Math 202B: Lecture 5

***All problems in this lecture are due 01/20 at 23:59***

***No lecture on 01/16***

By now you should have a reasonably good feeling for subalgebras. Unfortunately, this is only half the battle. Let $\mathcal{A}$ and $\mathcal{B}$ be algebras, and let $\Phi \colon \mathcal{A} \to \mathcal{B}$ be an algebra homomorphism. Since $\mathcal{A}$ and $\mathcal{B}$ are vector spaces, and since $\Phi$ is a linear transformation, $\mathrm{Ker}(\Phi)$ is a subspace of $\mathcal{A}$ and $\mathrm{Im}(\Phi)$ is a subspace of $\mathcal{B}.$ Since $\Phi$ is not just a linear map but an algebra homomorphism, one hopes that its kernel and image will have additional structure.

Problem 5.1. Prove that $\mathrm{Im}(\Phi)$ is a subalgebra of $\mathcal{B}.$

It is not necessarily true that $\mathrm{Ker}(\Phi)$ is a subalgebra of $\mathcal{A}$ , because it is not necessarily true that $\Phi(I_\mathcal{A})=0_\mathcal{B}.$ However, the other subalgebra properties, namely closure under conjugation and closure under multiplication, do hold for $\mathrm{Ker}(\Phi).$ In fact, $\mathrm{Ker}(\Phi)$ is not just closed under multiplication – it has the stronger property that $AK \in \mathrm{Ker}(\Phi)$ for any $A \in \mathcal{A}$ and any $K \in \mathrm{Ker}(\Phi).$ This observation leads to the consideration of subspaces of $\mathcal{A}$ which have this absorption property.

Definition 5.1. A subspace $\mathcal{J}$ of $\mathcal{A}$ is said to be an ideal if it is closed under conjugation and absorbs multiplication, in the sense that $AJ$ and $JA$ lie in $\mathcal{J}$ for any $J \in \mathcal{J}$ and all $A \in \mathcal{A}.$

Subalgebras of $\mathcal{A}$ are smaller algebras embedded in $\mathcal{A}.$ They inherit the algebra structure of $\mathcal{A}$ under restriction. Ideals on the other hand are special subspaces of $\mathcal{A}$ which can be used to produce smaller algebras by collapsing rather than restricting. More precisely, since an ideal $\mathcal{J}$ in $\mathcal{A}$ is a subspace we can form the quotient vector space $\mathcal{A}/\mathcal{J}$ whose points are translations of $\mathcal{J},$

$[A]=\{A+J \colon J \in \mathcal{J}\}.$

The zero vector in the quotient space is $[0_\mathcal{A}]=\mathcal{J},$ and linear combinations are defined by

$\alpha_1[A_1]+\alpha_2[A_2]=[\alpha_1 A_1 + \alpha_2 A_2].$

The linear transformation

$\Pi \colon \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{J}$

defined by $\Pi(A) = [A]$ has kernel $\mathcal{J},$ so by the rank-nullity theorem

$\dim \mathcal{A}/\mathcal{J}=\dim \mathcal{A}-\dim \mathcal{J}.$

The fact that $\mathcal{J}$ is not just a subspace but an ideal allows us to go further and put an algebra structure on $\mathcal{A}/\mathcal{J}$ by defining conjugation as $[A]^*=[A^*]$ and multiplication as $[A][B]=[AB].$

Problem 5.2. Check that $\mathcal{A}/\mathcal{J}$ really is an algebra.

We conclude that in our quest to understand the structure of a given algebra $\mathcal{A}$ , we have to understand not just its subalgebras but also its ideals. Thankfully, in many ways ideals are simpler and easier to understand than subalgebras. For example, recall that we have decided to view the set of all subalgebras of $\mathcal{A}$ as a poset under inclusion, making it an induced subposet of the lattice of subspaces of $\mathcal{A}$ , but not an induced sublattice: the max of two subalgebras of $\mathcal{A}$ is not just the span of their union, but the algebra generated by their union. Concerning the poset of ideals of $\mathcal{A}$ , this is an induced sublattice of the lattice of all subspaces without needing any additional constructions.

