Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 1.

## Rings and Ideals

Exercise

**Proof.**If $x_{n}=0$ then $(1+x)(1−x+x_{2}−x_{3}+...+(−1)_{n−1}x_{n−1})=1$, so $1+x$ is a unit. If $a$ is a unit and $x$ is a nilpotent, then $a_{−1}x$ is a nilpotent, hence $a+x=a(1+a_{−1}x)$ is a unit.

$□$

Exercise

1. | $f$ is a unit in $A[x]$ $⟺$ $a_{0}$ is a unit in $A[x]$, and $a_{1},…,a_{n}$ are nilpotent. |

2. | $f$ is nilpotent $⟺$ $a_{0},…,a_{n}$ are nilpotent. |

3. | $f$ is a zero-divisior $⟺$ there exists $a=0∈A$ such that $af=0$. |

4. | $f$ is said |

**Proof.**

1. | $⟹$: Suppose $g=b_{0}+b_{1}x+⋯+b_{m}x_{m}$ such that $fg=1$, then $a_{0}b_{0}=1$ hence $a_{0},b_{0}$ are both units. We Prove $a_{n}b_{m−r}=0$ for all $0≤r≤m$ by induction on r. If this is true for all $r_{′}<r$, consider the $(n+m−r)$-th coefficient of $fg$, we have $0=∑_{i=0}a_{n−r+i}b_{m−i}$, so $a_{n}b_{m−r}=∑_{i=0}a_{n−r+i}b_{m−i}$. Multiply $a_{n}$ to both side, then by induction hypothesis we get $a_{n}b_{m−r}=0$. In partial, $a_{n}b_{0}=0$, hence $a_{n}$ is nilpotent, and so is $a_{n}x_{n}$. By Ex.1.1, $f−a_{n}x_{n}$ is a unit, so repeat the proof above we have $a_{1},…,a_{n}$ are nilpotent. $⟸$: Repeat Ex.1.1 for $n$ times. |

2. | $⟹$ If $f_{k}=0$, then consider $nk$-th coefficient of $f_{k}$ we have $a_{n}=0$. Then $f−a_{n}x_{n}$ is also nilpotent, hence $a_{0},a_{1},…,a_{n}$ are all nilpotent. $⟸$: $f$ is sums of $n$ nilpotents, also a nilpotent. |

3. | $⟹$: Choose $g=b_{0}+b_{1}x+⋯+b_{m}x_{m}$ of least degree $m$ such that $fg=0$, then $a_{n}b_{m}=0$, so $a_{n}g$ is of at most degree $m−1$. But $a_{n}gf=0$, by the choice of $g$ we have $a_{n}g=0$. Then we prove $a_{n−r}g=0$ by induction on $r$. For $r=1$ it’s proved above, and if it is true for $0,1,…,r−1$, then we have $a_{n−r}b_{m}=0$, so similarly $a_{n−r}g=0$. So we have $b_{m}f=0$. $⟸$: Trivial. |

4. | $⟹$: if all coefficients of $f$ are in an ideal $a=(1)$, then obviously all coefficients of $fg$ are also in $a$, so $fg$ is not primitive. $⟸$ Left $f=a_{0}+a_{1}x+⋯+a_{n}x_{n}$ and $g=b_{0}+b_{1}x+⋯+b_{m}x_{m}$ both primitive. Suppose $fg=c_{0}+c_{1}x+⋯+c_{n+m}x_{n+m}$ is not primitive, that is, $(c_{0},c_{1},…,c_{n+m})=c=(1)$. Let $p$ be a prime ideal contains $c$, and let $i,j$ be the least number that $a_{i}∈/p,b_{j}∈/p$ (cause $f,g$ are primitive, these number exist), then $a_{i}b_{j}=c_{i+j}−∑_{k=0}a_{i}b_{i+j−k}−∑_{k=0}a_{i+j−k}b_{k}∈p$, contradiction. $□$ |

Exercise

**Proof.** Just consider $A[x_{1},x_{2},…,x_{n}]$ as a polynomial ring over $A[x_{1},x_{2},…,x_{n−1}]$. Repeat the proof above, we have: if $f,g∈A[x_{1},x_{2},…,x_{n}]$, then

1. | $f$ is a unit $⟺$ the constant term of $f$ is a unit, all other coefficients are nilpotent. |

2. | $f$ is nilpotent $⟺$ all coefficients of $f$ are nilpotent. |

3. | $f$ is a zero-divisior $⟺$ there exists $a=0∈A$ such that $af=0$ |

4. | $fg$ is primitive $⟺$ $f,g$ are both primitive. $□$ |

Exercise

**Proof.**Let $f∈A[x]$ belong to Jacobson radical, then for all $g∈A[x]$, $1−fg$ is a unit. In particular, $1−xf$ is a unit, hence any coefficient of $f$ is nilpotent, and $f$ is nilpotent, i.e. $f$ belongs to nilradical of $A[x]$.

