then for every ideal $a⊂A$, one has $a_{ec}=a$.

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Exercise

**Proof.**For each prime ideal $p$ of $A$, $(f_{∗})_{−1}(p)≃Spec(k_{p}⊗_{A}B)$, where $k_{p}$ is the residue field of $A_{p}$. Since $k_{p}⊗_{A}B$ is finitely generated $k$-algebra, it is Noetherian. Therefore $Spec(k_{p}⊗_{A}B)$ is Noetherian, so is $(f_{∗})_{−1}(p)$.

$□$

### Nullstellensatz, strong form

Exercise

**Proof.** If $f_{n}∈a$, then $f(x)_{n}=0$ for all $x∈V$, so $f(x)=0$. Hence $r(a)⊆I(V)$.

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Exercise

1. | $M$ is free. |

2. | $M$ is flat. |

3. | the mapping of $m⊗M$ into $A⊗M$ is injective. |

4. | $Tor_{1}(k,M)=0$. |

**Proof.**

i) $⟹$ ii): | Clearly $M=A_{n}$ is flat. |

ii) $⟹$ iii): | The map $m↪A$ is injective, tensor with $M$. |

iii) $⟹$ iv): | We have $0→m⊗M→A⊗M→k⊗M→0$exact, so by definition $Tor_{1}(k,M)=0$. |

iv) $⟹$ i): | Let $x_{1},…,x_{n}$ be elements in $M$ whose images in $M/mM$ (as a $k$-vector space) forms a $k$-basis. Then $x_{1},…,x_{n}$ generate $M$ (If they generate $N$, then $N+mM=M$). Let $F=A_{n}$ with the canonical basis $e_{1},…,e_{n}$, and define $ϕ:F→M$ by $ϕ(e_{k})=x_{k}$. Then $ϕ$ is surjective. Let $E=kerϕ$, tensor the exact sequence $0→E→F→M→0$ with $k$, notice that $Tor_{1}(k,M)=0$, then we get $0→k⊗_{A}E→k⊗_{A}F1⊗ϕ k⊗_{A}M→0$Since $k⊗_{A}F$ and $k⊗_{A}M$ are vector spaces of the same dimension, it follows that $1⊗ϕ$ is an isomorphism, hence $k⊗_{A}E=0$. Since $E$ is a submodule of $F$, and $A$ is Noetherian, $E$ is finitely generated. By Nakayama’s Lemma $E=0$ ($k⊗_{A}E=0⟹mE=0$), so $ϕ$ is an isomorphism and $M≃A_{n}$. $□$ |

Exercise

1. | $M$ is a flat $A$-module; |

2. | $M_{p}$ is a free $A_{p}$-module, for all prime ideals $p$; |

3. | $M_{m}$ is a free $A_{p}$-module, for all maximal ideals $m$. |

In other words, flat = locally free.

**Proof.**Since each $M_{p}$ is finitely generated over $A_{p}$, by the last Exercise, ii) or iii) is equivalent to that $M_{p}$ (respectively, $M_{m}$) is flat. Since flatness is a local property, the result follows immediately.

$□$

Exercise

**Proof.** Definte a submodule $N⊆M$ to be irreducible, if for any submodules $E,F⊆M$, $N=E∩F$ implies $N=E$ or $N=F$.

Similar to (7.11) and (7.12), we have

引理

**Proof.**Suppose not, then there is a maximal submodule $N$ of $M$ respecting to this property. Since $N$ is reducible, there is $N=E∩F$ with $N⊂E,F$. Hence $E$ and $F$ is a finite intersection of irreducible submodules, so is $N$. Contradiction.

$□$

引理

**Proof.**Suppose $N$ is an irreducible submodule of $M$. Passing to the quotient module $M/N$, we may assume $N=0$. Let $x∈A,v=0∈M$ such that $xv=0$, consider the submodules $(0:x)⊆(0:x_{2})⊆…$. Since $M$ is Noetherian, the chain is stationary, hence exists $n≥1$ such that $(0:x_{n})=(0:x_{n+1})=…$. It follows that $x_{n}M∩Av=0$, for if $x_{n}u=yv$, then $x_{n+1}u=xyv=0$, so $u∈(0:x_{n+1})=(0:x_{n})$ and $x_{n}u=0$. Since $0$ is irreducible and $Av=0$, there must be $x_{n}M=0$, which implies that $x∈r_{M}(0)$.

