If $R'$ is a faithfully flat $R-$algebra, then $(B/f(A))\otimes_R R' \cong B'/f' (A')$

Question

Let $R$ be a commutative ring, $R \to A \to B$ be an sequence of $R-$algebras, $f:A \to B$ a homomorphism and $R'$ a faithfully flat $R-$algebra. Now, tensor the initial sequence by $\otimes_R R'$, getting this: $$ R' \to A\otimes_R R' \to B\otimes_R R'. $$ For better notations, call $A' := A \otimes_R R' $ and $B' := B \otimes_R R'$. Of course we can denote $f':A' \to B'$ as $f' = f \otimes_R \mathrm{id}_{R'}$. While I was reading a paper, as usual, the writer tells that $$ (B/f(A))\otimes_R R' \cong B'/f'(A')$$ giving no clue whatsoever on how to prove this. It is not hard to prove that $(R/I)\otimes_R R' \cong R'/IR'$, but I could not achieve the result on the text; I've tried to create a bilinear and well defined map $B/f(A) \times R' \to B' /f'(A')$ to be able to use the universal property of tensor produts, so I've define $(\overline{b},r') \mapsto \overline{b \otimes r'}$, but I am a little afraid to just set this, since quotients tends to be a little hard on its properties. Anyways, if someone has some tip on solving this, it would be very appreciated! There is only one "rule" (If I can call it this way): Apparently, this is supposed to be achieved using only algebra, without category theory; but, of course if one has a "categorical" proof, I will be glad to read! Thanks in advance for the reading!

Answer 1

Sometimes the "better notations" are actually confusing, since you might forget what the objects are. We can first define an $R$-bilinear map $B\times R'\to (B\otimes_{R} R')/f'(A')$ by $(b,r')\to (b\otimes r')+f'(A')$. It is immediate that if $b=f(a)\in f(A)$ then $b\otimes r'\in f'(A')$, and so the pair $(b,r')$ is mapped to zero. If so, there is a well defined bilinear map:

$(B/f(A))\times R'\to (B\otimes_R R')/f'(A')$

And now use the universal property. Show that you get a homomorphism of $R$-algebras.

Similarly, you can define the inverse map. First, we can define an $R$-bilinear map $B\times R'\to (B/f(A))\otimes_R R'$ by $(b,r')\to (b+f(A))\otimes r'$. Use the universal property to get a homomorphism of $R$-modules:

$\phi:B'=B\otimes_R R'\to (B/f(A))\otimes R'$

Which satisfies $f(b\otimes r')=(b+f(A))\otimes r'$. A general element in $f'(A')$ is a finite sum $\sum\limits_{i=1}^n (f(a_i)\otimes r_i')$, and so $f'(A')$ is contained in the kernel of $\phi$. So we get a well defined homomorphism on $B'/f'(A')$.

Answer 2

By exactness properties of the tensor product, tensoring $f(A) \to B \to B / f(A) \to 0$ with $R'$, you get $f(A) \otimes_R R' \to B \otimes_R R' \to (B / f(A)) \otimes_R R' \to 0$ is exact. Thus $(B / f(A)) \otimes_R R' \cong (B \otimes_R R') / (f(A) \otimes_R R')$, or in your notation, $(B/f(A)) \otimes_R R' \cong B' / (f(A) \otimes_R R')$. Now you just need to show that $f(A) \otimes_R R' \cong (f \otimes \mathrm{id}_{R'})(A \otimes_R R')$, but this should be an easy check.

Answer 3

Here is the “categorical” proof.

Since $\_ \otimes_R R’$ is left adjoint, it preserves colimits; in particular, it preserves cokernels: $$ \operatorname{coker}(f) \otimes_R R’ \cong \operatorname{coker}(f \otimes_R R’). $$