I am trying to understand localization of a commutative ring, and when it can be $0$. I confused with some examples and came up to some questions, whose partial answers I couldn't convinced myself. (Definition and notation of localization is recalled after questions.)

Claim 1. If $0\in S$ then $R_S=0$

for consider $(1,1)$ and $(0,1)$. Then $(1,1)\sim (0,1)$ because taking $t=0$ in definition, we have $(1.1-0.1)0=0$. So multiplicative identity is $0$, which forces $R_S=0$.

Claim 2. Even if $0\notin S$, it can happen that $R_S=0$.

Proof: We give example: Consider $\mathbb{Z}/12$, and write its elements as integers $0,1,\cdots,11$, with operations mod 12. Take $S=\{1,3,6,9\}$. Then $S$ is multiplicatively closed in the ring.

In $R_S$, the (image of) $6$ becomes unit. Since $2.6=0$ we get that $2=0$ in $R_S$.

Next, since $6$ is unit and $6=2.3$, and as $2=0$, we get that $6$ is equal to $0$.

Thus, in $R_S$ we have shown that a unit is zero, which forces that $R_S=0$.

Question 1. If $0\in S$ then $R_S=0$; is this correct (with proof given in claim 1)?

Question 2. The above example shows $R_S$ can be zero even if $0\notin S$. Is this assertion and example correct?

Question 3. What is necessary and sufficient condition on $S$ for $R_S$ to be non-zero?

Let $R$ be a commutative ring with unity $1$ different from zero. Let $S$ be a multiplicative subset of $R$ containing $1$ (so $x,y\in S$ implies $xy\in S$).

On $R\times S$, we define $(r,s)\sim (r',s')$ if there is $t\in S$ such that $(rs'-r's)t=0$.

Let $R_S=(R\times S)/\sim$ denote the localization of $R$ at $S$. Write equiv. class of $(r,s)$ by $r/s$.

In the localization $R_S$, the element $1/1$ is multiplicative, and $0/1$ is additice identity.

  • 1
    $\begingroup$ Your $S$ is not multiplicatively closed: $6\times 6$ is zero in $Z/12$, so at least $S$ should contain $0$ to be multiplicatively closed if it will contain $6$. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '18 at 6:30
  • $\begingroup$ @Mariano: Sorry (for silly mistake!) Thanks for noticing me this. $\endgroup$ – Beginner Mar 3 '18 at 6:31

Your example $2$ is incorrect. The set $S=\{1,3,6,9\}$ is not multiplicatively closed: $6\times 6=0$ in $\Bbb Z/12\Bbb Z$ but $0\notin S$.

In fact $R_S=0$ iff $0\in S$. To prove this, note that $1/1=0/1$ in $R_S$ iff there exists $s\in S$ with $s(1-0)=0$.


Question 1 In Claim 1 you're actually proving that any couple of $(r,s),(p,t)$, $r,p\in R, s,t\in S$ are equivalent wich means that $R_{S}$ is consisted in just one class.

Question 2 The thing with Claim 2 is that you have "Zero divisors" ($3\cdot 4=12\equiv 0$(mod$12$), $6\cdot 2=12\equiv0$(mod$12$), $9\cdot6=36\equiv0$(mod$12$)), and again any $(r,s),(p,t)$ are equivalent.

Question 3 You can see that $R_{S}=0$ if and only if $0\in S$. To check that is enough to observe that $(1,1)\sim(0,1)$ which means that exist $s\in S$ such that $(1\cdot1-0\cdot1)\cdot s=0$, $1\cdot s=0$ and this only happens if $s=0$

There is a canonical morphism $\alpha:R\to R_{S}$ given by $\alpha(a)=\frac{a}{1}$. You may prove (and should try to do it in order to convince yourself) that $\alpha$ is one-to-one if and only if $S$ has no zero divisors and so if $R\neq0$ then $R_{S}\neq0$.

  • $\begingroup$ I couldn't understand how injectivity criterian for $R\rightarrow R_S$ is useful to decide whether $R_S$ is zero or not? $\endgroup$ – Beginner Mar 3 '18 at 7:16
  • $\begingroup$ @Beginner Sorry, i got a confusion about the criterion from my notes, but it's corrected now. $\endgroup$ – Alberto Macías Mar 3 '18 at 21:29

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