$□$

$□$

Exercise

1. | $p$ belongs to $0$ in $M$; |

2. | there exists $x∈M$ such that $Ann(x)=p$; |

3. | there exists a submodule of $M$ isomorphic to $A/p$. |

**Proof.** Suppose $0=⋂_{i=1}Q_{i}$ is a minimal primary decomposition of $0$ in $M$, where $Q_{i}$ is $p_{i}$-primary.

ii) $⟹$ iii): | If $Ann(x)=p$, then $Ax≅A/p$. |

iii) $⟹$ ii): | If $M_{′}⊆M$ and $M_{′}≅A/p$, then $M_{′}$ is finitely generated since $A$ is Noetherian. Let $x_{1},…,x_{n}$ generate $M_{′}$, then $p=Ann(M_{′})=⋂_{i=1}Ann(x_{i})$. Hence $p=Ann(x_{i})$ for some $i$. |

ii) $⟹$ i): | $p=Ann(x)=r(0:x)$ belongs to $0$. |

i) $⟹$ ii): | Assume $p=p_{1}$. Choose $v∈/Q_{1}$ but $v∈⋂_{i=2}Q_{i}$, then $r(Ann(v))=p$. If $a_{1},…,a_{n}$ generate $p$ (since $A$ is Noetherian), then $a_{i}∈r(Ann(v))$, hence $a_{i}v=0$ for some integer $k_{i}$. Hence $p_{k_{1}+⋯+k_{n}}v=0$. Let $n$ be the minimal integer such that $p_{n}v=0$ but $p_{n+1}v=0$, and let $x=0∈p_{n}v$. Since $x∈⋂_{i=2}Q_{i}$, it can not belong to $Q_{1}$. Then $px=0$, and if $ax=0∈Q_{1}$, then $a∈r_{M}(Q_{1})=p$. Hence $p=Ann(x)$. $□$ |

Deduce that there exists a chain of submodules $0=M_{0}⊂M_{1}⊂⋯⊂M_{r}=M$such that each quotient $M_{i}/M_{i−1}$ is of the form $A/p_{i}$, where $p_{i}$ is a prime ideal of $A$.

**Proof.** Suppose not. Let $Σ$ be the set of submodules $M_{′}⊆M$, so that there is no such chain with $M_{0}=M_{′}$ and $M_{r}=M$, then $0∈Σ$. Since $M$ is finitely generated over the Noetherian ring $A$, it is a Noetherian module. So $Σ$ has a maximal element, say $N$.

$□$

Exercise *irreducible* ideals. Prove that $r=s$ and that (possible after re-indexing the $c_{j}$) $r(b_{i})=r(c_{i})$ for all $i$.

State and prove an analogous result for modules.

**Proof.** We can regard $A$ as $A$-module; then ideals correspond to submodules, irreducible ideals correspond to irreducible submodules. So it’s enough to prove the statement for modules:

If $A$ is a ring and $M$ is a $A$-module, and $N$ is a submodule of $M$. Let $N=i=1⋂r P_{i}=j=1⋂s Q_{j}$be two minimal decompositions of $N$ as intersections of irreducible submodules of $M$; then $r=s$. If $M$ is Noetherian, then $r_{M}(P_{i})=r_{M}(Q_{i})$ for all $i$ (maybe after re-indexing the $Q_{i}$’s).

Clearly $P$ is a irreducible submodule containing $N$ if and only if $P/N$ is a irreducible submodule of $M/N$. So pass to the quotient module, we may assume $N=0$.

Let $N_{i}$ denote $⋂_{j=i}P_{j}$. First we prove the hint in the book: for each $i=1,…,r$, there exists $j$ such that $0=N_{i}∩Q_{j}$.

Let $1≤i≤r$ fixed. We have $⋂_{j=1}(N_{i}∩Q_{j})=0$ (since $⋂Q_{j}=0$). Therefore $j=1⋂s ((N_{i}∩Q_{j})+P_{i})=P_{i},$for if $(U+V)∩W=0$, then $(U+W)∩(V+W)=(U∩V)+W$. Since $P_{i}$ is irreducible, there exists $j$ so that $(N_{i}∩Q_{j})+P_{i}=P_{i}$, so $N_{i}∩Q_{j}⊆P_{i}$. Thus $N_{i}∩Q_{j}=0$ for $N_{i}∩P_{i}=0$.

Now we have $0=Q_{j_{1}}∩P_{2}∩⋯∩P_{r}$ for some $j_{1}$. This is also a irreducible decomposition of $0$, hence apply the result again we have $0=Q_{j_{1}}∩Q_{j_{2}}∩⋯∩P_{r}$ for some $j_{1},j_{2}$. Proceeding in this way, then there exists $j_{1},…,j_{r}$ such that $0=Q_{j_{1}}∩Q_{j_{2}}∩⋯∩Q_{j_{r}}$Since the decomposition $0=⋂_{j=1}Q_{j}$ is minimal, ${j_{1},…,j_{r}}={1,…,s}$. Hence $r≥s$. For the same reason $s≥r$, hence $s=r$.

