Sketch 4

Localization can make things bigger or smaller

Huh...

I want to know if localizations can be thought of as adding inverses or not. Initially, I thought it was just adding inverses, but I'm starting to realize that in some cases it can be thought of more realistically as identifying elements with 1.

Date Started: March 7, 2024
Date Finished: Unfinished

All Localizations of $\mathbb{Z}/6\mathbb{Z}$, $\mathbb{Z}/8\mathbb{Z}$, and $\mathbb{Z}/12\mathbb{Z}$

To study all the localizations of $\mathbb{Z}/6\mathbb{Z}$, we should look at all multiplicative subsets of $\mathbb{Z}/6\mathbb{Z}$. Here they are:

I have drawn the in this lattice to show inclusions. Now in localizing with respect to all of these multiplicative sets, we can notice some patterns. It so happens that all prime ideals of

Localizing $\mathbb{Z}/6\mathbb{Z}$ at $\{1\}$, $\{1,5\}$

Intuitively, we should have $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{1\}}=\mathbb{Z}/6\mathbb{Z}}}$, since $1$ is already invertible. We can look at this in two ways. Firstly, we can see via the construction. If we now consider elements $s^{-1}m$ where $s\in \{1\}$ and $m\in \mathbb{Z}/6\mathbb{Z}$, we have a bijection mapping $1^{-1}m$ in the localization to $m$ in $\mathbb{Z}/6\mathbb{Z}$ and vice versa. The result is also clear via the universal property: for any morphism from $\mathbb{Z}/6\mathbb{Z}$ to an object (ring or module) $M$ which sends 1 to an invertible element, it commutes obviously with the identity map $id:\mathbb{Z}/6\mathbb{Z}\to \mathbb{Z}/6\mathbb{Z}$. Since 5 is already invertible mod 6, we can also see pretty immediately that $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{1,5\}}=\mathbb{Z}/6\mathbb{Z}}}$.

Localizing $\mathbb{Z}/6\mathbb{Z}$ at $\{2,4\}$, $\{1,2,4\}$, or $\{1,2,4,5\}$

We start with $\{2,4\}$, I will claim that $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{2,4\}}=\mathbb{Z}/3\mathbb{Z}}}$ together with the natural quotient map. We can indeed see this via the universal property. For any map $\phi$ mapping $\mathbb{Z}/6\mathbb{Z}$ to an object $M$ where $\phi (2)$ and $\phi (4)$ are units, we have a map from $\mathbb{Z}/3\mathbb{Z}$ to $M$ mapping $0$ to $0$, $1$ to $\phi (2)$, and $2$ to $\phi (4)$ which is clearly a valid homomorphism making the diagram commute. Now making the conjecture that iterated localization works (I'll show this later but have also shown it on my MATH 250 problem set), we must have that $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{2,4\}}=(\mathbb{Z}/6\mathbb{Z}{)}_{\{1,2,4\}}=(\mathbb{Z}/6\mathbb{Z}{)}_{\{1,2,4,5\}}=\mathbb{Z}/3\mathbb{Z}}}$.

Localizing $\mathbb{Z}/6\mathbb{Z}$ at $\{3\}$, $\{1,3\}$, or $\{1,3,5\}$

We start with $\{3\}$, I will claim that $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{3\}}=\mathbb{Z}/2\mathbb{Z}}}$ together with the natural quotient map. We can indeed see this via the universal property. For any map $\phi$ mapping $\mathbb{Z}/6\mathbb{Z}$ to an object $M$ where $\phi (3)$ is a unit, we have a map from $\mathbb{Z}/3\mathbb{Z}$ to $M$ mapping $0$ to $0$, $1$ to $\phi (3)$ which is clearly a valid homomorphism making the diagram commute. Now via the conjecture that iterated localization works, we must have that $\boxed{{\displaystyle (\mathbb{Z}/6\mathbb{Z}{)}_{\{3\}}=(\mathbb{Z}/6\mathbb{Z}{)}_{\{1,3\}}=(\mathbb{Z}/6\mathbb{Z}{)}_{\{1,3,5\}}=\mathbb{Z}/2\mathbb{Z}}}$.

Localizing $\mathbb{Z}/6\mathbb{Z}$ at $\{1,2,3,4,5\}$

In this case, we can again localize iteratively and find that this is just $\boxed{{\displaystyle 0}}$. Alternatively, notice that in any map, if $3$ goes to a unit, then 2 must go to zero (or else the unit is a zerodivisor, which is not possible) but then zero must be a unit, which is only possible if we consider the zero ring.

Commentary

We must first acknowledge some properties of $\mathbb{Z}/6\mathbb{Z}$: all nonzero prime ideals and maximal, and it turned out in this case that all localizations ended up also being localizations at prime ideals except for the one at $\{1,2,3,4,5\}$. Somewhat surprisingly to me (at least when I realized this), our ring actually shrinks here when we localize. This is unlike the case below with $\mathbb{Z}$ where we will see that we can localize $\mathbb{Z}$ to get $\mathbb{Q}$. Notice also that many of our localizations gave fields because we ended up quotienting by a maximal ideal. There is this idea that if we localize some element $x$, all elements in $y\in \text{Ann}(x)$ must go to zero (for example in our construction we can simply multiply $y=y\cdot 1=y\cdot \frac{x}{x}$ to get 0), so we essentially quotient by $\text{Ann}(x)$. In all cases, we can see how localization precisely leaves what we want. When localizing at $\{2,4\}$, it just leaves the set $\{0,2,4\}$, and kills 3, and when localizing at $\{3\}$, it just leaves $\{0,3\}$ and kills $2$ and $4$.

In the future, I'll also only consider multiplicative subsets which do not contain units. This is because adding in units does not change anything by iterated localization, and no two non-units can ever multiply to a unit, so any time we have units in a multiplicative subset, we can just take out all units and what we are left with should still be a multiplicative subset.

Now to study localization in $\mathbb{Z}/8\mathbb{Z}$, notice the scarcity of multiplicative subsets. Nilpotents can't be in multiplicative subsets, so we're just left localizing things which are already units, since all nonunits are nilpotent in $\mathbb{Z}/8\mathbb{Z}$. In fact, this is what happens in any object of the form $\mathbb{Z}/p^k\mathbb{Z}$.

Localizing $\mathbb{Z}/8\mathbb{Z}$ at a subset of $\{1,3,5,7\}$

This is now obviously just $\boxed{{\displaystyle \mathbb{Z}/8\mathbb{Z}}}$.

All Localizations of $\mathbb{Z}$

Localizations of $\mathbb{Z}[x]$

Localizations of $\mathbb{R}[x]$

Localizations of $\mathbb{C}[x]$