$□$

Exercise

1. | $f$ is a unit of $A[[x]]$ $⟺$ $a_{0}$ is a unit of $A$. |

2. | If $f$ is nilpotent, then $a_{n}$ is nilpotent for all $n≥0$. |

3. | $f$ belongs to Jacabson radical of $A[[x]]$ $⟺$ $a_{0}$ belongs to Jacabson radical of $A$. |

4. | The contraction of a maximal ideal $m$ of $A[[x]]$ is a maximal ideal of $A$, and $m$ is generated by $m_{c}$ and $x$. |

5. | Every prime ideal of $A$ is the contraction of a prime ideal of $A[[x]]$. |

**Proof.**

1. | $⟹$: Trivial. $⟸$: Let $g=∑_{n=0}b_{n}x_{n}$ where $b_{0}=a_{0},b_{n}=−a_{0}∑_{i=1}a_{i}b_{n−i}$, then $gf=1$, so $f$ is a unit. |

2. | Induction on $n$. if $a_{0},…,a_{n−1}$ is nilpotent, then so is $f_{n}=∑_{m=0}a_{n+m}x_{m}$ (Cause $f−a_{0}−a_{1}x−⋯−a_{n−1}x_{n−1}=x_{n}f_{n}$). So there exists $k$ such that $f_{n}=0$, hence $a_{n}=0$. |

3. | Suppose $f$ belongs to Jacabson radical of $A[[x]]$, then for all $g$, $1−fg$ is a unit. In particular for all $b∈A$, $1−bf$ is a unit, so by i) $1−a_{0}b$ is a unit, so $a_{0}$ belongs to Jacabson radical of $A$; and vice versa. |

4. | If $m_{c}⊆a=(1)$, then $a+(x)$ is a ideal of $A[[x]]$, which contains all series of constant term $∈a$, so $a+(x)⊃m$, hence $a+(x)=m$ and $a=m_{c}$. |

5. | Let $p$ be a prime ideal of $A$, then $p$ is the contraction of $p+(x)$. So it is sufficient to prove $p+(x)$ is a prime ideal of $A[[x]]$. If $f=∑_{n=0}a_{n}x_{n},g=∑_{n=0}b_{n}x_{n}$, and $fg∈p+(x)$, i.e. $a_{0}b_{0}∈p$, then either $a_{0}$ or $b_{0}$ belongs to $p$. So either $f$ or $g$ belongs to $p+(x)$, hence $p+(x)$ is a prime ideal of $A[[x]]$. $□$ |

Exercise

**Proof.**If the Jacobson radical is not contained in the nilradical, then there exists a nonzero idempotent $e$ belongs to the Jacobson radical, so $1−e=1−1e$ is a unit. but $e_{2}=e$, hence $(1−e)e=0$, so $e=0$, contradiction.

$□$

Exercise

**Proof.**Let $p$ be a prime ideal, then we have $A/p$ is a intergal domain, and for all $xˉ$ there exists $n>1$ such that $xˉ_{n}=xˉ$, hence $(xˉ_{n−1}−1)xˉ=0$. If $xˉ=0$, then $xˉ_{n−1}−1=0$, so $xˉ$ is a unit. Hence $A/p$ is a field, and $p$ is maximal.

$□$

Exercise

**Proof.**Forall chains of prime ideals $p_{1}⊇p_{2}⊇…$, if we let $p=⋂_{n=1}p_{n}$, then $p$ is an ideal. Suppose $xy∈p$, then for all $n$ we have $xy∈p_{n}$, hence either $x$ or $y$ belongs to $p_{n}$, so at least one of them belongs to infinite $p_{n}$, hence belongs to all $p_{n}$, i.e. either $x$ or $y$ belongs to $p$. So all chains of prime ideals of $A$ has a lower bound, so by Zorn’s Lemma there exists a minimal elements along them.

