Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 1

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.

The statement is true. If $M$ is generated by $m_1,\dots ,m_r$, there is a surjection $R^r\twoheadrightarrow M$. The right exactness of the tensor product of modules gives a surjectionA finite direct sum of Noetherian modules is Noetherian, and a quotient of a Noetherian module is Noetherian. Hence $M{\otimes }_{R}N$ is Noetherian.

The statement is true. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

The statement is true. Fix an isomorphism $\varphi :M{\otimes }_{R}N\to R^n$, and write the inverse image of the first standard basis vector asDefineSince $\psi ((n_j{)}_j)=1$, this is a split surjection. Tensor its splitting with $M$. The resulting split surjection has the formThus $M$ is a direct summand of a finite free module, so it is a projective module. Symmetry gives the same conclusion for $N$. This is projectivity of factors of a nonzero finite free tensor product.

The statement is false. Give $\mathbb{C}[X,Y]$ its $\mathbb{C}[T]$-algebra structure byand use the quotient mapsThe induced maps send $T$ to $X$ and $Y$, respectively, so both polynomial rings are free of rank one, hence flat modules, over $\mathbb{C}[T]$.

Their tensor product over the middle ring isHere $T$ acts by zero, so this is the torsion module $\mathbb{C}[T]/(T)$. It is not flat: tensoring the injection $\mathbb{C}[T]\stackrel{T}{\to }\mathbb{C}[T]$ with it produces the zero map on a nonzero module.