1 Rings and Ideals

Highlights: nilradical, Jacobson radical, Zariski topology

Convention.

• All rings are commutative with $1_A$.
• $A=\{0\}$ iff $1_A=0_A$.
• Ring homomorphism $f:A\to B$ always satisfies $f(1_A)=1_B$

Proposition

Let $\alpha \subseteq A$ be an ideal. Then there exists an bijection:

\{\beta\subseteq A:\beta\mbox{ is an ideal, }\alpha\subseteq\beta\}\longleftrightarrow\{\mbox{ideals of }A/\alpha\}. $$ ^9eaa9e

Local

Definition

We say $A$ is a local ring, if it has a unique maximal ideal $m_A$. Furthermore, $A/m_A$ is called residue field.

Example.

• all fields
• $k[[x]]$ with $k$ field. Recall that it has the unique maximal ideal $(x)$.
• ${\mathbb{Z}}_p={\underleftarrow{\lim }}_{n⩾1}\mathbb{Z}/p^n\mathbb{Z}$, where $m_A=(p)$ and ${\mathbb{Z}}_p/p{\mathbb{Z}}_p\simeq {\mathbb{F}}_p$.

Proposition

Let $A$ be a ring, and let $m\subseteq A$ be an ideal.

• If $\forall x\in A\setminus m$ satisfies $x\in A^{\times }$, then $A$ is local and $m=m_A$.
• Suppose $m$ is maximal and for all $x\in m$ there is $1+x\in A^{\times }$, then $A$ is local.

\begin{proof} i) For any ideal $I⊊A$ and any $i\in I$, $i\notin A^{\times }$. Hence $I\subseteq m$ and so $m$ is the unique maximal ideal.

ii) Let $y\in A\setminus m$, then $m+(y)=A$ and so $1=m_0+ay$ for some $m_0\in M$. Then $ay\in A^{\times }$ and so for $y$. By i) we finish the proof. \end{proof}

Definition

We say a ring is semi-local, if it has a finite number of maximal ideals.

Example.

• If $A_1,A_2$ are local, then $A_1\times A_2$ is semi-local, as it has only two maximal ideals.
• $\mathbb{C}[x]/(x-1)(x-2)$. By ^9eaa9e, there is a bijection between ideals of $\mathbb{C}[x]/(x-1)(x-2)$ and ideals of $\mathbb{C}[x]$ containing $(x-1)(x-2)$. Hence, all maximal ideals of $\mathbb{C}[x]/(x-1)(x-2)$ are $(x-1)$ and $(x-2)$.

Nilradical

Proposition

Let $\mathcal{N}=\{x\in A:x\text{mbox}isnilpotent\}$. Then $\mathcal{N}$ is an ideal and $A/\mathcal{N}$ has no nontrivial nilpotent element. We say $\mathcal{N}$ is the nilradical, denoted by $\mathrm{Nil}(A)$.

\begin{proof} It is easy to verify $\mathcal{N}$ is an ideal: for any $a,b\in \mathcal{N}$, we have $a+b,ab,ra\in \mathcal{N}$ for any $r\in A$.

Let $\overline{x}\in A/\mathcal{N}$. If $\overline{x}$ is nilpotent, then ${\overline{x}}^n=0$ and so $x^n\in \mathcal{N}$. Then $(x^n{)}^m=0$ and $x\in \mathcal{N}$. It deduces that $\overline{x}=0$. \end{proof}

Proposition

For any ring $A$, there is

\mathcal N(A)=\cap_{\mathcal{P}\subseteq A\mbox{ is a prime ideal}} \mathcal P. $$ ^bgejpl

\begin{proof} Let ${\cap }_{\mathcal{P}\subseteq A\text{mbox}isprime}\mathcal{P}:=\mathbb{P}$. For any $x\in \mathcal{N}$ and any prime ideal $\mathcal{P}$, we have $x^n=0\in \mathcal{P}$ and so $x\in \mathcal{P}$. Hence $\mathcal{N}(A)\subseteq \mathbb{P}$.