$□$

Exercise

**Proof.** $⟹$: Let $b$ be the intersection of all prime ideals containing $a$. If $x∈b$, then in $A/a$, $xˉ$ belongs to all prime ideals, so $xˉ_{n}=0$ for some $n>0$, i.e. $x∈r(a)=a$. so $a=b$ is a intersection of prime ideals.

$□$

Exercise

1. | $A$ has exactly one prime ideal. |

2. | every element of $A$ is either a unit ot nilpotent. |

3. | $A/R$ is a field. |

**Proof.**

i) $⟹$ ii): | If $a∈/R$ is not a unit, then there exists a prime ideal $p$ containing $a$, so $p$ is the onlyprime ideal. But then $R=p$, contradiction. |

ii) $⟹$ iii): | Every elements $∈/R$ is a unit, so every elements of $A/R$ is a unit, therefore $A/R$ is a field. |

iii) $⟹$ i): | $R$ is a maximal ideal and the intersection of all prime ideals. So the only prime ideal of $A$ is $R$ itself. $□$ |

Exercise *Boolean* if $x_{2}=x$ for all $x∈A$. In a Boolean ring $A$, show that

1. | $2x=0$ for all $x∈A$. |

2. | every prime ideal $p$ is maximal, and $A/p$ is a field with two elements. |

3. | every finitely generated ideal in $A$ is principal. |

**Proof.**

1. | $x+1=(x+1)_{2}=x_{2}+2x+1=3x+1$, so $2x=0$ for all $x∈A$. |

2. | Suppose $x∈p$. By $x(1−x)=x−x_{2}=0∈p$ we have $1−x∈p$. So $A=p∪(1−p)$, and $A/p$ only contains two elements, hence is a field, and $p$ is maximal. |

3. | For any $a_{1},a_{2},…,a_{n}∈A$, let $a=1−∏_{i=1}(1−a_{i})$, then $a_{i}a=a_{i}−a_{i}(1−a_{i})∏=a_{i}$, so $(a_{1},a_{2},…,a_{n})=(a)$. $□$ |

Exercise

**Proof.**Let $m$ be the only maximal ideal of $A$. For any idempotent $e$, if $e$ is a unit then $e=e_{−1}e_{2}=e_{−1}e=1$. If $e$ is not unit, then $e∈m=R$ (the Jacabson radical of $A$), so $1−e$ is a unit. but $(1−e)_{2}=1−2e+e_{2}=1−e$ is also idempotent, so $1−e=1$, therefore $e=0$.

$□$

Exercise

Let $m$ be a maximal ideal of $A$ containing $a$, and let $K_{1}=A/m$. Then $K_{1}$is an extension field of $K$ in which each $f∈Σ$ has a root. Repeat the construction with $K_{1}$ in place of $K$, obtaining a field $K_{2}$, and so on. Let $L=⋃_{n=1}K_{n}$, Then $L$ is a field in which each $f∈Σ$ splits completely into linear factors. Let $Kˉ$ be the set of all elements of $L$ which are algebraic over $K$. Then $Kˉ$ is an algebraic closure of $K$.

**Proof.** Let $1∈a$, then $1$ can be written in a finite sum of finite products of $f(x_{f})$-s. Choose one form containing least $f$-s, suppose it contains $a_{1}=f_{1}(x_{f_{1}}),a_{2}=f_{2}(x_{f_{2}}),…,a_{n}=f_{n}(x_{f_{n}})$. then $(a_{1},a_{2},…,a_{n−1})=(1)$ but $(a_{1},a_{2},…,a_{n−1})+(a_{n})=(1)$.

Hence $(a_{1},a_{2},…,a_{n−1})(a_{n})=(a_{1},a_{2},…,a_{n−1})∩(a_{n})$, so $a_{n}∈(a_{1},a_{2},…,a_{n−1})(a_{n})$, i.e. exists $b∈(a_{1},a_{2},…,a_{n−1})$ such that $ba_{n}=a_{n}$. Obviously $A$ is an intergal domain, so $b=1$; but then $(a_{1},a_{2},…,a_{n−1})=(1)$, contradiction.

$□$

Exercise

**Proof.** For every chain in $Σ$ ${a_{α}}_{α}$, it has a upperbound $⋃_{α}a_{α}$. So by Zorn’s Lemma $Σ$ has maximal elements.

