Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 4

The inverse limit is the submodule of the direct product consisting of compatible families:Its projection to $N_i$ sends $(x_j)$ to $x_i$; these projections satisfy the universal property of an inverse limit.

Solved by gpt-5.6-sol high.

Two decreasing filtrations of a module $(M_n)$ and $(M_n^{\prime })$ are equivalent when each contains a fixed shift of the other: there are $a,b\ge 0$ such thatfor every $n\ge 0$.

For an $I$-filtration, $I{M}_n\subseteq M_{n+1}$, so $I^n{M}_0\subseteq M_n$. If it is stable from $s$ onward, thenIt is therefore equivalent to the I-adic filtration. Any two stable $I$-filtrations are consequently equivalent to each other.

Solved by gpt-5.6-sol high.

No. Give $R=k[x]$ the x-adic filtration $R_n=(x^n)$, take $M=R$, and let $N=(x)$. At $n=1$ the intersection filtration giveswhereas the induced filtration of $N$ givesThus intersection with a submodule need not equal the induced filtration.

Solved by gpt-5.6-sol high.

Yes. Scalar multiplication in the quotient module gives directlyHence the displayed filtration is precisely the induced filtration.

Solved by gpt-5.6-sol high.

Work modulo $I$. Putand let $A$ be the image of $I^{\prime }/I$ in $C$. The separating condition $I^{\prime }=I^{\prime }\cap (I+J)$ modulo $I$ says that $I^{\prime }/I\to C$ is injective, so we may regard $A$ as a submodule of $C$.

Since $r\in \sqrt{(I:I^{\prime })}$, some power $r^q$ annihilates $A$. Apply the Artin-Rees lemma to $A\subseteq C$ and the principal ideal $(r)$. There is $s$ such that for every $m\ge s$,For $m\ge s+q$, the right side is zero. Pulling the equality $A\cap r^{m}C=0$ back to $R$ givesas required.

Solved by gpt-5.6-sol high.