On the other hand, if $f\in \mathbb{P}$ and $f^n\ne 0$ for any $n$, define

$$\Sigma =\{\alpha \subseteq A:\alpha \text{mbox}isanidealand{f}^n\notin \alpha ,\forall n⩾1\}.$$

Notice that $\Sigma \ne \varnothing$ as $(0)\in \Sigma$. By Zorn’s lemma, there exists a maximal element $\beta \in \Sigma$.

We claim that $\beta$ is a prime ideal. Otherwise, there exist $x,y\notin \beta$ such that $xy\in \beta$. Since $\beta +(x)$ and $\beta +(y)$ are not contained in $\Sigma$, there are $t$ and $s$ such that $f^t\in \beta +(x)$ and $f^s\in \beta +(y)$. It deduces that $f^{t+s}\in (\beta +(x))(\beta +(y))\subseteq \beta$, which is a contradiction. Hence $\beta$ is a prime ideal and $f\notin \beta$. It contradicts with $f\in \mathbb{P}$. \end{proof}

Jacobson Radical

Definition

Define the Jacobson radical as the intersection of all maximal ideals. Remark that $\mathrm{Jac}(A)\supseteq \mathrm{Nil}(A)$.

Example.

• $\mathrm{Jac}(\mathbb{Z})=(0)=\mathrm{Nil}(\mathbb{Z})$
• $\mathrm{Jac}(k[[x]])=(x)$ and $\mathrm{Nil}(k[[x]])=(0)$

Proposition

Let $R=\mathrm{Jac}(A)$. Then $x\in R$ iff $1-xy\in A^{\times }$ for all $y\in A$.

\begin{proof} "→" If $1-xy\notin A^{\times }$, then there exists a maximal ideal $m$ such that $1-xy\in m$. Since $x\in m$, we have $1\in m$, which is impossible.

"←" If $x\notin R$, then there is a maximal ideal $m$ such that $x\notin m$. Then $m+(x)=A$ and so $1=m_0+x{y}_0$ for some $m_0\in m$ and $y_0\in A$. It deduces that $1-x{y}_0\in m$ and so $1-x{y}^0\notin A^{\times }$, contradiction. \end{proof}

Some Propositions

Recall that if $\alpha$ and $\beta$ are coprime, i.e., $\alpha +\beta =A$, then $\alpha \beta =\alpha \cap \beta$, where $\alpha \beta =\{{\sum }_{i=1}^n{a}_i{b}_j:a_i\in \alpha ,b_i\in \beta \}$. We have the following theorem.

the Chinese remainder theorem

Let ${\alpha }_1,\cdots ,{\alpha }_n$ be ideals. Consider $\varphi :A\to {\prod }_{i=1}^n(A/{\alpha }_i)$.

• If ${\alpha }_i,{\alpha }_j$ are pairwise coprime, then $\prod {\alpha }_i=\cap {\alpha }_i$.
• $\varphi$ is surjective if ${\alpha }_i,{\alpha }_j$ are pairwise coprime.
• $\varphi$ is injective if $\cap {\alpha }_i=(0)$.

\begin{proof} i) is a generalization of the above “recall” part.

ii) By the Chinese remainder theorem, if ${\alpha }_i,{\alpha }_j$ are coprime, then $A/({\cap }_{i=1}^n{\alpha }_i)\simeq {\prod }_{i=1}^n(A/{\alpha }_i)$ and so $\varphi$ can be identified as a quotient map.

iii) By ii). \end{proof}

Proposition

• Let $p_1,\cdots ,p_n$ be prime ideals, and let $\alpha$ be an ideal with $\alpha \subseteq {\cup }_{i=1}^n{p}_i$. Then $\alpha \subseteq p_i$ for some $i$.
• Let ${\alpha }_1,\cdots ,{\alpha }_n$ be ideals, and let $p$ be a prime ideal such that $p\supseteq {\cap }_{i=1}^n{\alpha }_i$. Then $p\supseteq {\alpha }_i$ for some $i$. Furthermore, if $p=\cap {\alpha }_i$, then $p={\alpha }_i$ for some $i$.