$□$

Exercise

1. | if $a$ is the ideal generated by $E$, then $V(E)=V(a)=V(r(a))$. |

2. | $V(0)=X,V(1)=∅$. |

3. | if $(E_{i})_{i∈I}$ is any family of subsets of $A$, then $V(i∈I⋃ E_{i})=i∈I⋂ V(E_{i})$ |

4. | $V(a∩b)=V(ab)=V(a)∪V(b)$ for any ideals $a,b$. |

**Proof.**

1. | If $E⊆p$ then also is $a=(E)$. If $a⊆p$ and $x_{n}∈a⊆p$ then by definition $x∈p$, so $r(a)⊆p$. The inverse is trivial. |

2. | Trivial. |

3. | Trivial. |

4. | If $a∩b⊆p$ then $a⊆p$ or $b⊆p$. So $V(a∩p)⊆V(a)∪V(b)$. $⊇$-s are trivial. $□$ |

Exercise

**Proof.**

1. | In $Spec(Z)$ there is countable infinite closed points $(p_{i})$, and a generic point $0$. |

2. | Cause $R$ is a field, it have only one prime ideal $0$. So $Spec(R)$ is a trivial topology space with only one point. |

3. | In $C[x]$ a prime ideal is $0$ or $(x−z)$ with $z∈C$, it’s similar to $Spec(Z)$ but with uncountable infinite points. |

4. | In $R[x]$ a prime ideal is $0,(x−r)$ or $(x_{2}+px+q)$ with $p_{2}−4q<0$, actually it’s isomorphic to $Spec(C[x])$ |

5. | In $Z[x]$ there are three sorts of prime ideals: $0$ or $(p)$; $(F(x))$ where $F$ is a irreducible polynomial over $Z[x]$ (or equivalentlly irreducible polynomial over $Q[x]$); $(p,F(x))$ where $F(x)$ is a monic irreducible polynomial over $Z/pZ$. For the Zariski topology, there is a closed base of it: ${(p,F(x))∣F(x)irreducible overZ_{p}[x]}$ for all $p$; ${(p,G(x))∣G(x)dividesF(x)overZ_{p}[x]}∪{(F(x))}$ for all irreducible $F(x)∈Z[x]$; ${(p,F(x))}$ for all $F(x)$ irreducible over $Z_{p}[x]$. $□$ |

Exercise

1. | $X_{f}∩X_{g}=X_{fg}$. |

2. | $X_{f}=∅⟺f$ is nilpotent. |

3. | $X_{f}=X⟺f$ is a unit. |

4. | $X_{f}=X_{g}⟺r((f))=r((g))$. |

5. | $X$ is quasi-compact (every open covering of $X$ has a finite sub-covering) |

6. | More generating, each $X_{f}$ is quasi-compact. |

7. | An open subset of $X$ is quasi-compact if and only if it is a finite union of sets $X_{f}$. The sets $X_{f}$ are called basic open sets of $X=Spec(A)$. |

**Proof.** For any open set $X−V(E)$ of $Spec(A)$ (that is, all prime ideals which doesn’t contain $E$), and a prime ideal $p⊇E$, we must find a $f$ such that $p∈X_{f}$ and $X_{f}⊆X−V(E)$. Just select an element $f∈p−E$.

1. | $X_{f}∩X_{g}$ is the complement of $V(f)∪V(g)=V(fg)$, i.e.$X_{f}∩X_{g}=X_{fg}$. |

2. | $X_{f}=∅⟺f∈p$ for all prime ideals $p⟺f$ is contained in the nilradical, i.e. $f$ is nilpotent. |

3. | $X_{f}=X⟺f∈/p$ for all prime ideals $p⟺f$ is a unit. |

4. | If $r((f))=r((g))$ then $V(f)=V(r(f))=V(r(g))=V(g)$, so $X_{f}=X_{g}$. On the other hand $r(f)$ is the nilradical of $A/(f)$, hence is intersection of all ideals which contains $(f)$. So $X_{f}=X_{g}$ implies $r(f)=r(g)$. |

5. | It’s sufficient to show: if $X$ is covered by some $X_{f}$ for $f∈F$, then this open covering has a finite sub-covering. But $⋃_{f∈F}X_{f}=X$ means the ideal generated by $F$ is $(1)$, so $1=∑_{i}a_{i}f_{i}$ for some (finite many) $a_{i}∈A,f_{i}∈F$. so these $X_{f}$ also cover $X$. |

