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If $R$ is a local ring with maximal ideal $m$ then show that $R_m \cong R$.

I know that maximals ideals are prime and every element in $R \setminus m$ is a unit, but how can I use this to solve the question?

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    Check that $ R $ satisfies the universal property of localization. Intuitively, localization at $ m $ adds inverses for every element in $ R $ that is not in $ m $; however, if $ R $ is already local with maximal ideal $ m $, then anything not in $ m $ is already invertible, which means localizing at $ m $ gets you nothing new. – 2017-02-11
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    @Starfall What is the universal property ? – 2017-02-11

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Since $R$ is local, every element of $R\setminus M$ is an unit.

For any $\frac{r}{s}\in R_M$, map $\frac{r}{s}\mapsto rs^{-1}$. (If $s$ were not already a unit of $R$, this would not be possible.)

You can work to confirm that this map is

  1. well-defined
  2. a ring homomorphism
  3. a bijection
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    Is the map from R_m to R? – 2017-02-11
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    @Bob yes, $R_M$ to $R$. You don't have to find the inverse to prove it is a bijection. Just show it is one-to-one and onto. – 2017-02-11
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    @Bob you can show $r\mapsto r/1$ is the inverse, also. The idea is that every $r/s$ is equivalent to $rs^{-1}/1$, and $Rs^{-1}=R$, multiplying by a unit doesn't change much – 2017-02-11
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    For well defined: I show that if $r/u=r'/u'$ then $ru^{-1}=r'u'^{-1}$? – 2017-02-11
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    @bob You are correct for the well-defined comment. – 2017-02-11
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    For Injectivity, show the kernel is zero. Surjectiveity is obvious if you think about it! – 2017-02-11
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    You got Injectivity. Can't you see what happens with $r/1$ for surjectivity? – 2017-02-11
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    @bob yes, except for one missing prime on an r. I don't understand your confusion on surjectivity. If you know $r/1$ maps to $r$ for any $r\in R$, that is already surjectivity. – 2017-02-11
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    @Bob No, because every $r$ has a preimage in $R_M$... That is the definition of surjectivity. Dunno what u you are thinking of. – 2017-02-11
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    I get it now thanks. – 2017-02-11