Prime ideals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 12 April 2012

Prime ideals

Let $R$ be a commutative ring.

• A prime ideal is an ideal $𝔭$ of $R$ such that $R/𝔭$ is an integral domain.
• A maximal ideal is an ideal $𝔪$ of $R$ such that $R/𝔪$ is a field.
• The nilradical of $R$ is the ideal $\mathrm{nil}\left(R\right)=\left\{nilpotent\; elements\; ofR\right\}.$
• The radical of an ideal $𝔞$ is the ideal $\sqrt{𝔞}$ of $R$ corresponding to the ideal $\mathrm{nil}(R/𝔞)$ in $R/𝔞.$
• A reduced ring is a ring $R$ such that $\sqrt{0}=0.$
• A radical ideal is an ideal $𝔞$ such that $\sqrt{𝔞}=𝔞.$

Let $R$ be a ring and let $𝔞$ be an ideal of $R$.

1. $\mathrm{nil}\left(R\right)=\left\{nilpotent\; elements\; ofR\right\}=\sqrt{0}=\bigcap _{𝔭\in \mathrm{Spec}\left(R\right)}𝔭.$
2. $\sqrt{a}= {r\in R|r^k\in 𝔞for\; somek\in {\mathbb{Z}}_{>0}} =\bigcap _{\underset{𝔭\supseteq 𝔞}{𝔭\in \mathrm{Spec}\left(R\right)}}𝔭.$

		Proof.
	b. Follows from a. and the correspondence between ideals of $R$ containing $𝔞$ and ideals of $R/𝔞.$ a. If $x\in \sqrt{0}$ then $x^k=0$ for some $k\in {\mathbb{Z}}_{>0}.$ If $𝔭$ is a prime ideal then $0\in 𝔭$ and so $x\in 𝔭.$ So $x\in \bigcap 𝔭.$ If $y\notin \sqrt{0}$ then $${y^k|k\in {\mathbb{Z}}_{>0}}$$ is a multiplicative subset of $R$ that does not contain $0$. By Zorn's lemma there exists an ideal $𝔭$ of $R$ maximal with respect to the condition that $$𝔭\cap {y^k|k\in {\mathbb{Z}}_{>0}} =\varnothing .$$ Let $x,z\in R$ such that $x\notin 𝔭$ and $z\notin 𝔭.$ Then $$𝔭+\left(x\right)\ne 𝔭and𝔭+\left(z\right)\ne 𝔭.$$ So $y^m=p_1+r_{1}x$ and $y^n=p_2+r_{2}z$ for some $p_1,p_2\in 𝔭,r_1,r_2\in R,$ and some $m,n\in {\mathbb{Z}}_{>0}.$ So $$y^{m+n}\in 𝔭+\left(xz\right).$$ So $xz\notin 𝔭.$ $\square$

Let $R$ be a commutative ring.

• The Krull dimension of $R$ is the maximal length of a chain of prime ideals of $R$.
• The prime spectrum of $R$ is the set $$\mathrm{Spec}\left(R\right)=\left\{prime\; ideals\; ofR\right\}.$$
• If $S\subseteq R$ is a subset of $R$ define $V\left(S\right)=\left\{prime\; ideals\; containingS\right\}.$
• If $Y\subseteq \mathrm{Spec}\left(R\right)$ define $I\left(Y\right)=\bigcap _{𝔭\in Y}𝔭.$
• The Zariski topology on $\mathrm{Spec}\left(R\right)$ is given by $$\left\{closed\; sets\; of\mathrm{Spec}\right(R\left)\right\}= {V\left(S\right)|S\subseteq R} .$$

Let $X=\mathrm{Spec}\left(R\right).$ If $E$ is a subset of $R$ then $$V\left(E\right)=V(⟨E⟩)=V\left(\sqrt{⟨E⟩}\right),$$ where $⟨E⟩$ is the ideal generated by $E$. Also $$\begin{array}{rl}V\left(0\right)=X,& V\left(1\right)=\varnothing ,\\ V\left(\bigcup _i{M}_i\right)=V\left(\sum _i{M}_i\right)=\bigcap _{i}V\left(M_i\right),& V(𝔞\cap 𝔞\text{'})=V(𝔞𝔞\text{'})=V\left(𝔞\right)\cup V(𝔞\text{'}),\end{array}$$ for any family $\left\{M_i\right\}$ of subsets of $R$ and any ideals $𝔞$ and $𝔞\text{'}$ of $R.$

