An Analogy Between Integer Factorization and Decomposition in Noetherian Spaces

 

Definition. A topology space $X$$X$ is called Noetherian if it satisfies the descending chain condition for closed subsets: for any for any sequence $Y_1\supseteq Y_2...$$Y_1\supseteq Y_2...$ of closed subset $Y_i$$Y_i$ of $X$$X$, there is an integer $m$$m$ such that $Y_m=Y_{m+1}=...$$Y_m=Y_{m+1}=...$

Remark. By definition, for a Noetherian ring $R$$R$, $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$​​ is Noetherian.

Recall tha a topology space $Y$$Y$ is called irreducible if $Y=Y_1\cup Y_2⟹Y_1=Y$$Y=Y_1\cup Y_2\implies Y_1=Y$ or $Y_2=Y$$Y_2=Y$ for any closed set $Y_1,Y_2$$Y_1,Y_2$​.

This property looks like prime number and we know that on affine scheme, $V(I)$$V(I)$ is irreducible iff $I$$I$ is prime ideal.

A subset of a topological space is called irreducible if it is irreducible resepct to the subspace topology.

Recall that irreducible component of a topological space $X$$X$ is the maximal irreducible set, whcih is closed, since

$U$$U$ is irreducible iff $\bar{U}$$\overline U$ is irreducible.

Proposition. In a Noetherian space X, every closed subset $Y$$Y$ can be expressed as finite union $\begin{array}{c}Y=\bigcup _{i=1}^n{Y}_i\end{array}$$Y=\bigcup_{i=1}^n Y_i\\$ of irreducible component, and this decomposition is unique.

The proof will looks similar to the proof of the fundamental theorem of arithmetic. Readers should image that $\cup$$\cup$ as product since $V(ab)=V(a)\cup V(b)$$V(ab)=V(a)\cup V(b)$​ and treat irreducible component as prime number. Recall that $p$$p$ is prime if $p|ab$$p|ab$ implies $p|a$$p|a$ or $p|b$$p|b$.

Let's draw a parallel proof of this proposition and the fundamental theorem of arithmetic.

Proof.

Let us proof the existence of the decomposition first. Let $S\subseteq \mathbb{N}$$S\subseteq\mathbb N$ be the set of the natural number which is not prime and not product of prime.

Let us proof the existence of the decomposition first. Let $S$$S$ be the set of nonempty closed subsets of $Y$$Y$ whcih can not be written as a finite union of irreducibe components.

Since $\mathbb{N}$$\mathbb N$ is well ordered, then we could find the minimal of $S$$S$, denote as $m$$m$

Since $X$$X$ is Noetherian space, by axiom of choice, we can find a minimal element of $S$$S$ denote as $T$$T$.

Then we could find two nonprime numbers $a,b$$a,b$ and $ab=m$$ab=m$. But since $a,b<m$$a,b< m$, $a,b$$a,b$ is the product of prime number, hence $m$$m$ is the product of prime number, therefore $S=\mathrm{\varnothing }$$S=\empty$.

Then we could find two proper closed subset of $T$$T$ such that $T_1\cup T_2=T$$T_1\cup T_2=T$. If we could not, then $T=\mathrm{\varnothing }\cup T$$T=\empty\cup T$ is the unique way to decompose $T$$T$ to two closed subsets, but that means $T$$T$ is irreducible but $T$$T$ is not irreducible since $T$$T$ is not finite union of irreducible components. But $T_1,T_2\subset T⟹T_1,T_2$$T_1,T_2\subset T\implies T_1,T_2$ is union of irreducible components, hence $T$$T$ is union of irreducible components, hence $S=\mathrm{\varnothing }$$S=\empty$.

Now let us proof the uniqueness.

Let $p_1...p_n=q_1...q_m$$p_1...p_n = q_1...q_m$ be two prime factorizations. Then $p_1$$p_1$ divides the right side, hence $p_1|q_i$$p_1|q_i$ for some $i$$i$. Since both are prime, $p_1=q_1$$p_1=q_1$. Cancel them and continue by induction.

Let $Y=Y_1\cup Y_2\cup ...\cup Y_n=Z_1\cup Z_2\cup ...\cup Z_m$$Y=Y_1\cup Y_2\cup...\cup Y_n=Z_1\cup Z_2\cup...\cup Z_m$, then $\begin{array}{c}{Y}_i=Y_i\cap Y=Y_i\cap \bigcup _{j=1}^m{Z}_j=\bigcup _{j=1}^m(Y_i\cap Z_j)\end{array}$$Y_i=Y_i\cap Y=Y_i\cap\bigcup_{j=1}^m Z_j=\bigcup_{j=1}^m (Y_i\cap Z_j)\\$. But $Y_i$$Y_i$ is irreducible, hence $Y_i=Y_i\cap Z_j$$Y_i=Y_i\cap Z_j$ for a $j$$j$ hence $Y_i\subseteq Z_j$$Y_i\subseteq Z_j$, but $Y_i$$Y_i$ is irreducible component, which means it is maximal, hence $Y_i=Z_j$$Y_i=Z_j$. $◻$$\square$