What is $R[U^{-1}]$ of $\mathbb{Z}/6\mathbb{Z}$, where $U=\{1,3\}$?
I have $R[U^{-1}] = \{0/1,1/1,2/1,3/1,4/1,5/1,0/3,1/3,2/3,3/3,4/3,5/3\}$.
I know I need $\mathbb{Z}/2\mathbb{Z}$ but how can I eliminate some of the elements?
What is $R[U^{-1}]$ of $\mathbb{Z}/6\mathbb{Z}$, where $U=\{1,3\}$?
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commutative-algebra
1 Answers
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I assume the equivalence relation on the "fractions" you write is the usual
$r_1/u_1 = r_2/u_2 \Leftrightarrow$ there is $t\in U$ such that $t(r_1u_2-r_2u_1) = 0.$
Then you should see that because $3\cdot 2 = 0$ in $R$, $2/1 = 0/1$ and $r/3 = r/1$ for all $r\in R$.
In other words, 3 becomes a unit which forces $2$ to become $0$ (which further forces $3=1$) in the localised ring.
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0What happens to 5/1 and 5/3? – 2017-02-18
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0What do you think? Should they not become $1 = 1/1$? Can you formally see that with the relation, or from the fact that the map $R \rightarrow R[U^{-1}]$ is a ring homomorphism? – 2017-02-18
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0$5/1=1/1$ because $3(5.1-1.1)=3(4)=(3.2)(2)=0$? – 2017-02-18
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0For example, yes. I think it is easier though to realise that we have a ring homomorphism $f$ from $R$ to $R[U^{-1}]$, $f(r) = r/1$, which sends 2 to 0 and 3 to 1. That forces $f(4) = 0$ and $f(5)=1$, and the rest follows from that. – 2017-02-18
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0Quick question: For R=Z/10Z and U={1,5} we would also have 2/1=0/1 since 5(2.1-0.1)=10=0 and 3/1=1/1 etc So we get Z/2Z for R[U^-1]? – 2017-04-28
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0Yes. More generally, if we localise $\mathbb{Z}/n$ at $U = \lbrace 1, m \rbrace$ for a divisor $m|n$, and call $k= m/n$, the "localisation" will be $\mathbb{Z}/k$ if $k$ and $m$ are coprime, the zero ring otherwise. – 2017-04-28