Remark. We cannot change $n$ to $\infty$. Here are two counterexamples.

• $A=\mathbb{C}[x_1,\cdots ,x_n,\cdots ]$. Let $p_i=(x_1\cdots x_i)$, then $\alpha =(x_1\cdots x_n\cdots )\subseteq {\cup }_{i=1}^{\infty }{p}_i$ is not contained in $p_i$ for all $i$.
• $A=\mathbb{Z}$. Let ${\alpha }_i=(p_i)$, then $\cap {\alpha }_i=(0)$ and $(0)⊉{\alpha }_i$ for all $i$.

\begin{proof} i) Consider the contrapositive statement. If $\alpha ⊈p_i$ for all $i$, then $\alpha ⊈{\cup }_{i=1}^n{p}_i$. We prove it by induction. The case of $n=1$ is trivial. Suppose it is true for $n-1$. By induction hypothesis, we have $\alpha ⊈{\cup }_{j\ne i}{p}_j$ for each $1⩽i⩽n$. Take $x_i\in \alpha$ such that $x_i\notin p_j$ for all $j\ne i$. If for some $i$ there is $x_i\notin p_i$, then $x_i\notin {\cup }_{i=1}^n{p}_i$ and we have done. Otherwise $x_i\in p_i$ for all $i$. Define $y:={\sum }_{i=1}^n{x}_1\cdots {\hat{x}}_i\cdots x_n$, then $y\in \alpha$ and $y\notin {\cup }_{i=1}^n{p}_i$.

ii) If not, we have $p⊉{\alpha }_i$ for all $i$. Then there exists $x_i\in {\alpha }_i$ such that $x_i\notin p$. Consider $y\in x_1\cdots x_n$, then $y\in {\cap }_{i=1}^n{\alpha }_i$ and so $y\in p$, which is a contradiction. \end{proof}

Ideal Quotient

Definition

Let $\alpha ,\beta \subseteq A$ be ideals. Define their ideal quotient as

$$(\alpha :\beta )=\{x\in A:x\beta \subseteq \alpha \}.$$

This is an ideal of $A$.

In particular, if $\alpha =(0)$, then $(0:\beta )$ is called the annihilator of $\beta$, denoted by $\mathrm{Ann}(\beta )$.

Example. Let $A=\mathbb{Z}$. Then $((m):(n))=(m/(m,n))$.

Radical

Definition

Let $\alpha \subseteq A$ be an ideal. Define the radical of $\alpha$ as

$$r(\alpha )=\{x\in A:x^n\in \alpha \text{mbox}forsomen\}.$$

Indeed, consider $\varphi :A\to A/\alpha$, then $r(\alpha )={\varphi }^{-1}(\mathrm{Nil}(A/\alpha ))$.

Text-Exercise. Let $P$ be a prime ideal, then $r(P^n)=P$ for all $n⩾1$.

\begin{proof} It is easy to show $P\subseteq r(P^n)$. For any $y\in r(P^n)$, there exists $m$ such that $y^m\in P^n\subseteq P$ and so $y\in P$. Hence, $r(P^n)\subseteq P$.

