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Let $R$ be a commutative ring with identity.

The following is a statement I came across about the submodule $Rt$ generated by a decomposable tensor $t=m\otimes n$ being free, given that $Rm$ and $Rn$ are free. I am not sure if the converse is true but I would be interested in seeing a counterexample.

Let $M, N$ be $R$-modules, and let $m$ be in $M$ and $n$ be in $N$. Suppose also that $Rm$ and $Rn$ are free.

Is $R(m \otimes n)$ free?

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3  
What does it mean for an element of a module to be "free in the module"? Also, what happened to $y$? I also don't know what you mean by "$\{m\otimes n\}$ is isomorphic to a free module"; to me, $\{m\otimes n\}$ looks like a set with a single element, so unless $m\otimes n = \mathbf{0}$, it has no hope of being a module... –  Arturo Magidin Sep 30 '11 at 18:14
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I think the condition is: "$R\cdot(m\otimes n)$ is a free $R$-module (of rank one)". –  Pierre-Yves Gaillard Sep 30 '11 at 18:19
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@Pierre-YvesGaillard: That is, "$m\otimes n$ is not a torsion element"? –  Arturo Magidin Sep 30 '11 at 18:34
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@Arturo. Yes, that's my guess. But the OP should answer this kind of queries... [I believe the word "torsion" is mainly used when $R$ is a PID.] (One could also say "the annihilator of $m\otimes n$ is $0$".) –  Pierre-Yves Gaillard Sep 30 '11 at 18:49
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@Pierre-YvesGaillard: Ah, you're right. Anderson and Fuller do restrict "torsion" to domains. I should do so as well... Thanks! –  Arturo Magidin Sep 30 '11 at 19:06

2 Answers 2

up vote 5 down vote accepted

I think this works. Let $A=k[x,y,z]/(x^2,xy,xz)$, let $M$ be the quotient of the free $A$-module generated by $e_1$ and $e_2$ subject to the relations $$xe_1=ye_2 \qquad ze_2=0$$ and let $N$ be the quotient of the free $A$-module generated by $f_1$ and $f_2$ subject to the relation $$yf_1=zf_2.$$ Then $e_1$ is free in $M$ and $f_1$ is free in $N$, yet $$x\cdot e_1\otimes f_1 = xe_1\otimes f_1 = ye_2\otimes f_1 = e_2\otimes yf_1 = e_2\otimes zf_2 = ze_2\otimes f_2 = 0.$$

Let me check using Macaulay2:

First, construct our base ring

i1 : R = QQ[x,y,z]/(x*z,x*y,x*x);

Next, $M$ as a quotient of the free module $F=R^2$

i2 : F = R^2;

i3 : M = F / (x*F_0 - y*F_1, z*F_1);

and then $N$, also as a quotient of $F$,

i4 : N = F / (y*F_0 - z*F_1);

The element $e_1$, the image of the first generator of $F$ in $N$ is free:

i5 : kernel map(M, R^1, {{1}, {0}})

o5 = image 0

Likewise, $f_1$,the image of the first generator of $F$ in $N$ is free:

                             1
o5 : R-module, submodule of R

i6 : kernel map(N, R^1, {{1}, {0}})

o6 = image 0
                             1
o6 : R-module, submodule of R

Finally, $e_1\otimes f_1$ is not free in $M\otimes N$:

i7 : kernel map(M**N, R^1, {{1}, {0}, {0}, {0}})

o7 = image | x |

                             1
o7 : R-module, submodule of R

This not only shows that $x$ kills $e_1\otimes f_1$ but that in fact it generates its (one-dimensional) annihilator.

N.B. I constructed this by first deciding the relations which define the modules, and then iteratively computing kernels using and adding relations to the ring until I got $e_1$ and $f_1$ to be free.

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2  
+1! Wonderful!!! Would the following variation work? Let $A$ be the quotient of $k[x,y,z]$ by the square of the ideal generated by $x,y$ and $z$. The freeness would be easier to check. –  Pierre-Yves Gaillard Oct 1 '11 at 0:04
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$$A:=k[x,y]/(x^2,xy,y^2),$$ $$xe_1=ye_2,\quad xe_2=0,\quad yf_1=xf_2,$$ $$x(e_1\otimes f_1)=xe_1\otimes f_1=ye_2\otimes f_1=e_2\otimes yf_1=e_2\otimes xf_2=xe_2\otimes f_2=0.$$ –  Pierre-Yves Gaillard Oct 1 '11 at 1:01
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@Pierre: probably :) Since I had Macaulay to do the computation for me, I did not worry much. –  Mariano Suárez-Alvarez Oct 1 '11 at 1:02
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The computation that $e_1\otimes f_1$ is killed by $x$ is independent of the relations in the ring (it works over the polynomial ring tout court, in fact) –  Mariano Suárez-Alvarez Oct 1 '11 at 1:21
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We have $x^2e_1=xye_2=yxe_2=0$. –  Pierre-Yves Gaillard Oct 1 '11 at 1:42

This is a minor complement to Mariano's answer. The goal is to make the computation as easy and visual as possible.

Let $K$ be a field, let $X,Y$ be indeterminates, and let $x,y$ be the canonical images of $X,Y$ in $$A:=\frac{K[X,Y]}{(X^2,XY,Y^2)}\quad. $$ Using the diagram $$ e_1\stackrel{y}{\to}e_2\stackrel{x}{\leftarrow}e_3\stackrel{y}{\to}e_4 $$ define the $A$-module $E$ as follows:

  • $\{e_1,e_2,e_3,e_4\}$ is a $K$-basis of $E$;

  • the first arrow means $ye_1=e_2$;

  • the absence of an $x$-arrow emanating from $e_1$ means $xe_1=0$;

and so on.

This is indeed an $A$-module because the arrows are uncomposable, and $Ae_3$ is free because two arrows emanate from $e_3$.

Let $F$ be the $A$-module attached in a similar way to the diagram $$ f_2\stackrel{x}{\leftarrow}f_3\stackrel{y}{\to}f_4\stackrel{x}{\leftarrow}f_5. $$ In particular $Af_3$ is free.

Now compute $$ x(e_3\otimes f_3)=xe_3\otimes f_3=ye_1\otimes f_3=e_1\otimes yf_3=e_1\otimes xf_5=xe_1\otimes f_5=0. $$ So $A(e_3\otimes f_3)$ is not free, although $Ae_3$ and $Af_3$ are.

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