## Prime ideals

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 $\mathrm{nil}\left(R\right)=\left\{nilpotent elements of\phantom{\rule{.5em}{0ex}}R\right\}.$
• The radical of an ideal $𝔞$ is the ideal $\sqrt{𝔞}$ of $R$ corresponding to the ideal $\mathrm{nil}\left(R/𝔞\right)$ 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. $\mathrm{nil}\left(R\right)=\left\{nilpotent elements of\phantom{\rule{.5em}{0ex}}R\right\}=\sqrt{0}=\bigcap _{𝔭\in \mathrm{Spec}\left(R\right)}𝔭.$
2. $\sqrt{a}= {r\in R|{r}^{k}\in 𝔞\phantom{\rule{.5em}{0ex}}for some\phantom{\rule{.5em}{0ex}}k\in {ℤ}_{>0}} =\bigcap _{\underset{𝔭\supseteq 𝔞}{𝔭\in \mathrm{Spec}\left(R\right)}}𝔭.$

 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 $k\in {ℤ}_{>0}.$ 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 $y k k ∈ ℤ > 0$ 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 $𝔭 ∩ y k k ∈ ℤ > 0 = ∅ .$ Let $x,z\in R$ such that $x\notin 𝔭$ and $z\notin 𝔭.$ Then $𝔭 + ( x ) ≠ 𝔭 and 𝔭 + ( z ) ≠ 𝔭 .$ So ${y}^{m}={p}_{1}+{r}_{1}x$ and ${y}^{n}={p}_{2}+{r}_{2}z$ for some ${p}_{1},{p}_{2}\in 𝔭,\phantom{\rule{.5em}{0ex}}{r}_{1},{r}_{2}\in R,$ and some $m,n\in {ℤ}_{>0}.$ So $y m + n ∈ 𝔭 + ( x z ) .$ 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 $Spec ( R ) = { prime ideals of R } .$
• If $S\subseteq R$ is a subset of $R$ define $V\left(S\right)=\left\{prime ideals containing\phantom{\rule{.5em}{0ex}}S\right\}.$
• 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 ${ closed sets of Spec ( R ) } = V ( S ) S ⊆ R .$

Let $X=\mathrm{Spec}\left(R\right).$ If $E$ is a subset of $R$ then $V ( E ) = V ( ⟨ E ⟩ ) = V ⟨ E ⟩ ,$ where $⟨E⟩$ is the ideal generated by $E$. Also $V ( 0 ) = X , V ( 1 ) = ∅ , V ⋃ i M i = V ∑ i M i = ⋂ i V ( M i ) , V ( 𝔞 ∩ 𝔞 ' ) = V ( 𝔞 𝔞 ' ) = V ( 𝔞 ) ∪ V ( 𝔞 ' ) ,$ 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 $V : { ideals of R } → { closed sets in Spec ( R ) } , and I : { subsets of Spec ( R ) } → { radical ideals of R } , I ( V ( 𝔞 ) ) = 𝔞 for an ideal 𝔞 of R , and V ( I ( S ) ) = S _ , for a subset S ⊆ Spec ( R ) ,$ and ${ closed sets in Spec ( R ) } ↔ { radical ideals of R } .$

Let $f:A\to B$ be a ring homomorphism. The map $f * : Spec ( B ) → Spec ( A ) 𝔟 ↦ f - 1 ( 𝔟 )$ 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 $\mathrm{im}{f}^{*}=V\left(\mathrm{ker}f\right).$ 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 $Spec ( A 𝔭 ) → Spec ( A ) is injective with image { prime ideals of A contained in 𝔭 } .$ 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 $Spec ( B ) → Spec ( A ) 𝔟 ↦ 𝔟 ∩ A$ 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 $𝔮\cap A=𝔭,$ $Spec ( B ) → Spec ( A ) 𝔮 ↦ 𝔭$
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 $Spec ( B ) → Spec ( A ) 𝔭 2 ∪ 𝔮 1 ↦ 𝔭 1$ then there exists a prime ideal ${𝔮}_{2}$ of $B$ such that $Spec ( B ) → Spec ( A ) 𝔮 2 ↦ 𝔭 2 ∪ ∪ 𝔮 1 ↦ 𝔭 1$ 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 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 $Spec(A) = {prime ideals x of A}.$
• The maximal ideal spectrum of $A$ is $Maxspec(A) = {maximal ideals x of A}$
• 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, $C(Y) = {f:Y→ℝ | f is continuous }, X = Spec(C(Y)), and |X| = Maxspec(C(Y)).$ For $y\in Y$ let Then $Y →∼ |X| y ↦ ker(evy) is a homeomorphism$ where the topology on $|X|$ has basic open sets

Let $A$ be a commutative ring.

• An element $f\in A$ is nilpotent if there exists $k\in {ℤ}_{>0}$ such that ${f}^{k}=0.$
• The radical of an ideal $𝔞$ of $A$ is $𝔞 = {f∈A | there exists k∈ℤ>0 such that fk∈𝔞} .$
• 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 $nil(A) = {nilpotent elements of A}.$

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 $V: {subsets of A} → {subsets of Spec(A)} S ↦ V(S) = {x∈Spec(A) | x⊇S} I: {subsets of Spec(A)} → {subsets of A} E ↦ I(E) = ⋂x∈Ex$

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

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

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

## Notes and References

Where are these from?

References?