\end{proof}

Proposition

For any ideal $\alpha \subseteq A$, there is

r(\alpha)=\cap_{\alpha\subseteq P,P\mbox{ is prime}}P. $$ ^8da6f1

\begin{proof} Note that

$$r(\alpha )/\alpha =\mathrm{Nil}(A/\alpha )={\cap }_{\alpha \subseteq P,P\text{mbox}isprime}(P/\alpha )=({\cap }_{\alpha \subseteq P}P)/\alpha$$

by ^bgejpl. Now we finish the proof. \end{proof}

Proposition

If $r(\alpha ),r(\beta )$ are coprime, then $\alpha ,\beta$ are coprime.

\begin{proof} It is easy to verify that $r(\alpha +\beta )=r(r(\alpha )+r(\beta ))$, and it deduces that $r(\alpha +\beta )=r(r(\alpha )+r(\beta ))=r(A)=A$. \end{proof}

Extension and Contraction

Definition

Define a ring homomorphism $f:A\to B$. For an ideal $\alpha \subseteq A$, $f(\alpha )$ is not necessarily an ideal. Define the extension ideal as

$${\alpha }^e=⟨f(\alpha )⟩\subseteq B.$$

For ideal $\beta \subseteq B$, define the contraction ideal as

$${\beta }^c=f^{-1}(\beta )\subseteq A.$$

Important Fact. Let $f:A\to B$ be a ring homomorphism, and let $\beta \subseteq B$ be a prime ideal. Then ${\beta }^c=f^{-1}(\beta )\subseteq A$ is prime. (Proof: $f$ induces a map $A/f^{-1}(\beta )\to B/\beta$, which is injective. Note that $\beta$ is prime $⟺$ $B/\beta$ is integral domain $⟹$ $A/f^{-1}(\beta )$ is integral domain $⟺$ $f^{-1}(\beta )$ is prime.)

Example. If $\alpha$ is prime, then ${\alpha }^e$ is not necessary prime. Define $f:\mathbb{Z}\to \mathbb{Z}[i]$. Note that $(2)\subseteq \mathbb{Z}$ is prime, but $2\mathbb{Z}[i]$ containing $(1+i)(1-i)$ is not prime.

Spectrum and Zariski Topology

Text-Ex. 1.15

Let $A$ be a ring. Define $X:=\{P:P\text{mbox}isaprimeidealofA\}$. For a subset $E\subseteq A$, let $V(E)=\{P:P\in X,P\supseteq E\}$. Then

• If $\alpha =⟨E⟩$, then $V(E)=V(\alpha )=V(r(\alpha ))$.
• $V(0)=X$, $V(1)=\varnothing$.
• Let $E_i\subseteq A$ with $i\in I$ be a system of subsets. Then ${\cap }_{i\in I}V(E_i)=V({\cup }_{i\in I}E)$.
• $V(\alpha )\cup V(\beta )=V(\alpha \beta )=V(\alpha \cap \beta )$ for any ideals $\alpha ,\beta$.

\begin{proof} i) Note that $E\subseteq \alpha \subseteq r(\alpha )$, and then $V(E)=V(\alpha )\supseteq V(r(\alpha ))$. For any $q\supseteq \alpha$, note that $r(\alpha )={\cap }_{P\supseteq \alpha }P$ by ^8da6f1 and so $q\supseteq r(\alpha )$. Hence $q\in V(r(\alpha ))$ and then $V(\alpha )=V(r(\alpha ))$. Now we finish the proof.

ii) and iii) are trivial.

iv) Since $\alpha \supseteq \alpha \cap \beta \supseteq \alpha \beta$, there is $V(\alpha )\subseteq V(\alpha \cap \beta )\subseteq V(\alpha \beta )$ and $V(\beta )\subseteq V(\alpha \cap \beta )\subseteq V(\alpha \beta )$. It remains to show $V(\alpha )\cup V(\beta )\supseteq V(\alpha \beta )$. For $P\in X$ with $P\supseteq \alpha \beta$, we have $P\supseteq \alpha$ or $P\supseteq \beta$ (otherwise, there exist $x\in \alpha \setminus P$ and $y\in \beta \setminus P$, which is impossible as $xy\in P$). \end{proof}

Remark. Notice that $X=\{P:P\text{mbox}isaprimeidealofA\}$ is a topological space, using $\mathcal{S}:=\{V(E):E\subseteq A\}$ as the set of closed subsets of $X$. Denote $X$ by $\mathrm{Spec}A$, call the topology the Zariski topology.

