Commutative Algebra and Algebraic Geometry (3) ideals and their geometry

Constructions of ideal (1)Lattice of idealsProduct, quotient, radicalGeometry of ideal (1)Geometry of Radical idealsGeometry of sum and intersectionGeometry of quotient ideals Geometry of the Chinese Remainder TheoremChinese Remainder Theorem for variety, a Categorical approachConstruction of ideals (2)Contraction and ExtensionGeometry of ideal (2)Geometry of contraction and extensionPrime and Maximal idealsGeometric meaning of prime and maximal ideals

In this essay, we first introduce some constructions of ideals, and their properties. Then interpret them in the world of geometry. Readers may already know the definition of an ideal. It is easy to see that $I\subseteq R$$I\subseteq R$ is an ideal if and only if $I$$I$ is a sub-$R$$R$-module of $R$$R$. We will use this point of view when we prove some propositions.

Constructions of ideal (1)

Lattice of ideals

Now let us consider the poset of the sub-$R$$R$-modules of $R$$R$, denoted as $(I,\subseteq )$$(I,\subseteq)$.

It is easy to see that

$$\begin{array}{}\text{(1)}& sup\{I,J\}=I+J:=\{a+b\mid a\in I,b\in J\}\end{array}$$

and

$$\begin{array}{}\text{(2)}& inf\{I,J\}=I\cap J\end{array}$$

Hence we get a lattice $(I,+,\cap )$$(I,+,\cap)$ and two most natural ways to construct new ideals from two.

Remark. The reason that I say $+,\cap$$+,\cap$ is the most natural way is they are the left and right adjoints of $\mathrm{\Delta }:I\to (I,I)$$\Delta:I\to (I,I)$.

For more details, see Math Essays: The Naturalness of Addition and Multiplication: A Categorical Perspective (marco-yuze-zheng.blogspot.com).

Product, quotient, radical

Definition. Let $I,J$$I,J$ be two ideals of $R$$R$. Define the product of $I$$I$ and $J$$J$ to be

$$\begin{array}{}\text{(3)}& I\cdot J:=(\{ab\mid a\in I,b\in J\})\end{array}$$

i.e. the submodule generated by $\{ab\mid a\in I,b\in J\}$$\{ab \mid a\in I, b\in J\}$.

Remark.

Notice that in general the set $\{ab\mid a\in I,b\in J\}$$\{ab \mid a\in I, b\in J\}$ is not an ideal. Consider $I=(x,y)=J\subset \mathbb{Q}[x,y]$$I=(x,y)=J\subset \mathbb{Q}[x,y]$.

Then $x^2,y^2\in \{ab\mid a\in I,b\in J\}$$x^2, y^2\in \{ab \mid a\in I, b\in J\}$ but $x^2+y^2\notin \{ab\mid a\in I,b\in J\}$$x^2 + y^2 \notin \{ab \mid a\in I, b\in J\}$.

Definition. Let $I,J$$I,J$ be two ideals of $R$$R$. The quotient of $I$$I$ by $J$$J$ is defined by

$$\begin{array}{}\text{(4)}& I:J:=\{a\in R\mid aJ\subseteq I\}\end{array}$$

Notice that we could view $a\in I:J$$a\in I:J$ as an $R-$$R-$module homomorphism from $J$$J$ to $I$$I$.

Since $r(a+b)=ra+rb=r(a)+r(b),r(sa)=rsa=sra=sr(a)$$r(a+b)=ra+rb=r(a)+r(b),r(sa)=rsa=sra=sr(a)$​. Then it is easy to see this forms a sub $R-$$R-$module of $R$$R$, i.e. an ideal.

Example. $(6):(2):=\{n\in \mathbb{Z}\mid n(2)\subseteq (6)\}=(3)$$(6):(2):=\{n\in\mathbb{Z} \mid n(2)\subseteq (6)\}=(3)$

Definition. We call

$$\begin{array}{}\text{(5)}& \sqrt{I}:=\{a\in R\mid \mathrm{\exists }n\in \mathbb{N},a^n\in I\}\end{array}$$

the radical of $I$$I$.

It is not obvious that $\sqrt{I}$$\sqrt{I}$ forms an ideal. So let us prove it.

Obviously, $0\in \sqrt{I}$$0\in \sqrt{I}$.

• If $a\in \sqrt{I}$$a\in \sqrt{I}$ then $ra\in \sqrt{I}$$ra\in \sqrt{I}$.

Suppose $a^n\in I$$a^n\in I$, then $(ra{)}^n=r^n{a}^n\in I$$(ra)^n=r^n a^n\in I$. Hence $ra\in \sqrt{I}$$ra\in \sqrt{I}$.

