How do we find localization of a ring?

Question

Let $R = \mathbb Z/6 \mathbb Z$ and let $S=\{1,2,4\}$. Show $ S^{-1}R \cong \mathbb Z/3 \mathbb Z $.

I understand that in $S^{-1}R$ the elements of $S$ must be invertible but I'm not getting a way to prove above isomorphism. Any ideas?

Answer 1

We want to find $(\mathbb{Z}/6\mathbb{Z})_2$. Now $$(\mathbb{Z}/6\mathbb{Z})_2\cong(\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})_2\cong (\mathbb{Z}/3\mathbb{Z})_2\times(\mathbb{Z}/2\mathbb{Z})_2\cong \mathbb{Z}/3\mathbb{Z}.$$ In the last step $(\mathbb{Z}/2\mathbb{Z})_2\cong 0$ because we are inverting $0$ and $(\mathbb{Z}/3\mathbb{Z})_2\cong \mathbb{Z}/3\mathbb{Z}$ because we are inverting something which is already invertible.

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Answer to comment: take $M,N$ two $A$-modules, $S\subseteq A$ multiplicative. Now $$S^{-1}(M\oplus N)\cong (M\oplus N)\otimes S^{-1}A\cong (M\otimes S^{-1}A)\oplus (N\otimes S^{-1}A)\cong S^{-1}M\oplus S^{-1}N.$$ Notice that $\oplus$ and $\times$ are the same since we are dealing with a finite number of modules.

If you don't know about tensor products yet, just show that the natural map $$S^{-1}(M\oplus N)\to S^{-1}M\oplus S^{-1}N$$ given by $\frac{(m,n)}{s} \mapsto (\frac{m}{s},\frac{n}{s})$ is a well-defined isomorphism of $A$-modules.

Answer 2

Overkill but:

1. Since $\frac31=\frac01$, we know the ring has characteristic $3$
2. The localization is a (finite) field since $R$ is a finite von Neumann regular ring.
3. There are only 18 fractions possible, many of which are equivalent. That narrows everything down to $F_3$ and $F_9$.
4. Since everything in $R$ satisfies $x^3=x$, the same is true for $S^{-1}R$, and then it is impossible for any element to have multiplicative order $8$ (which would have to happen in $F_9$.

So the only remaining possibility is $F_3=\mathbb Z/3\mathbb Z$.

Answer 3

According to the following questions, we can also construct $S^{-1} R$ as $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} \quotient{R[x_s \ : \ s \ \in \ S]}{\langle s x_s - 1 \ : \ s \ \in \ S\rangle}$.

• Can the localization $U^{-1} R$ also be constructed as $R[x_s : s \in U]/\langle s x_s - 1 : s \in U\rangle$?
• Localization as a quotient ring

This means that we have the following isomorphisms: $$ \begin{array}{rcl} \{1, 2, 4\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right) & \cong & \quotient{\left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right)[x_1, x_2, x_4]}{\langle x_1 - 1, 2x_2 - 1, 4x_4 - 1\rangle} \\ & \cong & \quotient{\left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right)[x, y, z]}{\langle x - 1, 2y - 1, 4z - 1\rangle} \\ & \cong & \quotient{\mathbb{Z}[x, y, z]}{\langle 6, x - 1, 2y - 1, 4z - 1\rangle} \\ & \cong & \quotient{\mathbb{Z}[y, z]}{\langle 6, 2y - 1, 4z - 1\rangle} \\ \end{array} $$ But now, consider the ideals $I = \langle 6, 2y - 1, 4z - 1\rangle$ and $J = \langle 3, y - 2, z - 1 \rangle$ of $\mathbb{Z}[y, z]$. We claim that $I = J$.

Indeed, we have:

• $3 = 3 - 6y + 6y = 3(1 - 2y) + 6y \in I$
• $y - 2 = 4y - 3y - 2 = 2(2y - 1) - 3y \in I$
• $z - 1 = 4z - 3z - 1 = 4z - 1 - 3z \in I$

This shows that $J \subseteq I$.

On the other hand, we also have:

• $6 = 2 \cdot 3 \in J$
• $2y - 1 = 2y - 4 + 3 = 2(y - 2) + 3 \in J$
• $4z - 1 = 4z - 4 + 3 = 4(z - 1) + 3 \in J$

This shows that $I \subseteq J$. Hence $I = J$ and we may continue our calculation: $$ \begin{array}{rcl} \quotient{\mathbb{Z}[y, z]}{\langle 6, 2y - 1, 4z - 1\rangle} & \cong & \quotient{\mathbb{Z}[y, z]}{\langle 3, y - 2, z - 1 \rangle} \\ & \cong & \quotient{\mathbb{Z}}{\langle 3 \rangle} \\ \end{array} $$ Putting everything together, we obtain: $$ \{1, 2, 4\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right) \cong \quotient{\mathbb{Z}}{\langle 3 \rangle} $$ The following questions are similar:

• Localization at a prime ideal in $\mathbb{Z}/6\mathbb{Z}$
• How to compute localization of module $\mathbb Z/n\mathbb Z$ at the prime ideal $(p)$?