Text-Ex. 1.16

Draw picture for $\mathrm{Spec}A$.

• $\mathrm{Spec}(\mathbb{Z})=\{(0),(p):p\text{mbox}isaprime\}$.
  • The set of closed subsets is $\mathcal{S}=\{V(E):E\subseteq \mathbb{Z}\}=\{V((n)):n\in \mathbb{Z}\}\cup \varnothing \cup \mathrm{Spec}(\mathbb{Z})$, where $V((n))=\{(p_1),\cdots ,(p_k)\}$ with $n={\prod }_{i=1}^k{p}_i^{n_i}$.
  • The set of open subsets is $\mathcal{T}=\mathrm{Spec}(\mathbb{Z})\cup \varnothing \cup \{\mathrm{Spec}(\mathbb{Z})\setminus \{(p_1),\cdots ,(p_k)\}\}$.
  • The closure of $(0)$ is $\mathrm{Spec}(\mathbb{Z})$, and so $(0)$ is called “fat point”.
• $\mathrm{Spec}(\mathbb{R})=\{(0)\}$.
• $\mathrm{Spec}(\mathbb{C}[x])=\{(0),(x-a{)}_{a\in \mathbb{C}}\}$.
  • This is called the $1$-dimensional affine space over $\mathbb{C}$.
  • Any non-empty open subset is dense.
• $\mathrm{Spec}(\mathbb{R}[x])=\{(0),(x-\alpha {)}_{\alpha \in \mathbb{R}},(x-z)(x-\overline{z}{)}_{z\in \mathbb{C}\setminus \mathbb{R}}\}$.
• $\mathrm{Spec}(\mathbb{Z}[x])=\{(0),(p),(f(x)),(p,g(x)):p\text{mbox}isprime,f\text{mbox}isirreducible,\overline{g}\in {\mathbb{F}}_p[x]\text{mbox}isirreducible\}$.

Remark.

• $\mathrm{Spec}(\mathbb{C}[x])$ can be seen as a plane + a fat point, and $\mathrm{Spec}(\mathbb{R}[x])$ can be seen as a upper half plane + a fat point.
• The Red Book of Varieties and Schemes, page 75, gives us a picture of $\mathbb{Z}[x]$.

Definition

Let $X$ be a set. A basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called the basis element) such that

• for any $x\in X$, there exists $B_x\in \mathcal{B}$ such that $x\in B_x$;
• if $x\in B_1\cap B_2$ where $B_1,B_2\in \mathcal{B}$, then there exists some $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.

Lemma

If $\mathcal{B}$ is a basis for a topology, then $\mathcal{T}:=\{{\cup }_{i\in I}{B}_i:B_i\in B\}$.

Text-Ex. 1.17

Let $A$ be a ring and let $X=\mathrm{Spec}(A)$. For $f\in A$, define $X_f=\mathrm{Spec}(A)\setminus V(f)$. Then $\{X_f:f\in A\}$ forms a basis of Zariski topology. Also:

• $X_f\cap X_g=X_{fg}$;
• $X_f=\varnothing$ iff $f$ is nilpotent;
• $X_f=X$ iff $f\in A^{\times }$;
• $X_f=X_g$ iff $r((f))=r((g))$;
• $X$ is quasi-compact, i.e., any open covering has a finite subcover;
• each $X_f$ is quasi-compact;
• an open subset of $X$ is quasi-compact iff it is a finite union of some $X_f$.

\begin{proof} i) By ^vf0aa5 $V(fg)=V(f)\cup V(g)$. It deduces that $X_f\cap X_g=X_{fg}$.