If in addition $M$ is Noetherian, then each $P_{i}$ and $Q_{j}$ is primary. Intersecting all submodules with the same radical, then we get two minimal primary decompositions of $N$. By the uniqueness theorem of primary decompositions, the set of the radical ideals is uniquely determined by $N$. Let ${p_{1},…,p_{l}}$ denote this set (i.e. the set of prime ideals of $N$), then we must show that, for each $i$, the number of occurrences of $p_{i}$ (in the radical of irreducible submodules) is uniquely determined.

$□$

Exercise

1. | Show that a subset $E$ of $X$ belongs to $F$ if and only if $E$ is a finite union of sets of the form $U∩C$, where $U$ is open and $C$ is closed. |

2. | Suppose that $X$ is irreducible and let $E∈F$. Show that $E$ is dense in $X$ (i.e. that $E=X$) if and only if $E$ contains a non-empty open set in $X$. |

**Proof.**

1. | Let $F_{′}$ be the set of subsets $E⊆X$ which is a finite union of sets of the form $U∩C$. Then clearly $F_{′}⊆F$; Conversely, if $E=⋃_{i=1}(U_{i}∩C_{i}),F=⋃_{j=1}(U_{j}∩C_{j})$, then $E∩F=i=1⋃n j=1⋃m ((U_{i}∩U_{j})∩(C_{i}∩C_{j}))$Hence $E∩F∈F_{′}$. Also the complement of $U_{i}∩C_{i}$ is $U_{i}∪C_{i}∈F_{′}$, hence $E_{c}=i=1⋂n (U_{i}∩C_{i})_{c}∈F_{′}$Hence $F_{′}$ contains all open subsets and is closed under intersection and complement. Therefore $F⊆F_{′}$, so $F=F_{′}$. |

2. | If $E∈F$, then by i), $E=⋃_{i=1}(U_{i}∩C_{i})$ for some open subsets $U_{i}$ and closed subsets $C_{i}$. If $E=X$, then $X=E=i=1⋃n (U_{i} ∩C_{i}).$Since $X$ is irreducible, there exists $i$ that $X=U_{i} ∩C_{i}$, so $U_{i}=∅$ and $C_{i}=X$. Hence $U_{i}⊆E$, and $E$ contanins a non-empty open set in $X$. Conversely, if there is a non-empty open set $U_{i}⊆X$, then $U_{i}∪U_{i} =X$. Hence $U_{i} =X$. If $E⊇U_{i}$, then $E=X$. $□$ |

Exercise

**Proof.** $⟹$: If $E∈F$, then for each $X_{0}$, let $F_{0}$ be the collection of subsets in $X_{0}$, then clearly $E∩X_{0}∈F_{0}$. Hence if $E∩X_{0} =X_{0}$, by Exercise 20, $E∩X_{0}$ contains a non-empty open subset of $X_{0}$.

$⟸$: Assume $E∈/F$. Let $Σ$ be the collection of closed subsets $C⊆X$ such that $C∩E∈/F$. Then $X∈Σ$, $Σ=∅$. Since $X$ is Noetherian, $Σ$ has a minimal element, say $X_{0}$. If $C_{1},C_{2}⊂X_{0}$, then $C_{1}∩E,C_{2}∩E∈F$, hence $(C_{1}∪C_{2})∩E∈F$, so in particular $C_{1}∪C_{2}=X_{0}$, show that $X_{0}$ is irreducible.

Let $F=E∩X_{0} =E∩X_{0}$. If $F=X_{0}$, then $E∩F∈F$. But $E∩F=E∩E∩X_{0}=E∩X_{0}∈/F$, contradiction. Hence $E∩X_{0} =X_{0}$.

If $U$ is a non-empty open subset of $X_{0}$ such that $U⊆E∩X_{0}$, then $U=X_{0}$, or else $E∩X_{0}=X_{0}∈F$. Let $C=X_{0}−U⊂X_{0}$, then $C$ is closed in $X_{0}$, hence closed in $X$. By the minimality of $X_{0}$, we have $E∩C∈F$. There exists open subset $V$ such that $U=V∩X_{0}$, so $E∩U=U=V∩X_{0}∈F$. Hence $E∩X_{0}=(E∩U)∪(E∩C)∈F$, contradiction.

