How to Prove an Element Is Nonzero in a Tensor Product

To prove that an element of a tensor product is nonzero, first check what the tensor product is taken over. For vector spaces over a field, linear functionals give a reliable test: a nonzero tensor can be detected by evaluating it against a pair of functionals. For modules over a general ring, the same test may fail, so you need to use the module relations or find a suitable map out of the tensor product.

The test for vector spaces

Let $V$ and $W$ be vector spaces over a field $k$, and let $x \in V \otimes_k W$. Every pair of linear functionals $\alpha:V\to k$ and $\beta:W\to k$ defines a linear map

$$\alpha\otimes\beta:V\otimes_k W\to k, \qquad v\otimes w\mapsto \alpha(v)\beta(w).$$

The key fact is

$$x\ne 0 \quad\Longleftrightarrow\quad (\alpha\otimes\beta)(x)\ne 0 \text{ for some }\alpha,\beta.$$

The reverse implication follows immediately: a linear map sends zero to zero. For the forward implication, write $x$ as a finite sum and choose a basis $e_1,\ldots,e_r$ for the span of its first factors. Collect terms to write

$$x=\sum_{j=1}^r e_j\otimes z_j$$

for some $z_j\in W$. If every $z_j$ were zero, then $x$ would be zero. So when $x\ne0$, some $z_j\ne0$. Choose a functional $\beta$ with $\beta(z_j)\ne0$, and choose $\alpha$ that takes value $1$ on $e_j$ and $0$ on the other $e_i$. Then $(\alpha\otimes\beta)(x)=\beta(z_j)\ne0$. These functionals exist because a linearly independent set can be extended to a basis.

This also proves the familiar special case: over a field, $v\otimes w\ne0$ whenever both $v$ and $w$ are nonzero. Choose functionals that are nonzero on each factor and evaluate their tensor product.

A worked example

Take $V=W=\mathbb{Q}^2$ and define

$$x=(1,0)\otimes(1,1)+(0,1)\otimes(2,3).$$

Let $\alpha(a,b)=a$ and $\beta(c,d)=c$, the functionals that select the first coordinate. Applying their tensor product gives

$$\begin{aligned} (\alpha\otimes\beta)(x) &=\alpha(1,0)\beta(1,1)+\alpha(0,1)\beta(2,3)\\ &=1\cdot1+0\cdot2\\ &=1. \end{aligned}$$

Since a linear map cannot send the zero vector to $1$, this proves $x\ne0$.

For finite-dimensional spaces, there is an equivalent coordinate method. If $e_1,\ldots,e_m$ and $f_1,\ldots,f_n$ are bases of $V$ and $W$, respectively, then the vectors $e_i\otimes f_j$ form a basis of $V\otimes_k W$. Expand $x$ in that basis. It is nonzero if and only if at least one coefficient is nonzero. In the example, the coefficient of $(1,0)\otimes(1,0)$ is $1$, which gives another proof.

What changes over a ring?

For modules over a commutative ring $R$, a map $M\otimes_R N\to R$ obtained from $R$-linear functionals may not detect every nonzero tensor. More fundamentally, nonzero factors can have zero tensor product. For example,

$$(\mathbb{Z}/2\mathbb{Z})\otimes_{\mathbb{Z}}(\mathbb{Z}/3\mathbb{Z})=0.$$

Both modules are nonzero, but their tensor product is zero. In fact, the tensor is generated by $\overline{1}\otimes\overline{1}$, and

$$\overline{1}\otimes\overline{1} =(3\overline{1})\otimes\overline{1} =\overline{1}\otimes(3\overline{1}) =\overline{1}\otimes\overline{0}=0.$$

The first equality uses $3\overline{1}=\overline{1}$ in $\mathbb{Z}/2\mathbb{Z}$; the second uses the balancing relation for tensor products.

A practical test over a ring is to construct a balanced bilinear map $b:M\times N\to P$. “Balanced” means $b(rm,n)=b(m,rn)$ for every $r\in R$. The universal property gives a map $\widetilde b:M\otimes_R N\to P$ with $\widetilde b(m\otimes n)=b(m,n)$. If $\widetilde b(x)\ne0$, then $x\ne0$. Another method is to use presentations of the modules: write down the generators and relations they induce in the tensor product, then check that the proposed element is not forced to be zero by those relations.

Common mistakes

• Checking only the factors: Over a field, nonzero factors make a pure tensor nonzero; over a general ring, this need not hold.
• Checking each summand separately: A sum of nonzero pure tensors can still be zero because the terms may cancel. Test the whole tensor.
• Using one particular functional as a universal test: A chosen pair may evaluate to zero even when the tensor is nonzero. Over a field, the criterion says that some pair detects it, not that every pair does.
• Ignoring the base ring: The notation $M\otimes_R N$ depends on $R$, and changing the base ring can change whether a tensor is zero.