ii) $X_f=\varnothing$ iff $V(f)=X$ iff $f\in {\cap }_{P\text{mbox}isprimeideal}P=\mathrm{Nil}(A)$ iff $f$ is nilpotent.

iii) $X_f=X$ iff $V(f)=\varnothing$ iff $f$ is not contained in any prime ideals. If $f$ is not contained in any prime ideals, then it is not contained in any maximal ideals as maximal ideal is prime. So it can not contained in any non-trivial ideals and $⟨f⟩=A$. On the other hand, if $f\in A^{\times }$, $⟨f⟩=A$ and so $X_f=X$.

iv) If $r((f))=r((g))$, then $V(f)=V(r(f))=V(r(g))=V(g)$ and $X_f=X_g$. If $V(f)=V(g)$, then $r((f))=r((g))$ by ^8da6f1.

v) Suppose $\{X_{f_i}:i\in I\}$ is an open cover of $X$, then ${\cup }_{i\in I}{X}_{f_i}=X_{(\{f_i:i\in I\})}=X$ and so $\{f_i:i\in I\}$ generate $A$. Therefore, there is a finite subset $J$ such that $1={\sum }_{j\in J}{g}_j{f}_j$ with $g_j\in A$. Now $\{f_j:j\in J\}$ generates $A$ and so $\{X_{{}_{f_j}}:j\in J\}$ is a finite subcover of $X$.

vi) Suppose $\{X_{g_i}:i\in I\}$ is an open cover of $X_f$, then

$${\cup }_{i\in I}(X_f\cap X_{g_i})={\cup }_{i\in I}{X}_{f{g}_i}=X_{\{f{g}_i:i\in I\}}=X_f$$

and so $V(\{f{g}_i:i\in I\})=V((f))$. It deduces that $r(⟨\{f{g}_i:i\in I\}⟩)=r((f))$. Since $f\in r((f))$, there exists $n\in {\mathbb{N}}_{+}$ and a finite set $J\subseteq I$ such that ${\sum }_{j\in J}f{g}_j=f^n$. Thus, $f^n$ can be generated by $\{f{g}_i:i\in J\}$ and $X_{f^n}={\cup }_{j\in J}{X}_{f{g}_j}$. For any $x\in r((f))$, $x^n\in r((f^n))$ and so $x\in r((f^n))$. Also, for any $x\in r((f^n))$, $x^m=a{f}^n=(a{f}^{n-1})f\in (f)$ yields $x\in r((f))$. Hence $r((f))=r((f^n))$ and then

$$X_f=X_{f^n}={\cup }_{j\in J}{X}_{f{g}_j}\subseteq {\cup }_{j\in J}{X}_{g_j}.$$

Therefore, $\{X_{g_j}:j\in J\}\subseteq \{X_{g_i}:i\in I\}$ is an open subcover of $X_f$ and so $X_f$ is quasi-compact.

vii) If an open set is a finite union of some $X_f$, then it is quasi-compact by vi). Otherwise, if an open set $O$ is quasi-compact, then there exists an open cover $\{X_{f_i}:i\in I\}$. Since $\{X_f:f\in A\}$ forms a basis of Zariski topology, we can take $I$ such that ${\cup }_{i\in I}{X}_{f_i}=O$. As it has a finite subcover, there is finite $J\subseteq I$ with $O={\cup }_{j\in J}{X}_{f_j}$ and so $O$ is a finite union of some $X_f$. \end{proof}

Text-Ex. 1.19

$\mathrm{Spec}A$ is irreducible iff $\mathrm{Nil}(A)$ is a prime ideal.

\begin{proof} Note that $\mathrm{Spec}A$ is irreducible iff $X_f\cap X_g=X_{fg}\ne \varnothing$ for any $X_f,X_g\ne \varnothing$ iff for any non-nilpotent $f,g$, $fg$ is also not nilpotent iff $\mathrm{Nil}(A)$ is a prime ideal. \end{proof}