# The exact sequence of tensor product

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.
• Can I rewrite $\displaystyle \bigoplus_{I}M$ by $M^{(I)}$? – Nguyễn Hoàng Hiệp May 10 '19 at 11:13
• What is the definition of $M^{(I)}$? – Christoph May 10 '19 at 13:25