• If $a,b\in \sqrt{I}$$a,b\in \sqrt{I}$ then $a+b\in \sqrt{I}$$a+b\in \sqrt{I}$.

Assume $a^n\in I$$a^n\in I$ and $b^m\in I$$b^m\in I$. Then consider $(a+b{)}^{n+m}$$(a+b)^{n+m}$.

$$\begin{array}{}\text{(6)}& (a+b{)}^{n+m}=\sum _{i=0}^{n+m}(\genfrac{}{}{0ex}{}{n+m}{i})a^i{b}^{n+m-i}\end{array}$$

Observe that for each $a^i{b}^{n+m-i}$$a^i b^{n+m-i}$, either $i\ge n$$i\ge n$ or $n+m-i\ge m$$n+m-i\ge m$. Hence each $a^i{b}^{n+m-i}\in I$$a^i b^{n+m-i}\in I$, this implies $(a+b{)}^{n+m}\in I$$(a+b)^{n+m}\in I$.

Note that $I\subseteq \sqrt{I}$$I\subseteq \sqrt{I}$, if $\sqrt{I}=I$$\sqrt{I}=I$, then we say $I$$I$ is a radical ideal. Notice that if $I=\sqrt{J}$$I=\sqrt{J}$, then $\sqrt{I}=I=\sqrt{J}$$\sqrt{I}=I=\sqrt{J}$.

That is, $\sqrt{I}$$\sqrt{I}$ is a closure operator. Hence we have the following Galois Connection.

Let $i:(\sqrt{I},\subseteq )\to (I,\subseteq )$$i:(\sqrt{I},\subseteq)\to (I,\subseteq)$ be the inclusion map from the poset of radical ideals to the poset of ideals.

Then obviously

$$\begin{array}{}\text{(7)}& \sqrt{I}\subseteq J⟺I\subseteq i(J)\end{array}$$

Proposition. If $I,J\in \sqrt{I},I\cap J\in (\sqrt{I},\subseteq )$$I,J\in \sqrt{I}, I\cap J\in (\sqrt{I},\subseteq)$.

Proof. If $a^n\in I\wedge a^m\in J$$a^n\in I \wedge a^m\in J$, then $a^{max\{n,m\}}\in I\cap J$$a^{\max\{n,m\}}\in I\cap J$. $◻$$\square$

Example.

• Let $(0)$$(0)$ be the zero ideal, $\sqrt{(0)}$$\sqrt{(0)}$ is called the nilradical ideal of $R$$R$. The elements of $\sqrt{(0)}$$\sqrt{(0)}$ are called nilpotent elements.

• If $R$$R$ is an integral domain, then $(0)$$(0)$ is a radical ideal.

• Consider $(p^n)\subseteq \mathbb{Z}$$(p^n)\subseteq \mathbb{Z}$, then $\sqrt{(p^n)}=(p)$$\sqrt{(p^n)}=(p)$.

• Let $I=I(X)$$I=I(X)$, then $I(X)$$I(X)$ is a radical ideal. Since $f^n(x)=0\mathrm{\forall }x\in X⟹f(x)=0\mathrm{\forall }x\in X$$f^n(x)=0 \forall x\in X \implies f(x)=0 \forall x\in X$.

Definition. If $\sqrt{(0)}=(0)\subseteq R$$\sqrt{(0)}=(0)\subseteq R$, then we call $R$$R$ reduced.

Lemma. An ideal $I\subseteq R$$I\subseteq R$ is radical if and only if $R/I$$R/I$ is reduced.

Proof. ${\overline{a}}^n=0⟺a\in I⟺\overline{a}=0◻$$\bar a^n=0 \iff a\in I \iff \bar a=0 \quad \square$​

Remark. We will show the geometry of nilradical ideal of $R$$R$ after we introduce $\mathrm{Spec}:\mathrm{CRing}\to \mathrm{Top}$$\mathrm{Spec}:\mathrm{CRing}\to\mathrm{Top}$.

Operations on ideals in PID.

Let $R$$R$ be a PID and $I=(a)=(p_1^{a_1}\cdots p_n^{a_n}),J=(b)=(p_1^{b_1}\cdots p_n^{b_n})$$I=(a)=(p_1^{a_1} \cdots p_n^{a_n}), J=(b)=(p_1^{b_1} \cdots p_n^{b_n})$.

