I'm trying to solve this question in Hungerford's Algebra

I didn't use the corollary:

And I used this map: $g:S^{-1}R_1\to S^{-1}R_2$, $g(r/s)=f(r)/f(s)$. I'm wondering how to prove using the corollary, is it hard?

Second, the map I have chosen indeed extends $R$? I identified the elements $r/1$ of $S^{-1}R$ as the elements of $r$, is that right?

Any help is welcome

Thanks a lot

  • 1
    $\begingroup$ There is absolutely no need to link to an (illegal) copy of the text in question. Please refer to this meta thread and also this one. $\endgroup$
    – user642796
    Jun 5, 2013 at 11:32
  • $\begingroup$ @ArthurFischer thank you for the links I wasn't aware that this could be problematic to MathExchange legally speaking. However, this book is allocated in another site. $\endgroup$
    – user42912
    Jun 5, 2013 at 23:30
  • $\begingroup$ Another thing to point out is EVERY post has problems with copyright laws in some degree, it's impossible to learn and share knowledge with this kind of laws. $\endgroup$
    – user42912
    Jun 5, 2013 at 23:33

1 Answer 1


You actually don't even need to use an explicit construction of the $F_i=\mathrm{Frac}(R_i)$. If $R_1\hookrightarrow R_2$ is an injection of domains, then, by considering the composite $R_1\hookrightarrow R_2\hookrightarrow F_2$, you get a ring map which, by injectivity, sends non-zero elements of $R_1$ to units (since every non-zero element of $F_2$ is a unit). So the universal property of $R_2\hookrightarrow F_2$ yields a unique ring map $\varphi:F_1\rightarrow F_2$ compatible with the original map $R_1\hookrightarrow R_2$. If the map $R_1\hookrightarrow R_2$ is in fact an isomorphism, then, by similar considerations, you get a unique ring map $\psi:F_2\hookrightarrow F_1$ compatible with $R_2\cong R_1$. Now consider $\psi\circ\varphi:F_1\rightarrow F_1$. It is compatible with $R_1\hookrightarrow F_1$ by construction, so, by the uniqueness part of the universal property of $R_1\hookrightarrow F_1$, it must be the identity. Similarly $\varphi\circ\psi$ is the identity, so $\varphi$ and $\psi$ are mutually inverse isomorphisms.

More generally, if $R$ is a ring, $S$ a multiplicative subset of $R$, and $S^{-1}R$ the localization, with $\alpha:R\rightarrow S^{-1}R$ the canonical map, then the only ring map $\varphi:S^{-1}R\rightarrow S^{-1}R$ such that $\varphi\circ\alpha=\alpha$ is $\varphi=\mathrm{id}_{S^{-1}R}$ (this follows from the uniqueness part of the universal property of $\alpha$). Put succinctly, localizations of $R$ have no non-identity $R$-algebra endomorphisms. This is a particular case of the fact that a localization map is an epimorphism in the category of commutative rings with identity (again this is because of the uniqueness clause in the universal property which characterizes $S^{-1}R$).

Of course, to prove the existence of $S^{-1}R$, one uses an explicit construction, most commonly as a set of equivalence class of fractions. Then, for a ring map $f:R\rightarrow R^\prime$ such that $f(S)$ is contained in the units of $R^\prime$, the unique map $\varphi:S^{-1}R\rightarrow R^\prime$ such that $\varphi\circ\alpha=f$ is given by the formula the OP gives: $\varphi(r/s)=f(r)f(s)^{-1}$. It is well-defined by the definition of the equivalence relation used in the construction of $S^{-1}R$ and the fact that $f(s)\in (R^\prime)^\times$. In the case $\varphi$ is an isomorphism of domains, once the maps between the fields of fractions are written down, it can be checked by hand (using fractions) that it is an isomorphism. So you don't have to use the universal property if you don't want to; the explicit construction as a ring of fractions is sufficient. Note also that, in the notation of the first paragraph, you apply the universal property of localization to the map $R_1\hookrightarrow R_2\hookrightarrow F_2$, not directly to $R_1\hookrightarrow R_2$, since there is no reason that non-zero elements in $R_1$ should be mapped to units in $R_2$; they are mapped to non-zero elements by injectivity, and because $R_2$ injects into $F_2$, the image of a non-zero element in $R_1$ under $R_1\hookrightarrow R_2\hookrightarrow F_2$ is a non-zero element of the field $F_2$, hence invertible, so you can apply the universal property.

  • $\begingroup$ Why does $f$ extend to an isomorphism? $\endgroup$
    – user42912
    Jun 6, 2013 at 17:30
  • $\begingroup$ In the last paragraph? It doesn't. There exists a unique $\varphi:S^{-1}R\rightarrow R^\prime$ such that $\varphi\circ\alpha=f$. I do not claim that this $\varphi$ is an isomorphism. When $R_1\cong R_2$ is an isomorphism of domains, you can check that $\varphi$ is an isomorphism because the map I called $\psi$ above is inverse to it and vice versa. $\endgroup$ Jun 6, 2013 at 17:38

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