Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 3

The tensor product $M{\otimes }_{R}N$ is the free abelian group on symbols $(m,n)$ modulo additivity in each variable and the balancing relation $(rm,n)=(m,rn)$. It has the universal property that balanced bilinear maps $M\times N\to P$ correspond uniquely to homomorphisms $M{\otimes }_{R}N\to P$.

The mapis well defined and surjective. Its inverse sends $m+IM$ to $(1+I)\otimes m$; elements of $IM$ map to zero because $(1+I)\otimes am=(a+I)\otimes m=0$ for $a\in I$. Hence

Let $k=R/\mathfrak{m}$. Associativity and part i giveIf $M{\otimes }_{R}N=0$, the tensor product on the right is zero. Two nonzero vector spaces over a field have nonzero tensor product, so one factor vanishes. Nakayama lemma then gives $M=0$ or $N=0$.

An $R$-module $F$ is flat when $-{\otimes }_{R}F$ preserves injections, equivalently all finite exact sequences. Since $-{\otimes }_{R}R$ is naturally the identity functor, $R$ is flat. A free module is a direct sum of copies of $R$, and tensor products commute with direct sums, so every free module is flat.

As an $R$-module,is free. Therefore the polynomial algebra $R[X]$ is a flat $R$-algebra.