I've been reading the Queen Mary Note of Maurice Auslander 's paper Representation dimension of artin algebras. However I met some difficulties that are not so relevant to the topic. The following is such one.

On page six he describes the image of a morphism between two commutative rings in the category of rings with identities.

Let R and T be commutative rings and $f:R\rightarrow T$ a ring homomorphism. Let $f(R)=S$. We have $f_1:T\rightarrow T\otimes_ST$ defined by $f_1(t)=1\otimes t$ and $f_2(t)=t\otimes 1$.

Let $T'=\{t\in T| f_1(t)=f_2(t)\}$. Clearly $T'$ contains S, so that $f$ maps $R$ into $T'$. Prove that the inclusion map $S\rightarrow T'$ is an epimorphism in the category of rings and the inclusion map $T'\rightarrow T$ is a strong monomorphism.

I tried to use the notion dominion and pushout to prove the epimorphism part but failed. enter image description here

  • $\begingroup$ Concerning the second part: Regular monomorphisms are strong. And $T' \to T$ is the equalizer of $f_1,f_2$. $\endgroup$ – HeinrichD Apr 6 '17 at 14:15
  • $\begingroup$ I am not confident that $S \to T'$ is always an epimorphism of comm. rings. Notice that $T'$ is the dominion aka epicenter of $S \hookrightarrow T$, whose elements have a matrix-like representation (see P. Mazet, Characterization des epimorphismes par generators and relateurs, Seminaire Samuel; or stacks.math.columbia.edu/tag/04VY), but these matrices have coefficients in $T$, not necessarily in $T'$. Also, in the literature on epimorphisms of commutative rings (which I am quite familiar with) I cannot find the statement that $S \to T'$ is an epimorphism. $\endgroup$ – HeinrichD Apr 6 '17 at 14:30
  • $\begingroup$ @HeinrichD I will post this paper if it helps $\endgroup$ – user12580 Apr 6 '17 at 14:38
  • $\begingroup$ If you mean a link to Auslander's paper, this would be appreciated. I could not find it online. $\endgroup$ – HeinrichD Apr 6 '17 at 14:44
  • $\begingroup$ @HeinrichD I find the paper through my university's data base, and I could not find it online either $\endgroup$ – user12580 Apr 6 '17 at 14:53

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