Prove that for all free right $R$-module $F$ and for all exact sequences of $R$-modules $$0\to M\xrightarrow{f}N\xrightarrow{g}P\to 0$$ then $$0\to F\otimes_RM\xrightarrow{1\otimes f}F\otimes_RN\xrightarrow{1\otimes g}F\otimes_RP\to 0$$ is an exact sequence.

I proved that Im$(1\otimes f)=$ Ker $(1\otimes g)$ and $1\otimes g$ is an epimorphism. How can I to prove $1\otimes f$ is monomorphism.

Thanks alot!


By choosing a basis $(e_i)_{i\in I}$ of $F$ we have $F=\bigoplus_{i\in I} Re_i$ and $$ F\otimes_R M = \left(\bigoplus_{i\in I} Re_i\right)\otimes_R M \cong \bigoplus_{i\in I} (Re_i \otimes_R M) \cong \bigoplus_{I} M $$ and similar for $F\otimes_R N$. Under this isomorphism the map $1\otimes f\colon F\otimes_R M\to F\otimes_R N$ becomes a map $\bigoplus_I M \to \bigoplus_I N$ given as $\bigoplus_I f$, that is, the $i$th summand of $\bigoplus_I M$ gets mapped into the $i$th summand of $\bigoplus_I N$ via $f$. Since $f$ is injective, this direct sum of injective maps is also injective.

You might as well say that the resulting sequence is just an $I$-indexed direct sum of copies of the original exact sequence and hence still exact, if you define "direct sum of exact sequences" properly.

  • $\begingroup$ Thankyou so much. I got it $\endgroup$ – Nguyễn Hoàng Hiệp May 10 '19 at 8:18
  • $\begingroup$ Can I rewrite $\displaystyle \bigoplus_{I}M$ by $M^{(I)}$? $\endgroup$ – Nguyễn Hoàng Hiệp May 10 '19 at 11:13
  • $\begingroup$ What is the definition of $M^{(I)}$? $\endgroup$ – Christoph May 10 '19 at 13:25
  • $\begingroup$ oh sorry, my bad. I got it, thankyou so much $\endgroup$ – Nguyễn Hoàng Hiệp May 10 '19 at 14:23

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