• $(a)\cap (b)=(p_1^{max(a_1,b_1)},\cdots ,p_n^{max(a_n,b_n)})$$(a)\cap(b)=(p_1^{\max (a_1,b_1)}, \cdots, p_n^{\max (a_n,b_n)})$

• $(a)+(b)=(p_1^{min(a_1,b_1)},\cdots ,p_n^{min(a_n,b_n)})$$(a)+(b)=\left(p_1^{\min (a_1,b_1)}, \cdots, p_n^{\min (a_n,b_n)}\right)$

• $(a)\cdot (b)=(ab)=(p_1^{(a_1+b_1)},\cdots ,p_n^{(a_n+b_n)})$$(a)\cdot(b)=(ab)=\left(p_1^{(a_1+b_1)}, \cdots, p_n^{(a_n+b_n)}\right)$

• $(a):(b)=(p_1^{max\{(a_1-b_1),0\}},\cdots ,p_n^{max\{(a_n-b_n),0\}})$$(a):(b)=\left(p_1^{\max\{(a_1-b_1),0\}}, \cdots, p_n^{\max\{(a_n-b_n),0\}}\right)$

• $\sqrt{(a)}=(p_1^{min(a_1,1)},\cdots ,p_n^{min(a_n,1)})$$\sqrt{(a)}=\left(p_1^{\min (a_1,1)}, \cdots, p_n^{\min (a_n,1)}\right)$

Definition. (Coprime ideals) Two ideals $I$$I$ and $J$$J$ are coprime if $(I)+(J)=(R)$$(I)+(J)=(R)$.

Lemma. For any two ideals $I$$I$ and $J$$J$ in $R$$R$, $I\cdot J\subseteq I\cap J$$I\cdot J\subseteq I\cap J$, and $\sqrt{I\cdot J}=\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$$\sqrt{I\cdot J}=\sqrt{I\cap J}=\sqrt{I}\cap\sqrt{J}$.

Proof. $I\cdot J\subseteq I\cap J$$I\cdot J\subseteq I\cap J$ since $I\cdot J\subseteq I$$I\cdot J\subseteq I$ and $I\cdot J\subseteq J$$I\cdot J\subseteq J$, $I\cap J$$I\cap J$ is the greatest lower bound.

For $\sqrt{I\cdot J}=\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$$\sqrt{I\cdot J}=\sqrt{I\cap J}=\sqrt{I}\cap\sqrt{J}$, we show a circular inclusion.

By $I\cdot J\subseteq I\cap J$$I\cdot J\subseteq I\cap J$ and $\sqrt{(-)}$$\sqrt{(-)}$ is monotone we get that $\sqrt{I\cdot J}=\sqrt{I\cap J}$$\sqrt{I\cdot J}=\sqrt{I\cap J}$. If $a\in \sqrt{I\cap J}$$a\in \sqrt{I\cap J}$, then $\mathrm{\exists }n,a^n\in I\cap J$$\exists n,a^n\in I\cap J$.

Hence $a^n\in I$$a^n\in I$ and $a^n\in J$$a^n\in J$, hence $a\in \sqrt{I}$$a\in \sqrt{I}$ and $a\in \sqrt{J}$$a\in \sqrt{J}$. Hence $a\in \sqrt{I}\cap \sqrt{J}$$a\in \sqrt{I}\cap \sqrt{J}$.

Finally, if $a\in \sqrt{I}\cap \sqrt{J}$$a\in \sqrt{I} \cap \sqrt{J}$, then $a^n\in I,a^m\in J$$a^n\in I, a^m\in J$, hence $a^{n+m}\in I\cdot J$$a^{n+m}\in I\cdot J$ and thus $a\in \sqrt{I\cdot J}$$a\in \sqrt{I\cdot J}$. $◻$$\square$

Geometry of ideal (1)

Geometry of Radical ideals

Proposition: Hilbert's Nullstellensatz.

Recall the Galois connection between ideals of $K[x_1,...,x_n]$$K[x_1,...,x_n]$ and subsets of ${\mathbb{A}}_K^n$$\mathbb{A}_K^n$.

$$\begin{array}{}\text{(8)}& I(X)\supseteq S⟺X\subseteq V(S)\end{array}$$

Now we know that the image of $V(-)$$V(-)$ is affine variety and the image of $I(-)$$I(-)$ is radical ideal.

In addition, when $K$$K$ is an algebraically closed field, the image of $I(-)$$I(-)$ is the whole $(\sqrt{I},\subseteq )$$(\sqrt{I},\subseteq)$.

Since $IV(J)=\sqrt{J}$$IV(J)=\sqrt{J}$ (We will prove this in the future).

