When Is an Element of a Tensor Product Zero?

The answer depends on what kind of objects you are tensoring. For vector spaces over a field, a single pure tensor is zero only when one of its factors is zero. But an element of a tensor product can also be a sum of pure tensors, and such a sum may vanish through the tensor-product relations. For modules over a general ring, even a pure tensor with two nonzero factors can be zero.

Pure tensors of vector spaces

Let $V$ and $W$ be vector spaces over the same field $F$. For $v\in V$ and $w\in W$,

$$v\otimes w=0 \quad\Longleftrightarrow\quad v=0\text{ or }w=0.$$

If either factor is zero, the tensor is zero by bilinearity. For the converse, suppose both factors are nonzero. Choose linear functionals $f:V\to F$ and $g:W\to F$ with $f(v)=1$ and $g(w)=1$. Such functionals can be defined on the spans of $v$ and $w$, then extended to bases of the spaces. The induced map on tensor products gives

$$(f\otimes g)(v\otimes w)=f(v)g(w)=1.$$

Since the image is not zero, $v\otimes w$ cannot be zero.

This result is specifically about a single pure tensor and vector spaces over a field. It does not say that every expression written as a sum of nonzero pure tensors is nonzero.

Sums of pure tensors

Every element of $V\otimes_F W$ is a finite sum of pure tensors. To decide whether such a sum is zero, choose bases $(e_i)$ of $V$ and $(f_j)$ of $W$. The tensors $e_i\otimes f_j$ form a basis of $V\otimes_F W$. Expand each factor in the chosen bases, combine terms with the same pair of indices, and check the resulting coefficients. The sum is zero exactly when every coefficient is zero.

For instance, take $V=W=F^2$ with standard basis $e_1,e_2$. Identify $e_i\otimes e_j$ with the matrix that has a $1$ in position $(i,j)$ and zeros elsewhere. Then

$$e_1\otimes e_2+e_2\otimes e_1 \quad\longleftrightarrow\quad \begin{pmatrix}0&1\\1&0\end{pmatrix}.$$

The matrix is not zero, so the tensor is not zero. In contrast, bilinearity gives

$$(e_1+e_2)\otimes e_1-e_1\otimes e_1-e_2\otimes e_1=0.$$

The individual displayed terms need not be zero for their sum to vanish.

An equivalent test for vector spaces is to apply all pairs of linear functionals: $x\in V\otimes_F W$ is zero if and only if $(f\otimes g)(x)=0$ for every $f\in V^*$ and $g\in W^*$. In practice, choosing bases usually makes the test more direct.

What changes for modules over a ring?

For modules over a commutative ring $R$, the tensor product is formed using additivity and the balancing relation

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

for $r\in R$, $m\in M$, and $n\in N$. Those relations can make a pure tensor vanish even when both factors are nonzero. Thus, for general modules, the implication

$$m\otimes n=0\quad\Longrightarrow\quad m=0\text{ or }n=0$$

is false.

A simple example is $\mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/3\mathbb{Z}$. Both modules contain nonzero elements, but their tensor product is zero. It is generated by $x=\overline{1}\otimes\overline{1}$. Since $2\overline{1}=0$ in the first factor, $2x=0$; since $3\overline{1}=0$ in the second, $3x=0$. Therefore $x=3x-2x=0$, and the generator is zero. Equivalently, $\mathbb{Z}/m\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}/\gcd(m,n)\mathbb{Z}$, which is zero when $m$ and $n$ are coprime.

For modules, there is no universal rule based only on whether each factor is nonzero. Use the module presentations and tensor-product relations, or apply a known theorem appropriate to the modules in question.

Common mistakes

• Treating every tensor like a product of numbers. The field-vector-space rule for pure tensors does not automatically hold for modules over a ring.
• Checking only the displayed terms. A sum can be zero because of additivity or balancing, even when its written terms are nonzero.
• Assuming a representation is unique. A tensor can have many different expressions as a sum of pure tensors. In a vector-space basis, the coefficients are unique; in an arbitrary expression, they are not.
• Confusing a zero tensor with a zero factor. A zero factor always makes a pure tensor zero, but the converse requires hypotheses such as working with vector spaces over a field.