Minimal submodule annihilating a given tensor

Question

Let $A$ be a principal ideal domain; let $M$ and $N$ be two finitely generated $A$-modules; let $x_1,\dots,x_n$ be elements of $M$; let $y_1,\dots,y_n$ be elements of $N$; and let $X$ be the submodule of $M$ generated by the $x_i$.

Assume that $\sum x_i\otimes y_i$ vanishes in $M\otimes_AN$, and let $\mathcal P$ be the poset (ordered by inclusion) of those submodules $P$ of $M$ such that $P$ contains $X$ and $\sum x_i\otimes y_i$ vanishes in $P\otimes_AN$.

Does $\mathcal P$ have necessarily a minimal element?

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Let us show that the answer is Yes if $A=\mathbb Z$.

Let us write $\otimes$ for $\otimes_{\mathbb Z}$.

Let $\mathcal T$ be a totally ordered subset of $\mathcal P$. By Zorn's Lemma it suffices to show $$ I:=\bigcap_{T\in\mathcal T}T\in\mathcal P. $$ For any module $Q$ such that $X\subset Q\subset M$ denote by $z_Q$ the element of $Q\otimes N$ defined by the expression $\sum x_i\otimes y_i$. (Here "$\subset$" means "is a not necessarily proper subset of".)

We have $N\simeq A/(a_1)\oplus\cdots\oplus A/(a_m)\oplus F$ with $a_j\ge2$ and $F$ free of finite rank.

Let $T$ be in $\mathcal T$.

The morphism $I\otimes N\to T\otimes N$ can be viewed as a "diagonal" morphism $$ (I\otimes F)\oplus\left(\bigoplus I/a_jI\right)\to(T\otimes F)\oplus\left(\bigoplus T/a_jT\right). $$ We have an element $z'_I$ corresponding to $z_I$ in the kernel of this morphism, and we must show $z'_I=0$.

Using our decomposition, let us write $$ z'_I=z_{I,F}+\sum\ (z_{I,j}+a_jI) $$ with $z_{I,j}\in I$.

By flatness of $F$ we have $z_{I,F}=0$, so that it suffices to prove $$ z_{I,j}\in a_jI\ \forall\ j. $$ Given an integer $j$ with $1\le j\le m$ there is a $u_{T,j}\in T$ such that $z_{I,j}=a_ju_{T,j}$.

But there are only finitely many elements $u\in M$ which satisfy $z_{I,j}=a_ju$.

Thus, by the pigeonhole principle, there are elements $u_j\in M$ such that the set $$ \{T\in\mathcal T\ |\ u_{T,j}=u_j\ \forall\ j\} $$ is co-cofinal in $\mathcal T$ (is it better to say "co-initial in $\mathcal T$"?).

This implies $u_j\in I$ and $z_{I,j}=a_ju_j\in a_jI$ for all $j$, as was to be shown.

Answer 1

Let us show that the answer is Yes for any principal ideal domain $A$.

Let us write $\otimes$ for $\otimes_A$.

Let $\mathcal Q$ be a totally ordered subset of $\mathcal P$. By Zorn's Lemma it suffices to show $$ I:=\bigcap_{Q\in\mathcal Q}Q\in\mathcal P. $$ For any module $Q$ such that $X\subset R\subset M$ denote by $z_R$ the element of $R\otimes N$ defined by the expression $\sum x_i\otimes y_i$. (Here "$\subset$" means "is a not necessarily proper subset of".)

We have $N\simeq A/(a_1)\oplus\cdots\oplus A/(a_m)\oplus F$ with $a_j\in A$, $a_j\ne0$, $a_j$ not a unit, and $F$ free of finite rank.

Let $Q$ be in $\mathcal Q$.

The morphism $I\otimes N\to Q\otimes N$ can be viewed as a "diagonal" morphism $$ (I\otimes F)\oplus\left(\bigoplus I/a_jI\right)\to(Q\otimes F)\oplus\left(\bigoplus Q/a_jQ\right). $$ We have an element $z'_I$ corresponding to $z_I$ in the kernel of this morphism, and we must show $z'_I=0$.

Using our decomposition, let us write $$ z'_I=z_{I,F}+\sum\ (z_{I,j}+a_jI) $$ with $z_{I,j}\in I$.

By flatness of $F$ we have $z_{I,F}=0$, so that it suffices to prove $$ z_{I,j}\in a_jI\ \forall\ j. $$ Given an integer $j$ with $1\le j\le m$ there is a $u_{Q,j}\in Q$ such that $$ z_{I,j}=a_ju_{Q,j}. $$ For $Q\in\mathcal Q$ we have $Q=T_Q\oplus F_Q$ with $T_Q$ torsion and $F_Q$ free.

The $T_Q$ forming a weakly decreasing family of artinian modules, we can assume $T_Q=T$ for all $Q\in\mathcal Q$.

It is enough to verify $u_{Q,j}\in I$.

We get $I=T\oplus F_I$ with $F_I$ free. Recall that we also have $Q=T\oplus F_Q$ with $F_Q$ free.

Write $I\ni z_{I,j}=t_{I,j}+f_{I,j}$ and $Q\ni u_{Q,j}=t_{Q,j}+f_{Q,j}$ (obvious notation). It suffices to check $f_{Q,j}\in I$.

Our equation $z_{I,j}=a_ju_{Q,j}$ becomes the system $$ t_{I,j}=a_jt_{Q,j},\quad f_{I,j}=a_jf_{Q,j}. $$ Let $R\in\mathcal Q$. We must show $f_{Q,j}\in R$. We have $$ f_{Q,j}=(f_{Q,j}-f_{R,j})+f_{R,j}\in T+f_{R,j}\subset R, $$ and the proof is complete.