By the property of Galois connection, for $f(a)\le b⟺a\le g(b)$$f(a)\le b \iff a\le g(b)$, we have

$$\begin{array}{}\text{(9)}& f(a)\le f(a)⟺a\le gf(a),g(b)\le g(b)⟺fg(b)\le b\end{array}$$

Hence

$$\begin{array}{}\text{(10)}& f(a)\le f(gf)(a)=fgf(a)=(fg)f(a)\le f(a)⟹fgf=f\end{array}$$

Similarly, $gfg=g$$gfg=g$.

Hence consider $f:A\to \mathrm{Im}f,g:B\to \mathrm{Im}g$$f:A\to \mathrm{Im} f, g:B\to \mathrm{Im} g$. Then $fg={\mathrm{Id}}_{\mathrm{Im}f},gf={\mathrm{Id}}_{\mathrm{Im}g}$$fg=\mathrm{Id}_{\mathrm{Im} f}, gf=\mathrm{Id}_{\mathrm{Im} g}$. i.e. $\mathrm{Im}f\cong \mathrm{Im}g$$\mathrm{Im} f\cong \mathrm{Im} g$ as a poset.

Hence we get an anti-isomorphism between the poset of radical ideals and algebraic varieties. $◻$$\square$

Geometry of sum and intersection

According to

$$\begin{array}{}\text{(11)}& I(X)\supseteq S⟺X\subseteq V(S)\end{array}$$

and since left adjoints preserve colimits, right adjoints preserve limits.

We have $I(X\cup Y)=I(X)\cap I(Y),V(I+J)=V(I)\cap V(J)$$I(X\cup Y)=I(X)\cap I(Y), V(I+J)=V(I)\cap V(J)$.

Hence the union of varieties corresponds to the intersection of ideals, and the sum of ideals corresponds to the intersection of varieties.

Geometry of quotient ideals

Let $Y,Z$$Y,Z$ be two subvarieties of $X$$X$. Then

$$\begin{array}{}\text{(12)}& I(Y-Z)=\{f\in A(X)\mid f(x)=0\mathrm{\forall }x\in Y-Z\}=\{f\in A(X)\mid f(x)g(x)=0\mathrm{\forall }x\in Y,g\in I(Z)\}\end{array}$$

But

$$\begin{array}{}\text{(13)}& \{f\in A(X)\mid f(x)g(x)=0\mathrm{\forall }x\in Y,g\in I(Z)\}=\{f\in A(X)\mid f\cdot I(Z)\subseteq I(Y)\}=I(Y):I(Z)\end{array}$$

Geometry of the Chinese Remainder Theorem

Proposition: Chinese Remainder Theorem. Let $I_1,...,I_n$$I_1,...,I_n$ be ideals in a ring $R$$R$ and consider the ring homomorphism

$$\begin{array}{}\text{(14)}& \phi :R\to R/I_1\times ...\times R/I_n,r⟼(\overline{r},...,\overline{r})\end{array}$$

• $\phi$$\varphi$ is injective iff ${\displaystyle \bigcap _{j=1}^n{I}_j=0}$$\displaystyle\bigcap_{j=1}^n I_j=0$.

• $\phi$$\varphi$ is surjective iff $I_1,...,I_n$$I_1,...,I_n$ are pairwise coprime.

Proof.

The first claim is obvious since $\mathrm{Ker}\phi ={\displaystyle \bigcap _{j=1}^n{I}_j}$$\mathrm{Ker}\varphi=\displaystyle\bigcap_{j=1}^n I_j$.

For the second one:

If $\phi$$\varphi$ is surjective, then $(0_1,0_2,1_i,...,0_n)\in \mathrm{Im}\phi$$(0_1,0_2,1_i,...,0_n)\in \mathrm{Im} \varphi$. Hence there exists $a\in R,a\equiv 1\mathrm{mod}{I}_i,a\equiv 0\mathrm{mod}{I}_j$$a\in R,a\equiv 1\mod I_i,a\equiv 0\mod I_j$.

Then $1=(1-a)+a\in I_i+I_j$$1=(1-a)+a\in I_i+I_j$. Since $a\equiv 1\mathrm{mod}{I}_i⟹1-a\equiv 0\mathrm{mod}{I}_i⟹1-a\in I_i$$a\equiv 1\mod I_i \implies 1-a\equiv 0\mod I_i \implies 1-a\in I_i$.

Hence $I_i+I_j=R$$I_i+I_j=R$.

If $I_1,...,I_n$$I_1,...,I_n$ are pairwise coprime, i.e. $I_i+I_j=R$$I_i+I_j=R$, then there exists $a_i+b_j=1,b_j=1-a_i$$a_i+b_j=1,b_j=1-a_i$.