6. | $X_{f}$ is covered by ${X_{g}∣g∈G}$ if and only if $f=∑_{i}a_{i}g_{i}$ for some $a_{i}∈A,g_{i}∈G$. |

7. | An open set of $X$ is then union of some $X_{f}$. If it’s quasi-compact, then it’s a finite union of some $X_{f}$. Conversely, finite union of quasi-compact sets is quasi-compact. $□$ |

Exercise

1. | The set ${x}$ is closed ($x$ is a “closed point”) in $Spec(A)⟺p_{x}$ is maximal; |

2. | ${x} =V(p_{x})$; |

3. | $y∈{x} ⟺p_{x}⊂p_{y}$; |

4. | $X$ is a $T_{0}$-space. |

**Proof.**

1. | $x$ is a closed point $⟺{x}=V(E)$ for some $E⟺p_{x}$ is maximal. |

2. | ${x} =⋂_{E⊆p_{x}}V(E)=V(p_{x})$. |

3. | $y∈{x} ⟺p_{y}∈V(p_{x})⟺p_{x}⊂p_{y}$. |

4. | If $x,y$ are distinct points of $X$, then either $p_{x}⊂p_{y}$ or $p_{y}⊂p_{x}$. Suppose the former, then $p_{y}∈{x} $, so $X−{x} $ is a neighborhood of $y$ which seperate $x,y$. $□$ |

Exercise

**Proof.**$Spec(A)$ is irreducible if and only if for every two prime ideal $a,b$, there is a prime ideal $p⊆a∩b$. This means minimal prime ideal $p$ of $A$ is unique, so it’s contained in every prime ideal. So $p⊆R$; and obviously $R⊆p$, so $R=p$ is a prime ideal.

$□$

Exercise

1. | If $Y$ is an irreducible subspace of $X$, then the closure $Yˉ$ of $Y$ in $X$ is irreducible. |

2. | Every irreducible subspace of $X$ is contained in a maximal irreducible subspace. |

3. | The maximal irreducible subspace of $X$ are closed and cover $X$. They are called the irreducible components of $X$. What are the irreducible components of a Hausdorff space? |

4. | If $A$ is a ring and $X=Spec(A)$, then the irreducible components of $X$ are the closed sets $V(p)$, where $p$ is a minimal prime ideal of $A$. |

**Proof.**

1. | Every open subset of $Yˉ$ is dense in $Y$, hence dense in $Yˉ$. |

2. | If $X_{1}⊆X_{2}⊆…$ are irreducible subspaces, then $⋃_{i}X_{i}$ is irreducible, so by Zorn’s lemma, every irreducible subspace of $X$ is contained in a maximal irreducible subspace. |

3. | If $Y$ is a maximal irreducible subspace, then $Y⊆Yˉ$ is irreducible, so $Y=Yˉ$, that means $Y$ is closed. All sets of one point ${x}$ are irreducible, hence contained in some maximal irreducible subspace. Hence maximal irreducible subspaces cover $X$. If $X$ is Hausdorff and $X$ is irreducible, then $X$ must contains only one point. So irreducible components of a Hausdorff space is all one-point subspace. |

4. | If $Y$ is a irreducible component of $X$, then $Y$ is closed and irreducible, so there is a unique minimal prime ideal $p$ in $Y$ and $Y=V(p)$. The converse is trivial. $□$ |

Exercise

1. | If $f∈A$ then $ϕ_{∗−1}(X_{f})=Y_{ϕ(f)}$, and hence that $ϕ_{∗}$ is continuous. |

2. | If $a$ is an ideal of $A$, then $ϕ_{∗−1}(V(a))=V(a_{e})$. |

3. | If $b$ is an ideal of $B$, then $ϕ_{∗}(V(b)) =V(b_{c})$. |

4. | If $ϕ$ is surjective, then $ϕ_{∗}$ is a homeomorphism of $Y$ onto the closed subset $V(Ker(ϕ))$ of $X$. (In particular, $Spec(A)≃Spec(A/R)$). |

5. | If $ϕ$ is injective, then $ϕ_{∗}(Y)$ is dense in $X$. More precisely, $ϕ_{∗}(Y)$ is dense in $X⟺Ker(ϕ)⊂R$. |