$□$

Exercise

**Proof.** $⟹$: If $E∩X_{0}=∅$, then it is a non-empty open subset of $X_{0}$.

$⟸$: Suppose $E$ is not open. Let $Σ$ be the collection of closed subsets $C⊆X$ such that $C∩E$ is not open in $C$. Since $X∈Σ$, it is not empty.

Let $X_{0}$ be a minimal element of $Σ$. If $C_{1},C_{2}⊂X_{0}$ closed, then there exists open subsets $U_{1},U_{2}$ of $X_{0}$ such that $U_{1}∩C_{1}=E∩C_{1},U_{2}∩C_{2}=E∩C_{2}$. Hence $((U_{1}−C_{2})∪(U_{2}−C_{1}))∩(C_{1}∪C_{2})=E∩(C_{1}∪C_{2})$. Therefore $C_{1}∪C_{2}=X_{0}$, or else $E∩X_{0}=(U_{1}−C_{2})∪(U_{2}−C_{1})$ is open in $X_{0}$. Thus $X_{0}$ is irreducible.

$□$

Exercise

**Proof.** Since constructible sets are closed under unions, by Exercise 20 it’s enough to show $f_{∗}(U∩C)$ is constructible, for $U$ open and $C$ closed. If $C=V(a)=Spec(B/a)$, then replace $B$ with $B/a$, it reduce to the case where $E=U$ is open in $Y$.

Since $Y$ is Noetherian, $E$ is quasi-compact, and therefore a finite union of the basic open sets $X_{g}≅Spec(B_{g})$. Hence it’s sufficient to show each $f_{∗}(Spec(B_{g}))$ is constructible. Replacing $B$ with $B_{g}$ then it reduce to the case where $E=Y$.

By Exercise 21, to show $f_{∗}(Y)$ is constructible, is equivalent to show that if $X_{0}$ is an irreducible subset of $X$ where $f_{∗}(Y)∩X_{0}$ is dense in $X_{0}$, then $f_{∗}(Y)∩X_{0}$ contains some non-empty open subset of $X_{0}$. Suppose $X_{0}=Spec(A/p)$ for some prime ideal $p$ of $A$, then $f_{∗}(Y)∩X_{0}=f_{∗}(f_{∗−1}(X_{0}))$, and $f_{∗−1}(X_{0})=Spec((A/p)⊗_{A}B)$. Again replacing $A$ with $A/p$, $B$ with $(A/p)⊗_{A}B$. Then we should show that if $A$ is a Noetherian integral domain, $B$ an $A$-algebra of finite type and $f:A→B$ injective, then $f_{∗}(Y)$ contains some open subset of $X$.

Since $B$ is Noetherian, $Y=⋃_{i=1}Y_{j}$ for irreducible closed subsets $Y_{j}$ (irreducible complements of $Y$). We have $X=f_{∗}(Y) =⋃_{i=1}Y_{j} $. Since $X$ is irreducible, there is some $Y_{j}$ dense in $X$. Suppose $Y_{j}=V(q)=Spec(B/q)$, and let $g$ denote the composite mapping $A→B→B/q$. Since $f_{∗}(Y_{j})=g_{∗}(Spec(B/q))$ is dense in $X$, it follows that $kerg⊆R(A)=0$, hence $g$ is also injective. Replacing $(B,f)$ with $(B/q,g)$, we may assume $A⊆B$ are Noetherian integral domains, the inclusion map $A↪B$ is of finite type. We have to show that $f_{∗}(Y)$ contains some non-empty open subset of $X$; and it is equivalent to that $f_{∗}(Y)$ contains some basic open subset $X_{s}$ of $X$. That is, there exists $s∈A$ such that all prime ideals not containing $s$ are contracted ideals.

$□$

Exercise

**Proof.** By Chapter 5, Exercise 10, $f_{∗}$ is an open mapping implies $f$ has the going-down property.

Conversely, suppose $f$ has the going-down property. To show that $f_{∗}$ is an open mapping, it’s enough to show that $f_{∗}(Y_{s})=f_{∗}(Spec(B_{s}))$ is open for every $s∈B$. Clearly $B→B_{s}$ has the going-down property, hence so does the composite mapping $A→B_{s}$. Repalcing $B$ with $B_{s}$ then it’s sufficient to show that $f_{∗}(Y)$ is open in $X$.

$□$

Exercise *flat* (i.e. $B$ is flat as an $A$-module). Then $f_{∗}:Spec(B)→Spec(A)$ is an open mapping.