Then we have

$$\begin{array}{}\text{(15)}& b_j\equiv 1\mathrm{mod}{I}_i,b_j\equiv 0\mathrm{mod}{I}_j\end{array}$$

Hence

$$\begin{array}{}\text{(16)}& B_i=\prod _{j\ne i}{b}_j\equiv 1\mathrm{mod}{I}_i,\prod _{j\ne i}{b}_j\equiv 0\mathrm{mod}{I}_j\mathrm{\forall }j\ne i\end{array}$$

Therefore we have

$$\begin{array}{}\text{(17)}& \phi (B_i)=(0_1,...,1_i,...,0_n)\end{array}$$

and $\phi (B_i)$$\varphi(B_i)$ generate $\mathrm{Im}\phi$$\mathrm{Im} \varphi$. $◻$$\square$

The Geometric meaning of the Chinese Remainder Theorem

Let $X\subseteq {\mathbb{A}}_K^n$$X\subseteq \mathbb{A}_K^n$ be an algebraic variety and $Y_1,...,Y_n$$Y_1,...,Y_n$ be some subvarieties, then consider

$$\begin{array}{}\text{(18)}& \psi :A(X)\to A(Y_1)\times ...\times A(Y_n)\end{array}$$

• $\psi$$\psi$ is injective iff ${\displaystyle \bigcap _{i=1}^{n}I(Y_i)=0⟺V(\bigcap _{i=1}^{n}I(Y_i))=\bigcup _{i=1}^n{Y}_i=V(0)=X}$$\displaystyle\bigcap_{i=1}^n I(Y_i)=0 \iff V(\bigcap_{i=1}^n I(Y_i))=\bigcup _{i=1}^n Y_i=V(0)=X$.

• $\psi$$\psi$ is surjective iff $I(Y_i)+I(Y_j)=A(X)⟺V(I(Y_i)+I(Y_J))=Y_i\cap Y_j=\mathrm{\varnothing }$$I(Y_i)+I(Y_j)=A(X) \iff V(I(Y_i)+I(Y_J))=Y_i\cap Y_j=\emptyset$.

Chinese Remainder Theorem for variety, a Categorical approach

Notice that the coordinate ring functor $A(-)$$A(-)$ is representable, i.e.

$$\begin{array}{}\text{(19)}& A(-)\cong {\mathrm{Hom}}_{K-\mathrm{Variety}}(-,{\mathbb{A}}_K^1)\end{array}$$

Hence it will map colimit to limit. Hence for ${\displaystyle X=\coprod _{i\in I}{Y}_i}$$\displaystyle X=\coprod_{i\in I} Y_i$,

$$\begin{array}{}\text{(20)}& A(X)\cong {\mathrm{Hom}}_{K-\mathrm{Variety}}(\coprod _{i\in I}{Y}_i,{\mathbb{A}}_K^1)\cong \prod _{i\in I}{\mathrm{Hom}}_{K-\mathrm{Variety}}(Y_i,{\mathbb{A}}_K^1)\cong A(Y_1)\times ...\times A(Y_n)\end{array}$$

Construction of ideals (2)

Contraction and Extension

Let $\psi :R\to S$$\psi:R\to S$ be a ring homomorphism.

Definition: Contraction of ideal.

For any ideal $I\subseteq S$$I\subseteq S$, ${\psi }^{-1}(I)$$\psi^{-1}(I)$ is an ideal in $R$$R$. Since if $a,b\in {\psi }^{-1}(I)$$a,b\in \psi^{-1}(I)$, $\psi (a)+\psi (b)=\psi (a+b)\in I$$\psi(a)+\psi(b)=\psi(a+b)\in I$. Hence $a+b\in {\psi }^{-1}(I)$$a+b\in \psi^{-1}(I)$. Similarly $a\in {\psi }^{-1}(I)$$a\in \psi^{-1}(I)$ then $\psi (a)\in I⟹\psi (r)\psi (a)=\psi (ra)\in I⟹ra\in {\psi }^{-1}(I)$$\psi(a)\in I\implies \psi(r)\psi(a)=\psi(ra)\in I \implies ra\in \psi^{-1}(I)$.

When $\psi$$\psi$ is clear, we denote ${\psi }^{-1}(I)$$\psi^{-1}(I)$ as $I^c$$I^c$, contraction of $I$$I$.

Remark: Spec as a functor.

