How to Test Whether an Element of a Tensor Product Is Zero

An element of a tensor product is usually written as a finite sum of pure tensors, such as $t=\sum_{i=1}^k m_i\otimes n_i$. The expression is not unique: additivity and scalar-balancing relations can make different-looking sums represent the same element. So you cannot decide whether $t=0$ just by looking at one chosen expression.

The practical method depends on what is known about the modules. Vector spaces have a coordinate test. For general modules, use presentations and the quotient relations, or find a map that detects the element.

What the tensor-product relations say

Assume $R$ is a commutative ring and $M,N$ are $R$-modules. The tensor product $M\otimes_R N$ is generated by symbols $m\otimes n$, subject to additivity in both entries and the balancing relation

$$(rm)\otimes n = m\otimes(rn).$$

In particular, $0\otimes n=m\otimes0=0$. But a sum of nonzero pure tensors may also vanish because of cancellation or scalar relations. Thus, to prove a sum is zero, reduce it using these relations or use an equivalent presentation of the tensor product.

For modules over a noncommutative ring, left- and right-module conventions matter: the balancing relation identifies $(mr)\otimes n$ with $m\otimes(rn)$. The rest of this article uses the commutative-ring convention.

Over a field: use coordinates

If $M$ and $N$ are vector spaces over a field $F$, choose bases. The tensors $e_i\otimes f_j$, where $e_i$ and $f_j$ are basis vectors, form a basis of $M\otimes_F N$. Therefore, collect the coefficients of each basis tensor. The original sum is zero exactly when every collected coefficient is zero.

For example, let $e_1,e_2$ and $f_1,f_2$ be basis vectors, and consider

$$t=2e_1\otimes f_1+3e_1\otimes f_2-e_2\otimes f_1.$$

Its coefficients in the tensor-product basis are $2$, $3$, and $-1$. Over a field of characteristic zero, these are not all zero, so $t\ne0$. In characteristic $2$, the coefficient $2$ becomes zero, but the coefficients of $e_1\otimes f_2$ and $e_2\otimes f_1$ remain nonzero, so $t$ is still nonzero. Always interpret coefficients in the field you are working over.

A pure tensor has an especially simple test over a field: $m\otimes n=0$ if and only if $m=0$ or $n=0$. This does not hold for arbitrary modules over arbitrary rings.

For general modules: use presentations

Suppose you know a module by generators and relations. Tensor products preserve cokernels, so relations in one module become relations in the tensor product. For example, if $M$ is presented as the cokernel of a matrix $A:R^p\to R^q$, then $M\otimes_R N$ is the cokernel of the induced map $N^p\to N^q$. In practice, this turns the question into checking whether a vector belongs to the submodule generated by the relations.

When the modules are finitely presented over a principal ideal domain such as $\mathbb Z$, Smith normal form can often reduce the presentation to a direct sum of cyclic modules. Over more general rings, there may be no simple normal form, and whether a computation is effective depends on the ring and the input data.

Worked example: cyclic groups

Compute $\mathbb Z/4\mathbb Z\otimes_{\mathbb Z}\mathbb Z/6\mathbb Z$. For any abelian group $N$, there is an isomorphism

$$(\mathbb Z/4\mathbb Z)\otimes_{\mathbb Z}N \cong N/4N.$$

Taking $N=\mathbb Z/6\mathbb Z$, the subgroup $4N$ consists of the residues $0,2,4$. Hence the quotient has two elements, and the tensor product is isomorphic to $\mathbb Z/2\mathbb Z$. Under this identification, $[a]\otimes[b]$ corresponds to the class of $ab$ modulo $4N$.

In particular, $[2]\otimes[1]=0$, since $2\in4N$. Yet $[2]$ is nonzero in $\mathbb Z/4\mathbb Z$, and $[1]$ is nonzero in $\mathbb Z/6\mathbb Z$. This shows why the pure-tensor test for vector spaces cannot be applied to arbitrary modules.

Use a map to prove nonzero

The universal property gives a useful one-way test. Any bilinear map $B:M\times N\to P$ induces a linear map $\widetilde B:M\otimes_R N\to P$ with $\widetilde B(m\otimes n)=B(m,n)$. If you find one for which $\widetilde B(t)\ne0$, then $t\ne0$.

For vector spaces, coordinate functionals give such detecting maps. For arbitrary modules, a particular map may fail to detect a nonzero tensor, so a zero image does not by itself prove $t=0$. You need a family of maps known to separate the elements, or a direct computation from a presentation.

Common mistakes

• Treating a sum as a unique expansion. Tensor expressions are not unique until you choose suitable bases or a presentation.
• Assuming nonzero factors give a nonzero pure tensor. This is true for vector spaces, but not for modules over general rings, as the cyclic-group example shows.
• Using one map as a two-way test. A nonzero image proves nonzero; a zero image proves nothing unless the map is known to be injective or part of a separating family.
• Ignoring the base ring. Relations, coefficients, and even whether an element is zero can change with the ring.