Problem 5.3. Given two ideals $\mathcal{J},\mathcal{K}$ in $\mathcal{A}$ , show that

$\mathrm{Span}(\mathcal{J} \cup \mathcal{K})= \mathcal{J}+\mathcal{K}=\{J+K \colon J \in \mathcal{J},\ K \in \mathcal{K}\}.$

As we saw in Lecture 4, we have a correspondence between partitions of $X$ and subalgebras of $\mathcal{F}(X)$ . The combinatorial objects which parameterize ideals of $\mathcal{F}(X)$ are simpler — they are just subsets of $\mathcal{X}$ . Given a subset $S \subseteq X,$ we define a corresponding ideal in $\mathcal{F}(X)$ by

$\mathcal{I}(S) = \{A \in \mathcal{F}(X) \colon A(x)=0 \text{ for all }x \in S\}.$

Thus $\mathcal{I}(S)$ is the set of all functions in $\mathcal{F}(X)$ which vanish on every point of $S.$ Hence, we call $\mathcal{I}(S)$ the vanishing ideal of $\mathcal{S}.$

Problem 5.4. Prove that $\mathcal{I}(S)$ really is an ideal in $\mathcal{F}(X).$ Moreover, show that for subsets $S \leq T$ of $X$ we have $\mathcal{I}(T) \leq \mathcal{I}(S).$

A nice feature of vanishing ideals in $\mathcal{F}(X)$ is that they are coordinate spaces relative to the elementary basis $\{E_x \colon x \in X\}.$

Problem 5.5. Show that elementary functions $\{E_x \colon x \in X\backslash S\}$ form a basis of $\mathcal{I}(S).$ Moreover, show that the quotient algebra $\mathcal{F}(X)/\mathcal{I}(S)$ is isomorphic to the function algebra $\mathcal{F}(S).$

We can now view $S \mapsto \mathcal{I}(S)$ as an order-preserving function from the lattice of subsets of $X$ to the lattice of ideals of $\mathcal{F}(X)$ .

Problem 5.6. Show that the mapping $S \mapsto \mathcal{I}(X)$ is injective.

With the above problems solved, we will have an isomorphism between the lattice of subsets of $X$ and the lattice of ideals of $\mathcal{F}(X)$ once we show surjectivity.

Theorem 5.2. The mapping $S \mapsto \mathcal{J}(S)$ is surjective.

Proof: Given an arbitrary ideal $\mathcal{J}$ of $\mathcal{F}(X),$ we need to construct a corresponding point set $S \subseteq X$ such that $\mathcal{I}(S) = \mathcal{J}.$ A natural candidate is the variety defined by $\mathcal{J}$ , which by definition is the set

$V(\mathcal{J}) = \{x \in X \colon A(x)=0 \text{ for all }A \in \mathcal{J}\}$

all points in $X$ at which every function $A \in \mathcal{J}$ vanishes. By construction, we have $\mathcal{J} \leq \mathcal{I}(V(\mathcal{J}))$ and we need to show that the reverse inclusion also holds

Let $S = V(\mathcal{J}).$ In order to show $\mathcal{I}(S) \leq \mathcal{J},$ it is sufficient to show that $E_x \in \mathcal{J}$ for all $x \in X\backslash S.$ This is because we know that the set $\{E_x \colon x \in X \backslash S\}$ is a basis of $\mathcal{I}(S).$ So, fix a point $x \in X \backslash S.$ Since $\mathcal{S}$ is the variety cut out by $\mathcal{J},$ there must exist a function $F_x \in \mathcal{J}$ which does not vanish at $x$ , else $x$ would be a point of $S.$ Furthermore, we can assume WLOG that $F_x(x)=1.$ We do not have any information about the values of the function $F_x$ at any other points. However, multiplying $F_x$ by the elementary function $E_x$ produces something very simple, namely $E_xF_x=E_x.$ Since $F_x \in \mathcal{J}$ and $\mathcal{J}$ is an ideal, we have $E_x \in \mathcal{J}.$ $\square$

The above proof showed that for any subset $S \subseteq X,$ we have

$\mathcal{I}(V(\mathcal{I}(S)))=\mathcal{I}(S).$

This is a very simple finite set version of Hilbert’s Nullstellensatz, a basic theorem of algebraic geometry on vanishing sets of polynomial ideals which will be discussed in Math 202C.

Math 202B: Lecture 5

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