If $\mathfrak{p}$$\mathfrak{p}$ is a prime ideal of $R$$R$, then ${\psi }^{-1}(\mathfrak{p})$$\psi^{-1}(\mathfrak{p})$ is a prime ideal as well. Since if $ab\in {\psi }^{-1}(\mathfrak{p})$$ab\in \psi^{-1}(\mathfrak{p})$, then $\psi (a)\psi (b)\in \mathfrak{p}$$\psi(a)\psi(b)\in \mathfrak{p}$.

This implies $\psi (a)\in \mathfrak{p}$$\psi(a)\in \mathfrak{p}$ or $\psi (b)\in \mathfrak{p}$$\psi(b)\in \mathfrak{p}$. Hence we have $a\in {\psi }^{-1}(\mathfrak{p})$$a\in \psi^{-1}(\mathfrak{p})$ or $b\in {\psi }^{-1}(\mathfrak{p})$$b\in \psi^{-1}(\mathfrak{p})$.

Hence if $ab\in {\psi }^{-1}(\mathfrak{p}),a\in {\psi }^{-1}(\mathfrak{p})$$ab\in \psi^{-1}(\mathfrak{p}),a\in \psi^{-1}(\mathfrak{p})$ or $b\in {\psi }^{-1}(\mathfrak{p})$$b\in \psi^{-1}(\mathfrak{p})$. i.e. ${\psi }^{-1}(\mathfrak{p})$$\psi^{-1}(\mathfrak{p})$ is a prime ideal.

Let $A$$A$ be a ring and define $\mathrm{Spec}(A)=\{\mathfrak{p}\subseteq A\mid \mathfrak{p}\text{ is prime}\}$$\mathrm{Spec}(A)=\{\mathfrak{p}\subseteq A \mid \mathfrak{p}\text{ is prime} \}$, we get a functor

$$\begin{array}{}\text{(21)}& \mathrm{Spec}:{\mathrm{CRing}}^{\mathrm{op}}\to \mathrm{Set},(\psi :R\to S)⟼{\psi }^{-1}:\mathrm{Spec}(S)\to \mathrm{Spec}(R)\end{array}$$

We will define the Zariski topology on $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$, and make $\mathrm{Spec}$$\mathrm{Spec}$ become a functor to $\mathrm{Top}$$\mathrm{Top}$.

Definition: extension of ideal.

Let $J$$J$ be an ideal in $R$$R$, then we call the ideal generated by $\psi (J)$$\psi(J)$, denoted as $J^e$$J^e$, the extension ideal of $J$$J$.

Remark. In general, $\psi (J)$$\psi(J)$ is not an ideal. For example, consider the inclusion map

$$\begin{array}{}\text{(22)}& i:\mathbb{Z}\to \mathbb{Z}[T]\end{array}$$

Then $\mathbb{Z}$$\mathbb{Z}$ is not an ideal of $\mathbb{Z}[T]$$\mathbb{Z}[T]$.

But if $\psi :R\to S$$\psi:R\to S$ is a surjective map, then $\psi (J)$$\psi(J)$ is an ideal.

Proposition. Galois connection between contraction and extension.

Let $\psi :R\to S$$\psi:R\to S$ be a ring homomorphism

• $I\subseteq (I^e{)}^c$$I\subseteq (I^e)^c$ for all $I\subseteq R$$I\subseteq R$

• $(J^c{)}^e\subseteq J$$(J^c)^e\subseteq J$ for all $J\subseteq S$$J\subseteq S$

i.e.

$$\begin{array}{}\text{(23)}& I^e\subseteq J⟺I\subseteq J^c\end{array}$$

Proof. $I\subseteq (I^e{)}^c$$I\subseteq (I^e)^c$ is obvious since $I\subseteq {\psi }^{-1}(\psi (I))$$I\subseteq \psi^{-1}(\psi(I))$. To see $(J^c{)}^e\subseteq J$$(J^c)^e\subseteq J$, notice that $\psi ({\psi }^{-1}(J))\subseteq J$$\psi(\psi^{-1}(J))\subseteq J$. $◻$$\square$

Corollary. $(J\cap J^{\prime }{)}^c=J^c\cap J^{\prime c},(I+I^{\prime }{)}^e=I^e+I^{\prime e}$$(J\cap J')^{c}=J^c\cap J'^c,(I+I')^e=I^e+I'^e$.

Proposition. $(ab{)}^e=a^e{b}^e$$(ab)^e=a^e b^e$.

