# Tensor is non-zero

Let $$A,B$$ be $$k$$-algebras, where $$k$$ is a field. Let $$a_1,a_2,\dots a_n$$ be elements of $$A$$, which are linearly independent over $$k$$. Similarly let $$b_1,b_2, \dots, b_n$$ be elements of $$B$$, which are linearly independent over $$k$$. Is $$\sum a_i \otimes_k b_i$$ non-zero in $$A \otimes_k B$$?

On first look the claim seems to be true. As far as I know $$\sum a_i \otimes_k b_i =0$$ if and only if we can bring $$\sum a_i \otimes_k b_i$$ to the $$0 \otimes_k 0$$ tensor using the tensor-defining relations:

$$(a+a',b) = (a,b) + (a',b); \quad \quad (a,b+b') = (a,b)+(a,b'); \quad \quad (ka,b) = (a,kb);$$

However, I feel the linear independence really prohibits us from bringing the sum to a simple tensor along either of the coordinates, unless $$n=1$$. But, I'm struggling to prove that this is indeed the case if $$n\ge2$$. While if $$n=1$$, $$a \otimes_k b = 0$$ implies that $$a=0$$ or $$b = 0$$, as $$A$$, $$B$$ are flat, given that $$k$$ is a field. Obviously this contadicts the linear independence.

Also I tried to use the flatness of $$A$$ and $$B$$, however to no avail.

Finally, if it makes the problem simplier we may assume that $$A$$ is an integral domain and $$B$$ is a finite field extension of $$k$$ with $$\{b_1, \dots, b_n\}$$ a basis of $$B$$ over $$k$$. In fact, this is the problem I initially wanted to solve, but I thought that the result might be somewhat generalized.

Let's back up to a special case where $$A, B$$ are vector spaces, with respective bases $$\{a_j\}, \{b_j\}$$.

Furthermore, let's define the tensor product $$A \otimes B$$ (I'll omit the subscript $$k$$ for brevity) as the dual of the space $${\tt Bil}(A, B)$$ of all bilinear forms on the direct sum $$A \oplus B$$. Namely, $$a \otimes b$$ is the linear functional mapping each bilinear form $$\psi( \cdot, \cdot)$$ on $$A \oplus B$$ to the scalar $$\psi(a, b)$$. [I find this, geometric, definition a lot friendlier to work with.]

How does $$\sum_{i}a_{i} \otimes b_{i}$$ act on a bilinear form $$\psi( \cdot, \cdot) \in {\tt Bil}(A, B)$$? By mapping that form to $$\sum_{i} \psi( a_{i}, b_{i} ).$$ Can we construct a $$\psi$$ so that the latter sum is not zero? If so, then the linear functional $$\sum_{i}a_{i} \otimes b_{i}$$ on $${\tt Bil}(A, B)$$ cannot be identically zero.

• I know that by the universal property of tensors a tensor is equal to zero if every bilinear map on $A \times B$ sends it to zero. However, I can't see a way to constuct a map as desired, given that the vector spaces are not explicitly given. Commented Apr 29, 2019 at 17:01
• Maybe something along the lines $\psi (a,b) = \sum c_i \sum d_i$, where $a=\sum c_i a_i$ and $b = \sum d_ib_i$? Commented Apr 29, 2019 at 17:08
• Yes. By using the multiplicative identity $1$ in the field $k$. Since $a_{i}, b_{i}$ are linearly independent sets, define a linear functional $f_{A}$ on $A$ by $f_{A}(a_{i}) = 1$ for all i, and a linear functional $f_{B}$ on $B$ by: $f_{B}(b_{i}) = 1$ for all i. Now, define a bilinear form $\psi$ on ${\tt Bil}(A, B)$ by: $\psi(a, b) = f_{A}(a) f_{B}(b)$.
– avs
Commented Apr 29, 2019 at 17:09
• I reckon we set $f_A(a_i) = 0$ if $f_A(a) = 0$ for some $i$, where $a_i$ has non-zero coefficient in the representation of $a$ Commented Apr 29, 2019 at 17:19
• Anyway, the answer seems fine. However I would want to wait a bit for some more algebraic answer, avoiding the use of bilinear forms. If nothing comes in short amount of time I will accept your answer. Commented Apr 29, 2019 at 17:21

We even do get somewhat more! If you have $$k$$ vector spaces $$V, W$$ with bases $$(v_i)_{i \in I}$$ and $$(w_j)_{j \in J}$$, then $$(v_i \otimes w_j)_{i,j}$$ is a basis of $$V \otimes W$$. Now this implies the statement, because in the context of your question $$a_1, \dots a_n$$ and $$b_1, \dots, b_n$$ lie in a basis of $$A$$ and $$B$$ respectively. Therefore by the upper statement $$(a_i \otimes b_j)_{i,j}$$ is a subfamily of a basis of $$A \otimes B$$ and therefore in particular linearly independent. Thus $$\sum_i a_i \otimes b_i \not = 0$$.

Here are some detailed hints for the first statement. But if you want to, give it a try yourself. Probably the most accessible setting is to try to construct an isomorphism $$\oplus_I k \times \oplus_J k \to \oplus_{I \times J}k$$ through the universal property of the tensor product.

Our aim is to construct an isomorphism $$V \otimes W \cong \oplus_{I \times J}k$$. Through the choice of bases we have an isomorphism $$V \times W \cong \oplus_I k \times \oplus_J k$$. Now we can define the map $$\oplus_I k \times \oplus_J k \to \oplus_{I \times J} k, (\sum_i \lambda_i \delta_i, \sum_j \mu_j \delta_j) \mapsto \sum_{i,j} \lambda_i \mu_j \delta_{i,j}.$$ whereby the $$\delta$$ -s denote the standard basis vectors. It is not too hard to see, that this map is bilinear. By the universal property of the tensor product this yields a unique liner map $$V \otimes W \to \oplus_{I \times J}k$$. To show, that this is truly an isomorphism we need an inverse map. Remember that we want, the family $$(a_i \otimes b_j)_{i,j}$$ to be a basis of $$V \otimes W$$. So what could be more natural than to try the map given by $$\oplus_{I \times J}k \to V \otimes W, \delta_{i,j} \mapsto v_i \otimes w_j?$$ To convince yourself, that this defines an inverse should be another good exercise.

Let $$k$$ be a field, and $$V$$ and $$W$$ two $$k$$-vector spaces, with bases $$e_i$$ and $$f_j$$ respectively. Then $$V\otimes_kW$$ is a $$k$$-vector space with basis $$e_i\otimes f_j$$. Thus the answer to your question is yes.

One way to see this result about bases is to use the universal property. A $$k$$-bilinear map $$V\times W\to X$$ is completely determined by its values on $$(e_i,f_j)$$, and we can choose these values freely.