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Give a commutative ring $R$ and a multiplicative subset $S$ of $A$, we have the normal localization map $\lambda_S: R \rightarrow S^{-1}R$. How does one prove that this is an epimorphism?

So given some $f, g: S^{-1}R \rightarrow C$ such that $f \circ \lambda_S = g \circ \lambda_S$, I can see that $f$ and $g$ agree on elements with denominator 1. But I need to show that they agree on all elements in $S^{-1}R$.

sqrt-3299
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2 Answers2

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If $s\in S$, then you already have that $f(\frac{s}{1}) = g(\frac{s}{1})$. Now note that this is a unit, hence its image is a unit in $C$. Units have unique inverses, and the image of the inverse is the inverse of the image, so $$f\left(\frac{1}{s}\right) = \left(f\left(\frac{s}{1}\right)\right)^{-1} = \left( g\left(\frac{s}{1}\right)\right)^{-1} = g\left(\frac{1}{s}\right).$$ Now, given an arbitrary element $\frac{r}{s}\in S^{-1}R$, we have $f(\frac{r}{s}) = f(\frac{r}{1}\frac{1}{s}) = f(\frac{r}{1})f(\frac{1}{s}) = g(\frac{r}{1})g(\frac{1}{s}) = g(\frac{r}{s})$.

But in fact you do not need to assume that $S$ includes $1$, or that $C$ has a $1$, or that the maps must be unital. The conclusion follows using a Zigzag: given $t\in S$, the canonical map $\lambda_S$ given by $\lambda_S(r) = \frac{rt}{t}$ is a well-defined ring homomorphism that does not depend on the choice of $t$, and we have $$\begin{align*} f\left(\frac{r}{s}\right) &= f\left(\frac{1}{s}\frac{rt}{t}\right)\\ &= f\left(\frac{1}{s}\right)f\left(\lambda_S(r)\right)\\ &= f\left(\frac{1}{s}\right)g\left(\lambda_S(r)\right)\\ &= f\left(\frac{1}{s}\right)g\left(\frac{rt}{t}\right)\\ &= f\left(\frac{1}{s}\right)g\left(\frac{srt}{st}\right)\\ &= f\left(\frac{1}{s}\right)g\left(\frac{st}{t}\right)g\left(\frac{r}{s}\right)\\ &= f\left(\frac{1}{s}\right)g\left(\lambda_S(s)\right)g\left(\frac{r}{s}\right)\\ &= f\left(\frac{1}{s}\right)f\left(\lambda_S(s)\right)g\left(\frac{r}{s}\right) \\ &= f\left(\frac{1}{s}\right)f\left(\frac{st}{t}\right)g\left(\frac{r}{s}\right)\\ &= f\left(\frac{st}{st}\right)g\left(\frac{r}{s}\right)\\ &= f\left(\frac{t}{t}\right)g\left(\frac{r}{s}\right)\\ &= f\left(\lambda_S(1)\right)g\left(\frac{r}{s}\right)\\ &= g\left(\lambda_S(1)\right)g\left(\frac{r}{s}\right)\\ &= g\left(\frac{t}{t}\right)g\left(\frac{r}{s}\right)\\ &= g\left(\frac{tr}{ts}\right) = g\left(\frac{r}{s}\right). \end{align*}$$

It's the same as the argument to show that two (not necessarily unital) ring homomorphisms from $\mathbb{Q}$ that agree on $\mathbb{Z}$ must in fact be equal. See this answer

(You can even avoid assuming that $R$ itself has a $1$ by replacing $\frac{1}{s}$ with $\frac{t}{st}$ throughout.)

Arturo Magidin
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  • Thank you for that! – sqrt-3299 Sep 10 '19 at 19:10
  • As long as we're generalizing, you don't even need to assume that you're dealing with a ring, i.e. a preadditive category with one object, but you might as well just work with an arbitrary preadditive category! (or a category, or...) – Kevin Carlson Sep 10 '19 at 21:44
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    @KevinCarlson: Or without going that far, note that we only use the multiplicative properties of $f$ and $g$, so this is about the multiplicative semigroup structure, not the ring structure. – Arturo Magidin Sep 10 '19 at 21:47
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How about lets just use universal property of localization.

Lets denote $k = f\lambda = g\lambda : A\to C$, then we can verify that $k$ maps every thing in $S$ to invertible element in $C$.

  • Basically for any $s\in S$, $\lambda(s)$ is invertible in $S^{-1}A$, so there is $t\in S^{-1}A$, such that $\lambda(s)t = 1$, then we get $1=f(\lambda(s)t) = f(\lambda(s))f(t) = k(s)f(t)$. So $f(t)$ is the inverse of $k(s)$ in $C$.

Then the universal property of localization $S^{-1}A$ shows, there is unique $h: S^{-1}A \to C$ such that

$$h\lambda = k$$

But we know $f$ and $g$ both satisfy the above condition, then the uniqueness of $k$ implies $f = g$.

Feng Guan
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