**Proof.**By Chapter 5, Exercise 11, $f$ is flat implies $f$ has the going-down property. By Exercise 24, $f$ is an open mapping.

$□$

### Grothendieck groups

Exercise *Grothendieck group* of $A$, and is denoted by $K(A)$. If $M$ is a finitely generated $A$-modyle, let $γ(M)$, or $γ_{A}(M)$, denote the image of $(M)$ in $K(A)$.

1. | Show that $K(A)$ has the following universal property: for each additive function $λ$ on the class of finitely generated $A$-modules, with values in an abelian group $G$, there exists a unique homomorphism $λ_{0}:K(A)→G$ such that $λ(M)=λ_{0}(γ(M))$ for all $M$. Proof. The map $λ:F(A)→G$ induces an homomorphism $λˉ:C→G$ such that $λ(M)=λˉ((M))$. The hypothesis that $λ$ is additive is equivalent to that $kerλˉ⊇D$. Hence $λˉ$ induces a homomorphism $λ_{0}:K(A)=C/D→G$, such that $λ_{0}(γ(M))=λ(M)$ for all $M$. Since $K(A)$ is generated by all $γ(M)$, $λ_{0}$ is unique. $□$ |

2. | Show that $K(A)$ is generated by the elements $γ(A/p)$, where $p$ is a prime ideal of $A$.
Since $K(A)$ is generated by all $γ(M)$, it follows that $K(A)$ is generated by the elements $γ(A/p)$. $□$ |

3. | If $A$ is a field, or MORE generally if $A$ is a principal ideal domain, then $K(A)≅Z$.
Let $L$ denote the field of fractions of $A$, then for each finitely generated $A$-module $M$, $L⊗_{A}M$ is a finite dimensional vector space over $L$. Let $r(M)$ denote its dimension. Since $L$ is flat, each short exact sequence $0→N→M→P→0$ tensor with $L$, gives a short exact sequence $0→L_{r(N)}→L_{r(M)}→L_{r(P)}→0$. Hence $r(M)=r(N)+r(P)$, so $r$ is a additive function. By i) $r$ induces a homomorphism $r_{0}:K(A)→Z$. Since $r(A_{n})=n$, it follows that $r_{0}$ is surjective. But $K(A)$ is generated by a single element $γ(A)$, hence $K(A)≅Z$. $□$ |

4. | Let $f:A→B$ be a
Clearly a $B$-module short exact sequence is also a $A$-module short exact sequence under the restriction of scalars. Hence $γ_{A}∘∣_{f}$ is additive for $B$-modules. By i), $γ_{A}∘∣_{f}$ induces a homomorphism $f_{!}:K(B)→K(A)$, such that $f_{!}(γ_{B}(N))=γ_{A}(N∣_{f})$ for all $B$-module $N$. If $g:B→C$ is another finite ring homomorphism, then $N∣_{g∘f}=(N∣_{g})∣_{f}$ for every $C$-module $N$. Since $(g∘f)∣_{!}(γ_{C}(N))=γ_{A}(N∣_{g∘f})=γ_{A}((N∣_{g})∣_{f})=f∣_{!}(γ_{B}(N∣_{g}))=f∣_{!}(g_{!}(γ_{C}(N)))$For every $C$-module $N$, it follows that $(g∘f)∣_{!}=f∣_{!}∘g∣_{!}$. $□$ |

Exercise *flat* $A$-modules. Repeating the construction of Exercise 26 we obtain a group $K_{1}(A)$. Let $γ_{1}(M)$ denote the image of $(M)$ in $K_{1}(A)$.

1. | Show that tensor product of modules over $A$ induces a commutative ring structure on $K_{1}(A)$, such that $γ_{1}(M)⋅γ_{1}(N)=γ_{1}(M⊗N)$. The identity element of this ring is $γ_{1}(A)$.
Let $0→M_{′′}→M→M_{′}→0$ be a short exact sequence where $M,M_{′},M_{′′}∈F_{1}(A)$. If $N∈F_{1}(A)$, then tensor the exact sequence with $N$, we have $0→N⊗M_{′′}→N⊗M→N⊗M_{′}→0$exact. Let $D_{1}$ denote the (additive) subgroup of $C_{1}$ generated by all elements of the form $(M_{′})+(M_{′′})−(M)$ for short exact sequences $0→M_{′′}→M→M_{′}→0$, then $D_{1}$ is also an ideal of the commutative ring $C_{1}$. So $K_{1}(A)=C_{1}/D_{1}$ is also a quotient ring of $C_{1}$, such that $γ_{1}(M)⋅γ$ |