6. | Let $ψ:B→C$ be another ring homomorphism, then $(ψ∘ϕ)_{∗}=ϕ_{∗}∘ψ_{∗}$. |

7. | Let $A$ be an intergal domain with just one non-zero prime ideal $p$, and let $K$ be the field of fractions of $A$. Let $B=(A/p)×K$. Define $ϕ:A→B$ by $ϕ(x)=(xˉ,x)$, where $xˉ$ is the image in $A/p$. Show that $ϕ_{∗}$ is bijective but not a homeomorphism. |

**Proof.**

1. | $f∈/ϕ_{−1}(q)⟺ϕ(f)∈/q$. |

2. | $a⊆ϕ_{−1}(p)⟺a_{e}⊆p$. |

3. | $a∈ϕ_{∗}(V(b)) $ if and only if for every $V(E)⊃ϕ_{∗}(V(b))$, we have $a∈V(E)$. But $V(E)⊃ϕ_{∗}(V(b))⟺b_{c}⊂E$, so it’s equivalent to $b_{c}⊂a$,i.e.$a∈V(b_{c})$. |

4. | If $ϕ$ is surjective, then $A/Ker(ϕ)≃B$, so we can identify $B$ with $A/Ker(ϕ)$. Now $ϕ:A→A/Ker(ϕ)$ is the quotient map, and prime ideals of $A/Ker(ϕ)$ is one-to-one corresponding with primes ideals of $A$ which containing $Ker(ϕ)$. Also $ϕ(V(a))⊆Spec(A/Ker(ϕ))$ is same as $V(aKer(ϕ))⊆V(Ker(ϕ))$, hence $ϕ_{∗}:Spec(A/Ker(ϕ))→V(Ker(ϕ))$ is a homeomorphism. |

5. | If $ϕ$ is injective, then $ϕ_{∗}(Y) =ϕ_{∗}(0) =V(0)=X$. So $ϕ_{∗}(Y)$ is dense in $X$. $ϕ_{∗}(Y)$ is dense $⟺ϕ_{∗}(Y) =V(0_{c})=V(Ker(ϕ))=X$, i.e. $Ker(ϕ)⊆R$, |

6. | Trivial. |

7. | There are two points in $0,p∈Spec(A)$, where $0$ is a closed point but $p$ isn’t. Also there are two points $0×K,(A/p)×0∈Spec(B)$. $ϕ_{∗}$ maps $0×K$ to $p$ and $(A/p)×0$ to $0$, so it’s bijective. But in $Spec(B)$ two points are both closed, so $ϕ_{∗}$ can’t be a homeomorphism. $□$ |

Exercise

Conversely, ler $A$ be any ring. Show that the following statements are equivalent:

1. | $X=Spec(A)$ is disconnected. |

2. | $A≃A_{1}×A_{2}$ where neither of the rings $A_{1},A_{2}$ is zero ring. |

3. | $A$ contains an idempotent $=0,1$. |

**Proof.** A prime ideal of $∏_{i=1}A_{i}$ is in the form $A_{1}×A_{2}×⋯×A_{i−1}×p_{i}×A_{i+1}×⋯×A_{n}$, where $p_{i}$ is a prime of $A_{i}$. So $Spec(A)$ is disjoint union of $X_{i}≃Spec(A_{i})$. Also $X_{i}=V(A_{1}×⋯×A_{i}^ ×⋯×A_{n})$ is closed.

Conversely, If $X=Spec(A)$ is disconnected, then $X=X_{1}∪X_{2}$ where $X_{1}∩X_{2}=∅$ and both $X_{i}$ is non-empty open (and closed) set. So $X_{1}=V(a),X_{2}=V(b)$ for some ideal $a,b⊆A$, and $a+b=(1),ab⊆R$. So select $a∈a,b∈b$ such that $a+b=1$, then there exists n s.t. $(ab)_{n}=0$. so $(a_{n}),(b_{n})$ are coprime and $(a_{n})(b_{n})=0$, and $A≃(A/(a_{n}))×(A/(b_{n}))$. So i) $⟺$ ii).