Proof. If $k\in (ab{)}^e$$k\in (ab)^e$, then $k=\sum s^{\prime }ab$$k=\sum s'ab$, and if $l\in a^e{b}^e$$l\in a^e b^e$, $\sum sa\sum tb=\sum stab$$\sum s a\sum t b=\sum s t ab$. $◻$$\square$

Geometry of ideal (2)

Geometry of contraction and extension

Let $f:X\to Y$$f:X\to Y$ be a morphism of varieties, and consider $f^{\ast }:A(Y)\to A(X),g⟼f^{\ast }(g)=g\circ f$$f^*:A(Y)\to A(X), g\longmapsto f^*(g)=g\circ f$.

• For any subvariety $Z\subseteq X$$Z\subseteq X$ we have

$$\begin{array}{}\text{(24)}& I(f(Z))=\{g\in A(Y)\mid g(f(x))=0\mathrm{\forall }x\in Z\}=\{g\in A(Y)\mid f^{\ast }(g)\in I(Z)\}=(f^{\ast }{)}^{-1}(I(Z))\end{array}$$

Hence taking the image of varieties corresponds to the contraction of ideals.

• For a subvariety $Z\subseteq Y$$Z\subseteq Y$, the zero locus of the extension $I(Z{)}^e$$I(Z)^e$ by $f^{\ast }$$f^*$ is

$$\begin{array}{}\text{(25)}& V(f^{\ast }(I(Z)))=\{x\in X\mid gf(x)=0\mathrm{\forall }g\in I(Z)\}\end{array}$$

But $gf(x)=0⟺f(x)\in \{y\in Y\mid g(y)=0\mathrm{\forall }g\in I(Z)\}⟺x\in f^{-1}\{y\in Y\mid g(y)=0\mathrm{\forall }g\in I(Z)\}=f^{-1}V(I(Z))=f^{-1}(Z).$$gf(x)=0 \iff f(x)\in \{y\in Y \mid g(y)=0 \forall g\in I(Z)\} \iff x\in f^{-1} \{y\in Y \mid g(y)=0 \forall g\in I(Z)\} = f^{-1} V(I(Z)) = f^{-1}(Z).$

Hence the extension of ideals corresponds to the inverse image. This shows us that the inverse image of closed set is still closed. Hence Zariski topology makes $f$$f$ become continuous.

Lemma. (Ideals in quotient rings.) Let $I$$I$ be an ideal in $R$$R$, consider the quotient map $\pi :R\to R/I$$\pi:R\to R/I$.

Then extension and contraction are a pair of inverses between ideals in $R/I$$R/I$, and $\{\text{ideals }J\text{ in }R,J\supseteq I\}$$\{\text{ideals } J \text{ in } R, J\supseteq I\}$.

Proof. Observe that $\pi$$\pi$ is surjective, hence $\pi (J)$$\pi(J)$ is an ideal.

• For any ideal $J\subseteq R/I$$J\subseteq R/I$ we have $(J^c{)}^e=\pi ({\pi }^{-1}(J))=J$$(J^c)^e=\pi(\pi^{-1}(J))=J$.

• For any ideal $J\subseteq R$$J\subseteq R$ with $J\supseteq I$$J\supseteq I$, we have $(J^e{)}^c={\pi }^{-1}(\pi (J))=J+I=J$$(J^e)^c=\pi^{-1}(\pi(J))=J+I=J$.

Prime and Maximal ideals

Definition. Let $R$$R$ be a ring and $\mathfrak{a}$$\mathfrak{a}$ an ideal. Then

• An ideal $\mathfrak{p}$$\mathfrak{p}$ is called prime if $ab\in \mathfrak{p}$$ab\in \mathfrak{p}$ implies $a\in \mathfrak{p}$$a\in \mathfrak{p}$ or $b\in \mathfrak{p}$$b\in \mathfrak{p}$.

• An ideal $\mathfrak{m}$$\mathfrak{m}$ is called maximal if $\mathfrak{m}\subseteq \mathfrak{a}$$\mathfrak{m}\subseteq \mathfrak{a}$ implies $\mathfrak{m}=\mathfrak{a}$$\mathfrak{m}=\mathfrak{a}$.

Proposition.

• An ideal $\mathfrak{p}$$\mathfrak{p}$ is prime iff $R/\mathfrak{p}$$R/\mathfrak{p}$ is an integral domain.

• An ideal $\mathfrak{m}$$\mathfrak{m}$ is a maximal ideal iff $R/\mathfrak{m}$$R/\mathfrak{m}$ is a field.

Proof.

The first claim is easy to verify. Since $\overline{a}\overline{b}=0\in R/\mathfrak{p}⟺ab\in \mathfrak{p}$$\bar a\bar b=0\in R/\mathfrak{p} \iff ab\in \mathfrak{p}$.