Let $R$ be a commutative ring. Then $$\begin{array}{rlcl}V:& \left\{ideals\; ofR\right\}& \to & \left\{closed\; sets\; in\mathrm{Spec}\right(R\left)\right\},and\\ I:& \left\{subsets\; of\mathrm{Spec}\right(R\left)\right\}& \to & \left\{radical\; ideals\; ofR\right\},\\ I\left(V\right(𝔞\left)\right)=\sqrt{𝔞}& for\; an\; ideal𝔞ofR,& and& V\left(I\right(S\left)\right)=\stackrel{\_}{S},for\; a\; subsetS\subseteq \mathrm{Spec}\left(R\right),\end{array}$$ and $$\left\{closed\; sets\; in\mathrm{Spec}\right(R\left)\right\}\leftrightarrow \left\{radical\; ideals\; ofR\right\}.$$

Let $f:A\to B$ be a ring homomorphism. The map $$\begin{array}{rccc}{f}^{*}:& \mathrm{Spec}\left(B\right)& \to & \mathrm{Spec}\left(A\right)\\ & 𝔟& \mapsto & f^{-1}\left(𝔟\right)\end{array}$$ is a continuous map. The reason that one uses the prime spectrum instead of the maximal spectrum is that the map $f^{*}$ is not well defined on maximal ideals: if $𝔟$ is a maximal ideal $f^{-1}\left(𝔟\right)$ may not be maximal.

Any ring homomorphism $f:A\to B$ can be factored as $A\to \mathrm{im}f\to B$ where the first map is surjective and the second is injective.

If $f:A\to B$ is surjective then $f^{*}:\mathrm{Spec}\left(B\right)\to \mathrm{Spec}\left(A\right)$ is injective with $\mathrm{im}{f}^{*}=V\left(\ker f\right).$ When $f:A\to B$ is injective, in other words, if $A\subseteq B$ is a ring extension, the situation is more subtle.

If $𝔭$ is a prime ideal of $A$ then $A\subseteq A_{𝔭}$ gives $$\mathrm{Spec}\left(A_{𝔭}\right)\to \mathrm{Spec}\left(A\right)is\; injective\; with\; image\left\{prime\; ideals\; ofAcontained\; in𝔭\right\}.$$ The best results are obtained when $A\subseteq B$ is an integral extension.

Let $A\subseteq B$ be an integral extension.

1. (Lying over.) The map $$\begin{array}{ccc}\mathrm{Spec}\left(B\right)& \to & \mathrm{Spec}\left(A\right)\\ 𝔟& \mapsto & 𝔟\cap A\end{array}$$ is surjective.
2. (Going up.) The Krull dimension of $A$ and $B$ are equal.

The names of these results comes from the fact that this theorem is often stated in the following form: Let $A\subseteq B$ be an integral extension.

1. (Lying over.) If $𝔭$ is a prime ideal of $A$, then there is a prime ideal $𝔮$ of $B$ such that $𝔮\cap A=𝔭,$ $$\begin{array}{ccc}\mathrm{Spec}\left(B\right)& \to & \mathrm{Spec}\left(A\right)\\ 𝔮& \mapsto & 𝔭\end{array}$$
2. (Going up.) If ${𝔭}_1\subseteq {𝔭}_2$ are prime ideals of $A$ and ${𝔮}_1$ is a prime ideal of $B$ such that $$\begin{array}{ccc}\mathrm{Spec}\left(B\right)& \to & \mathrm{Spec}\left(A\right)\\ & & {𝔭}_2\\ & & \cup \\ {𝔮}_1& \mapsto & {𝔭}_1\end{array}$$ then there exists a prime ideal ${𝔮}_2$ of $B$ such that $$\begin{array}{ccc}\mathrm{Spec}\left(B\right)& \to & \mathrm{Spec}\left(A\right)\\ {𝔮}_2& \mapsto & {𝔭}_2\\ \cup & & \cup \\ {𝔮}_1& \mapsto & {𝔭}_1\end{array}$$ If $Q_1$ is lying over $P_1$ and $P_1\subseteq P_2$ then there is an ideal $Q_2$ of $B$ lying over $P_2$ such that $Q_1\subseteq Q_2.$
3. (Incomparability.) If $P$ is an ideal of $A$ and $Q_1$ and $Q_2$ are prime ideals of $B$ lying over $P$ then $Q_1\subseteq Q_2.$
4. (Going down.) If $Q_2$ is lying over $P_2$ and $P_1\subseteq P_2$ then there is an ideal $Q_1$ of $B$ lying over $P_1$ such that $Q_1\subseteq Q_2.$

WHAT IS THE PROPER STATEMENT OF THESE IN TERMS OF $\mathrm{Spec}\left(R\right)$?