$□$

Exercise

1. | For each $f∈A$, the set $X_{f}$ is both open and closd in $X$. |

2. | Let $f_{1},…,f_{n}∈A$, Show that $X_{f_{1}}∪X_{f_{2}}∪⋯∪X_{f_{n}}=X_{f}$ for some $f∈A$. |

3. | The sets $X_{f}$ are the only subsets of $X$ which are both open and closed. |

4. | $X$ is a compact Hausdorff space. |

**Proof.**

1. | Obviously $X_{f}$ is open. By $f+(1−f)=1$ and $f(1−f)=0$ we have $X_{f}∩X_{1−f}=∅$ and $X_{f}∪X_{1−f}=X$. Show $X_{f}=V(1−f)$ is closed. |

2. | The complement of $X_{f_{1}}∪⋯∪X_{f_{n}}$ is $V(f_{1})∩⋯∩V(f_{n})=V((f_{1},f_{2},…,f_{n}))$. By Exercise 11, there exists $f$ such that $(f_{1},…,f_{n})=(f)$, So $X_{f_{1}}∪⋯∪X_{f_{n}}=X_{f}$. |

3. | If $V(a)$ is open, that is, $X$ is the disjoint union of $V(a)$ and some $V(b)$. So $a+b=(1)$ and $ab=R=0$. Let $f+g=1$ where $f∈a,g∈b$, then $V(a)⊂V(f),V(b)⊂V(g)$ and $V(f)∩V(g)=∅$. This means $V(a)=X_{g},V(b)=X_{f}$. |

4. | If $a$ and $b$ is two distinct point of $X$, we may assume $a⊆b$, then there exists $f∈a−b$. So $a∈/X_{f},b∈X_{f}$. Since $X_{f}$ is both open and closed, $a$ and $b$ are seperated by neighborhoods. $X$ is compact by Exercise 17. $□$ |

Exercise

1. | $L$ has a least element and a greatesr element (denoted by $0,1$ respectively). |

2. | Each of $∧,∨$ is distributive over the other. |

3. | Each $a∈L$ has a unique "complement" $a_{′}∈L$ such that $a∨a_{′}=1$ and $a∧a_{′}=0$ |

Let $L$ be a Boolean lattice. Define addition and multiplication in L by the rules $a+b=(a∧b_{′})∨(a_{′}∧b),ab=a∧b$Verify that in this way $L$ become a Boolean ring, say $A(L)$.

Conversely, starting from a Boolean ring $A$, define an ordering on a as follows: $a≤b$ means that $a=ab$. Show that, with respect to this ordering, $A$ is a Boolean lattice [The sup and inf are given by $a∨b=a+b+ab$ and $a∧b=ab$, and the complement by $a_{′}=1−a$.] In this way we obtain a one-to-one correspondence between Boolean rings and Boolean lattices.

**Proof.**Verify directly.

$□$

Exercise

**Proof.** If $L$ is a Boolean lattice then $A=A(L)$ is a Boolean ring. Let $X=Spec(A)$, then $X$ is Hausdorff, and there is a one-to-one correspondence between open-and-closed subsets of $X$ and elements of $A$: $X_{f}↔f$.

$□$

Exercise

Let $X$ be a compact Hausdorff space and let $C(X)$ denote the ring of all real-valued continuous functions on $X$ (add and multiply functions by adding and multiplying their values). For each $x∈X$, let $m_{x}$ be theset of all $f∈C(x)$ such that $f(x)=0$. The ideal $m_{x}$ is maximal, because it is the kernel of the (surjective) homomorphism $C(x)→R$ which takes $f$ to $f(x)$. If $Xˉ$ denotes $Max(C(x))$, we have therefore defined amapping $μ:X→Xˉ$, namely $x↦m_{x}$.

We shall show that $μ$ is a homeomorphism of $X$ onto $Xˉ$.

1. | Let $m$ be any maximal ideal of $C(x)$, and let $V=V(m)$ be theset of common zeros of the functions in $m$. That is, $V={x∈X:f(x)=0for allf∈m}$Suppose that $V$ is empty. Then for each $x∈X$ there exists $f_{x}∈m$ such that $f_{x}(x)=0$. Since $f_{x}$ is continus, there is an open neighborhood $U_{x}$ of $x$ in $X$ on which $f_{x}$ does not vanish. By compactness a finite number of neighborhoods, say $U_{x_{1}},…,U_{x_{n}}$, cover $X$. Let $f=f_{x_{1}}+⋯+f_{x_{n}}$Then $f$ doest not vanish at any point of $X$, hence is a unit in $C(x)$. But this contracdicts $f∈m$, hence $V$ is not empty. Let $x$ be a point of $V$. Then $m⊆m_{x}$, hence $m=m_{x}$ because $m$ is maximal. Hence $μ$ is surjective. |

