tensor product and exact sequence

Question

I have an exact sequence of $F$-algebras, where $F$ is a field: $$ 0 \to A \to \prod_i B_i \to \prod_{i} C_{i}, $$ where each $A,B_i, C_i$ are $F$-algebras. Let $K/F$ be a field extension and consider the sequence induced by the above $$ 0 \to A \otimes_F K \to \prod_i (B_i \otimes_F K) \to \prod_{i} (C_{i} \otimes_F K). $$ What can we deduce about this new sequence? Can we still say that this sequence is exact? Thank you

Answer 1

TL;DR:

The tensor product does not necessarily commute with the direct product. See the second edit.

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Since an $F$-algebra is also an $F$-vector space, we may view them as vector spaces first. Then it is easy to show (for example, c.f. this post), that for any exact sequence of $F$-vector spaces, after tensored with $K$, it is still exact.

Since being exact as $F$-vector spaces and as $F$-algebras mean the same thing: kernel equals image, we deduce that the sequence as $F$-algebras is exact as well.

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Remark: Actually this shows any module over a field is flat.

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Edit:

Thanks to @user26857 for pointing out a misunderstanding of the question.

I misunderstood the question. Actually the sequence in question involves the direct product. We shall show that $$\prod_i (B_i\otimes K)\cong(\prod_iB_i)\otimes K.$$

A quick way to do this is to show that $K\otimes M\cong\operatorname{Hom}(K^*, M)$ for every vector space $M$, where $K^*=\operatorname{Hom}(K,F)$ is the $F$-linear dual of $K$. Then it is easy to see that $\operatorname{Hom}$ commutes with the direct product, since it has a left adjoint.

Below is a proof expanding the above arguments into more basic (as it writes out basis elements) terms.

Consider the natural map sending $$(\prod_ib_i)\otimes k\in(\prod_iB_i)\otimes K$$ to $$\prod_i(b_i\otimes k)\in\prod_i (B_i\otimes K).$$

Let $\left\{k_\ell\right\}_\ell$ be a basis of $K/F$.

To show the map is surjective, it suffices to show that any element of the form $x:=\prod_i(b_{i}\otimes k_{\ell_i})$ is in the image. Now consider $$y=\sum_\ell k_\ell\otimes \prod_ib'_{i\ell},$$ where $b'_{i\ell}=\begin{cases}b_i & \ell_i=\ell\\0 & \text{otherwise}\end{cases}.$ We can show that $y$ is mapped to $x$ as desired.

To show the map is injective, suppose $z=\sum_j(\prod_ib_{ji})\otimes k_j$ is mapped to $0$. By rewriting $k_j$ as a linear combination of $k_\ell$ if necessary, we may rewrite it as $\displaystyle z=\sum_\ell(\prod_ib_{\ell i})\otimes k_\ell$. Its image under the map in question is $$\prod_i(\sum_\ell b_{\ell i}\otimes k_\ell).$$ By definition this means that $\sum_\ell b_{\ell i}\otimes k_\ell=0$ for each $i$. Since $k_\ell$ is a basis, this shows that $b_{\ell i}=0,\,\forall i,\ell$. So $z=0$ and the map is injective, which is what we wanted to show.

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Edit II:

I made a mistake again: the isomorphism $K\otimes M\cong\operatorname{Hom}(K^*, M)$ does not necessarily hold when $K/F$ is not a finite extension: an element in $\operatorname{Hom}(K^*, M)$ might send an infinite number of basis elements to non-zero elements in $M$, and this corresponds to no elements in $K\otimes M$.

About the expanded argument, the part about the surjectivity is not correct: $\sum_\ell k_\ell\otimes \prod_ib'_{i\ell}$ is not necessarily a finite sum.

But at least we know that when $K/F$ is finite extension or if the direct product $\prod_iB_i$ is for a finite index set, then the conclusion holds.

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If anything inappropriate or wrong occurs, please inform me. Thanks in advance.

Answer 2

If $A$ is an $F$-algebra and $K$ is a field extension, pick an $F$-basis $(e_i)_{i\in I}$ of $K$. Standard properties of the tensor product say that we have an isomorphism

$u_A:A\otimes_F K\simeq A^{(I)}$, sending $\displaystyle\sum_{i\in I} a_i\otimes e_i$ to $(a_i)_{i\in I}$.

Now if you have an $F$-linear map $f:A\to B$, you get a $K$-linear map $f_K:A\otimes_F K\to B\otimes_F K$ which sends $a\otimes \lambda$ to $ f(b)\otimes \lambda$. Composing by both isomorphisms, it corresponds to a map $f'_K= u_Bf_K uA^{-1}:A^{(I)}\to B^{(I)}$. Applying the definitions of the various maps shows that $f'_K$ sends $(a_i)_{i\in I}$ to $(f(a_i))_{i\in I}$.It is a map of $K$-algebras as soon as $f$ is.

Notice that $\ker(f'_K)=\ker(f)^{(I)}$ and $im(f'_K)=im(f)^{(I)}$ (obvious).

Now, if you have an exact sequence $A\overset{f}{\to} B\overset{g}{\to} C$, composing the extended sequences by the isomorphisms $u_A^{-1}$ and $u_B$ yields a sequence $A^{(I)}\overset{f'_K}{\to} B^{(I)}\overset{g'_K}{\to} C^{(I)}$, which happens to be exact as well, in view of the assumption and the description of the kernel and image above.

Now composing by isomorphisms preserves exacteness, so $A\otimes_F K\to B\otimes_F K\to C\otimes_F K$ is exact to.