Prime and maximal ideals

• An integral domain is a commutative ring $A$ such that
  1. if $a_1,a_2\in A, a_1\ne 0$ and $a_1{a}_2=0$ then $a_2=0.$
• A field is a commutative ring $A$ such that
  1. if $a\in A$ and $a\ne 0$ then there exists $a^{-1}\in A$ such that $a{a}^{-1}=1.$

Let $A$ be a commutative ring.

• A prime ideal of $A$ is an ideal $x$ of $A$ such that $\frac{A}{x}$ is an integral domain.
• A maximal ideal of $A$ is an ideal $x$ of $A$ such that $\frac{A}{x}$ is a field.
• The spectrum of $A$ is $$\mathrm{Spec}\left(A\right)=\left\{prime\; idealsxofA\right\}.$$
• The maximal ideal spectrum of $A$ is $$\mathrm{Maxspec}\left(A\right)=\left\{maximal\; idealsxofA\right\}$$
• The Krull dimension of $A$ is the maximal length of a chain of prime ideals in $A.$

HW: [AM, Ch.1, Ex.26] Let $Y$ be a compact Hausdorff topological space, $$\begin{array}{rcl}C\left(Y\right)& =& \{f:Y\to ℝ|fis\; continuous\},\\ X& =& \mathrm{Spec}\left(C\left(Y\right)\right),and\\ \left|X\right|& =& \mathrm{Maxspec}\left(C\left(Y\right)\right).\end{array}$$ For $y\in Y$ let $$\begin{array}{rrcl}{\mathrm{ev}}_y:& C\left(Y\right)& \to & ℝ\\ & f& \mapsto & f\left(y\right)\end{array}so\; that\; {\mathrm{ev}}_y\left(f\right)=f\left(y\right).$$ Then $$\begin{array}{rcl}Y& \stackrel{\sim }{\to }& \left|X\right|\\ y& \mapsto & \ker \left({\mathrm{ev}}_y\right)\end{array}is\; a\; homeomorphism$$ where the topology on $\left|X\right|$ has basic open sets $$\{x\in \left|X\right||f\notin x\},for\; f\in C\left(Y\right).$$

Let $A$ be a commutative ring.

• An element $f\in A$ is nilpotent if there exists $k\in {\mathbb{Z}}_{>0}$ such that $f^k=0.$
• The radical of an ideal $𝔞$ of $A$ is $$\sqrt{𝔞}=\{f\in A|there\; existsk\in {\mathbb{Z}}_{>0}such\; that{f}^k\in 𝔞\}.$$
• The niradical of $A$ is $\mathrm{nil}\left(A\right)=\sqrt{0}.$
• A radical ideal of $A$ is an ideal $𝔞$ of $A$ such that $\sqrt{𝔞}=𝔞.$
• The ring $A$ is reduced if $A$ satisfies $\sqrt{0}=0.$

HW: Show that $$\mathrm{nil}\left(A\right)=\left\{nilpotent\; elements\; ofA\right\}.$$

HW: Let $𝔞$ be an ideal of $A.$ Show that $\sqrt{𝔞}$ is the ideal of $A$ corresponding to the ideal $\mathrm{nil}\left(\frac{A}{𝔞}\right)$ of the ring $\frac{A}{𝔞}.$

Define $$\begin{array}{rrcl}V:& \left\{subsets\; ofA\right\}& \to & \left\{subsets\; of\mathrm{Spec}\left(A\right)\right\}\\ & S& \mapsto & V\left(S\right)=\{x\in \mathrm{Spec}\left(A\right)|x\supseteq S\}\\ I:& \left\{subsets\; of\mathrm{Spec}\left(A\right)\right\}& \to & \left\{subsets\; ofA\right\}\\ & E& \mapsto & I\left(E\right)=\bigcap _{x\in E}x\end{array}$$

HW: Let $S\subseteq A.$ Show that $$V\left(S\right)=V\left(⟨S⟩\right)=V\left(\sqrt{⟨S⟩}\right),$$ where $⟨S⟩$ is the ideal of $A$ generated by $S.$

HW: Let $E\subseteq \mathrm{Spec}\left(A\right).$ Show that $$I\left(E\right)=I\left(\stackrel{\_}{E}\right),$$ where $\stackrel{\_}{E}$ is the closure of $E$ in the Zariski topology on $\mathrm{Spec}\left(A\right).$

HW: Let $S\subseteq A$ and $E\subseteq \mathrm{Spec}\left(A\right).$ Show that $$I\left(V\left(S\right)\right)=\sqrt{⟨S⟩}andV\left(I\left(E\right)\right)=\stackrel{\_}{E}.$$

HW: Let $A$ be a commutative ring and let $X=\mathrm{Spec}\left(A\right).$ Show that $$\mathrm{nil}\left(A\right)=I\left(X\right).$$

Notes and References

Where are these from?

References

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