Is the Tensor Product of a Module Element and Zero Always Zero?

For modules, tensoring an element with zero always gives the zero element of the tensor product: $m\otimes 0=0$ and $0\otimes n=0$. The reason is that the tensor product is additive in each argument.

Here, $m$ and $n$ are elements of modules over the same ring, with the usual compatible module structure needed to form their tensor product. The result does not require either module to be free or finite-dimensional.

The key idea: additivity

A tensor product $M\otimes_R N$ is built so that the map sending a pair $(m,n)$ to $m\otimes n$ is additive in each variable. In particular,

$$m\otimes(n_1+n_2)=m\otimes n_1+m\otimes n_2.$$

The same property holds in the first variable. Since zero is the sum $0+0$, additivity forces a tensor with zero in either slot to be zero. This is a consequence of the defining relations of the tensor product, not an extra assumption about the elements.

Proof that a zero factor gives zero

Let $m\in M$. By additivity in the second argument,

$$m\otimes 0=m\otimes(0+0)=m\otimes 0+m\otimes 0.$$

The tensor product is an abelian group under addition. Subtract $m\otimes 0$ from both sides, or use cancellation in that group, to get

$$m\otimes 0=0.$$

The first argument works the same way. For $n\in N$,

$$0\otimes n=(0+0)\otimes n=0\otimes n+0\otimes n,$$

so cancellation gives $0\otimes n=0$. Therefore, if either factor of a pure tensor is zero, the pure tensor is zero.

This argument also applies when $R$ is not commutative, provided the tensor product is formed from a right $R$-module and a left $R$-module. Additivity in each variable is still part of the construction.

Worked example

Take the modules $M=N=\mathbb{Z}^2$ over the integers. There is a standard identification of $\mathbb{Z}^2\otimes_{\mathbb{Z}}\mathbb{Z}^2$ with the group of $2\times2$ integer matrices, where the pure tensor of column vectors corresponds to their outer product.

For $m=(2,-1)$ and $n=(0,0)$, the corresponding matrix is

$$m n^{\mathsf T}=\begin{pmatrix}2\\-1\end{pmatrix}\begin{pmatrix}0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$

Thus $m\otimes n$ is the zero tensor. The example makes the rule visible in coordinates, but the proof above is more general and does not depend on this matrix representation.

What the rule does not say

The implication works in one direction: a zero factor makes the pure tensor zero. The converse is not generally true. Even when both factors are nonzero, their tensor can be zero.

For example, take $M=\mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}/3\mathbb{Z}$, and let $x$ and $y$ be the residue classes of $1$. Both $x$ and $y$ are nonzero. But $3x=x$ in $M$, while $3y=0$ in $N$. The balancing relation for a tensor product over $\mathbb{Z}$ gives

$$x\otimes y=(3x)\otimes y=x\otimes(3y)=x\otimes 0=0.$$

So knowing that $m\otimes n=0$ does not by itself show that $m=0$ or $n=0$. Testing whether a tensor with two nonzero factors vanishes can require more information about the modules and their relations.

Common mistakes

• Confusing a pure tensor with a general tensor. An element of $M\otimes_R N$ can be a sum such as $m_1\otimes n_1+m_2\otimes n_2$. The zero-factor rule applies to each term that has a zero factor, but it does not automatically decide whether the whole sum is zero.
• Assuming nonzero factors guarantee a nonzero tensor. The example with $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ shows that this can fail.
• Treating $m\otimes 0$ as a new kind of zero. It is simply the zero element of the tensor product, whose addition is the group operation used for cancellation in the proof.

When the question is just whether tensoring an element with zero gives zero, bilinearity settles it. Questions about a general sum of pure tensors, or about a pure tensor with two nonzero factors, are different problems.