For the second one, notice that the $\pi :R\to R/\mathfrak{m}$$\pi: R\to R/\mathfrak{m}$ is a quotient map, hence contraction and extension are a pair of inverses.

Hence $(0)\subseteq R/\mathfrak{m}$$(0)\subseteq R/\mathfrak{m}$ is a maximal ideal. Then for any $a\ne 0\in R/\mathfrak{m},(a)=R=(1)$$a\ne 0\in R/\mathfrak{m},(a)=R=(1)$. Hence $\mathrm{\exists }b,ab=1$$\exists b, ab=1$. Hence $R/\mathfrak{m}$$R/\mathfrak{m}$ is a field. Conversely, it is clear that for a field, $(0)$$(0)$ is a maximal ideal. $◻$$\square$

Proposition. Every ring $R$$R$ contains a maximal ideal.

We need to apply Zorn's lemma. Let $\mathrm{\Sigma }:=(I,\subseteq )$$\Sigma:=(I,\subseteq)$ be the poset of ideals. $\mathrm{\Sigma }$$\Sigma$ is not empty since $(0)\in \mathrm{\Sigma }$$(0)\in \Sigma$.

Now let ${\mathrm{\Sigma }}^{\prime }\subseteq \mathrm{\Sigma }$$\Sigma' \subseteq \Sigma$ be a totally ordered subposet. Then consider $\mathfrak{b}:=\bigcup _{\mathfrak{a}\in {\mathrm{\Sigma }}^{\prime }}\mathfrak{a}$$\mathfrak{b}:=\bigcup_{\mathfrak{a}\in \Sigma'} \mathfrak{a}$. It is easy to see this forms a proper ideal.

By Zorn's lemma, $\mathrm{\Sigma }$$\Sigma$ contains a maximal element. $◻$$\square$

Corollary. For any ideal $\mathfrak{a}\subseteq R$$\mathfrak{a}\subseteq R$, there exists a maximal ideal $\mathfrak{m}$$\mathfrak{m}$ such that $\mathfrak{a}\subseteq \mathfrak{m}$$\mathfrak{a}\subseteq \mathfrak{m}$.

Proof. Consider the ring $R/\mathfrak{a}$$R/\mathfrak{a}$, there exists a maximal $\mathfrak{m}$$\mathfrak{m}$ ideal in $R/\mathfrak{a}$$R/\mathfrak{a}$.

Then it follows from the fact that contraction and extension are a pair of inverses between ideals in $R/I$$R/I$, and $\{\text{ideals }J\text{ in }R,J\supseteq I\}$$\{\text{ideals } J \text{ in } R, J\supseteq I\}$. $◻$$\square$

Geometric meaning of prime and maximal ideals

Let $Y$$Y$ be a non-empty subvariety of $X$$X$, corresponding to $I(Y)\subseteq A(X)$$I(Y)\subseteq A(X)$.

If $I(Y)$$I(Y)$ is not prime, then there exists $f,g\in A(X)$$f,g\in A(X)$ such that $fg$$fg$ vanishes on $Y$$Y$ but $f,g$$f,g$ do not.

Hence $V(f)\ne Y,V(g)\ne Y$$V(f)\ne Y, V(g)\ne Y$ but $V(f)\cup V(g)=V(fg)=Y$$V(f)\cup V(g)=V(fg)=Y$. Hence we can write $Y$$Y$ as a non-trivial union of two subvarieties. If a variety has such a property, then we call it irreducible, otherwise reducible.

Conversely, if $Y=Y_1\cup Y_2$$Y=Y_1\cup Y_2$ for two subvarieties $Y_1,Y_2$$Y_1,Y_2$, then we could find $f\in I(Y_1)-I(Y_2),g\in I(Y_2)-I(Y_1)$$f\in I(Y_1)-I(Y_2), g\in I(Y_2)-I(Y_1)$ and $fg$$fg$ vanishing on $Y$$Y$. Hence $I(Y)$$I(Y)$ is not prime.

Therefore, it is easy to see that prime ideals correspond to irreducible varieties and maximal ideals correspond to points.

Hence $A(Y)$$A(Y)$ is an integral domain $⟺$$\iff$ $Y$$Y$ is irreducible.

Example: The affine $n$$n$ space ${\mathbb{A}}_K^n$$\mathbb A_{K}^n$ is irreducible. Since $A({\mathbb{A}}_K^n)=K[T_1,...,T_n]$$A(\mathbb A_K^n)=K[T_1,...,T_n]$.