Can you give a simple proof that tensoring on the right by a vector space, preserves exactness of given exact vector spaces? (with the obvious identity map for the tensor multiplied on the right) I have seen proofs for general cases but for vector spaces, is there an easy proof?

  • 2
    $\begingroup$ It seems to me that the easiest proof would mimic the general fact that free modules are always flat. You start with an injection $V \to W$ and you tensor by some vector space of dim $n$, and now you have $V^n \to W^n$ defined on coordinates by the same map. Of course that's still injective. $\endgroup$
    – Hoot
    Aug 22, 2016 at 1:31
  • $\begingroup$ @Hoot would you write out a detailed answer. thanks. $\endgroup$
    – alireza
    Aug 22, 2016 at 1:34
  • $\begingroup$ If it seems to you that it should be trivial, why don't you try carrying out the proof? As mentioned, you need to use the fact that tensor products commute with direct sums and that tensoring with the base ring preserves exactness. $\endgroup$
    – Pedro
    Aug 22, 2016 at 1:42

1 Answer 1


Let $V$ be a vector space over a field $F$ with basis $v_i : i \in I$. If $W$ is another vector space with basis $w_j : j \in J$, then $V \otimes W$ is a vector space with basis $v_i \otimes w_j, (i,j) \in I \times J$. It follows immediately from here that if $v_1, ... , v_n$ are linearly independent in $V$, and $a_1, ... , a_n$ are nonzero elements of $W$, then $v_1 \otimes a_1, ... , v_n \otimes a_n$ is a linearly independent set in $V \otimes W$.

If $f: W_1 \rightarrow W_2$ is a linear transformation, then $$1_V \otimes f: V \otimes W_1 \rightarrow V \otimes W_2$$ is the linear transformation given on generators by $v \otimes w_1 \mapsto v \otimes f(w_1)$.

Now let $$W' \xrightarrow{f} W \xrightarrow{g} W''$$ be an exact sequence of vector spaces. We need to show that

$$V \otimes W' \xrightarrow{1 \otimes f} V \otimes W \xrightarrow{1 \otimes g} W''$$ is exact. It is immediate that $[1 \otimes g] \circ [1 \otimes f] = 0$. Now we just need to show that if $x \in V \otimes W$, and $[1 \otimes g](x) = 0$, then $x$ lies in the image of $ 1 \otimes f$. Write $x$ as $$x = \sum\limits_{i=1}^n v_i \otimes y_i$$ where $v_1, ... , v_n$ are linearly independent in $V$, and $y_1, ... , y_n$ are elements of $W$. Then

$$ 0 = [1 \otimes g](x) = \sum\limits_{i=1}^n v_i \otimes g(y_i)$$

By the remark in the first paragraph I wrote, applied to $W''$ instead of $W$, the only way the above sum can be zero is if all the coefficients $g(y_i)$ are zero. By the exactness of our original sequence, this implies that $y_i = f(z_i)$ for some $z_i \in W'$. Thus

$$x = \sum\limits_{i=1}^n v_i \otimes y_i = \sum\limits_{i=1}^n v_i \otimes f(z_i) = [1 \otimes f](\sum\limits_{i=1}^n v_i \otimes z_i)$$

  • 1
    $\begingroup$ so clear and lovely. $\endgroup$
    – alireza
    Aug 22, 2016 at 9:52

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