2. | By Urysohn’s lemma (this is the only non-trivial fact required in the argument) the continuous functions seperate the points of $X$. Hence $x=y⟹m_{x}=m_{y}$, and therefore $μ$ is injective. |

3. | Let $f∈C(X)$; let $U_{f}={x∈X:f(x)=0}$and let $Uˉ_{f}={m∈Xˉ:f∈m}$Show that $μ(U_{f})=Uˉ_{f}$. The open sets $U_{f}$ (resp. $Uˉ_{f}$) form a basis of the topology of $X$ (resp. $Xˉ$) and therefore $μ$ is a homeomorphism. Thus $X$ can be reconstructed from the ring of functions $C(X)$. |

**Proof.**$μ(U_{f})={m_{x}∈Xˉ:f(x)=0}={m_{x}∈Xˉ:f∈/m_{x}}$. But every $m$ is some $m_{x}$ cause of i). So $μ(U_{f})=μ(Uˉ_{f})$. If $x∈W$ and $W$ open, then by Urysohn’s lemma there exists a function $f:X→R$ such that $f(x)=0$ and for all $y∈/W$ we have $f(y)=0$, so $x∈U_{f}⊆W$. This shows $U_{f}$ form a basis of $X$. Also $Uˉ_{f}=Xˉ−V(f)$ forms a basis of $Xˉ$, so $μ$ is a homeomorphism.

$□$

### Affine algebraic varieties

Exercise

Consider the set of all polynomials $g∈k[t_{1},…,t_{n}]$ with the proerty that $g(x)=0$ for all $x∈X$. the set is an ideal $I(X)$ in the polynomial ring, and is called the ideal of the variety $X$. The quotient ring $P(X)=k[t_{1},…,t_{n}]/I(X)$is the ring of polynomial functions on $X$, because two polynomial $g,h$ define the same polynomial function on $X$ is and only if $g−h$ vanishes at every point of $X$, that is, if and only if $g−h∈I(X)$.

Let $ξ_{i}$ be the image of $t_{i}$ in $P(x)$. The $ξ_{i}(1≤i≤n)$ are the coordinate functions on $X$: if $x∈X$, then $ξ_{i}(x)$ is the $i$th coordinate of $x$. $P(X)$ is generated as a $k$-algebra by the coordinate functions, and is called the corrdinate ring (or affine algebra) on $X$.

As in Exercise 26, for each $x∈X$ let $m_{x}$ be the ideal of all $f∈P(x)$ such that $f(x)=0$; it is a maximal ideal of $P(x)$. Hence, if $Xˉ=Max(P(X))$, we have defined a mapping $μ:X→Xˉ$, namely $x→m_{x}$.

It’s easy to show that $μ$ is injective: if $x=y$, we must have $x_{i}=y_{i}$ for some $i$, and hence $ξ_{i}−x_{i}$ is in $m_{x}$ but not in $m_{y}$, so that $m_{x}=m_{y}$. What is less obvious (but still true) is that $μ$ is surjective. This is one form of Hilbert’s Nullstellensatz (See Chapter 7).

Exercise

Let $X,Y$ be affine algebraic varieties in $k_{n},k_{m}$ respectively. A mapping $ϕ:X→Y$ is said to be regular if $ϕ$ is the restriction to $X$ of a polynomial mapping from $k_{n}→k_{m}$.

If $η$ is a polynomial function on $Y$, then $η∘ϕ$ is a polynomial function on $X$. Hence $ϕ$ induces a $k$-algebra homomorphism $P(Y)→P(X)$, namely $η↦η∘ϕ$. Show that in this way we obtain a one-to-one correspondence between the regular mappings $X→Y$ and the $k$-algebra homomorphism $P(Y)→P(X)$.

**Proof.** Let $ψ_{i}(1≤i≤m)$ denote coordinates on $Y$. For any $k$-algebra homeomorphism $μ:P(Y)→P(X)$, let $fˉ _{i}=μ(ψ_{i})∈P(X)(1≤i≤m)$, and find any $f_{i}:k[t_{1},…,t_{n}]$ such that the image of $f_{i}$ in $P(X)$ is $fˉ _{i}$.

$□$