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$A$ is a commutative ring with unit and $\mathfrak a$, $\mathfrak b$ ideals. I have to show that $$A/\mathfrak a \otimes_{A} A/\mathfrak b \cong A/(\mathfrak{a+b}).$$ Any hint ?

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WLOG
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2 Answers2

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It may be easier to prove the more general fact: if $R$ is any $A$-ring (i.e. we have a chosen homomorphism $\varphi : A \to R$), then we have

$$ A/I \otimes_A R \cong R / IR $$

In the following, $a,a'$ are variables denoting elements of $A$, $i$ an element of $I$, and $r,r'$ elements of $R$.

This can be done by explicitly writing down the maps: in the forward direction, it's enough to define it on pure tensors

$$ a \otimes r \mapsto \varphi(a) r $$

and in the reverse direction

$$ r \mapsto 1 \otimes r $$

All that's left is to show the maps are well-defined, are homomorphisms, and are actually inverses.

e.g. the forward map followed by the reverse map is the map

$$ a \otimes r \mapsto 1 \otimes \varphi(a) r = a \otimes r $$

and is this the identity (the equality is one of the arithmetic properties of the tensor product).

To show the map in the forward direction is well-defined, we need to show four things:

  • $ i \otimes r$ maps to the zero element of $R / IR$
  • $a a' \otimes r$ and $a \otimes \varphi(a') r$ map to the same element of $R/IR$.
  • $(a+a') \otimes r$ maps to the sum of the images of $a \otimes r$ and $a' \otimes r$
  • $a \otimes (r+r')$ maps to the sum of the images of $a \otimes r$ and $a \otimes r'$

I'l let you work out what else needs to be shown.

  • Hurkyl how to prove injectivity in these cases ? – WLOG Apr 06 '14 at 20:03
  • @WLOG: You prove it's an isomorphism first. Injectivity then follows. When working with tensor products, it's often easier to show a map is an isomorphism directly, rather than going through the indirect route of showing that it's injective and surjective. –  Apr 06 '14 at 20:20
  • @Hurkyl How do we know $1\otimes\varphi(a)r=a\otimes r$? Thanks. – yoyostein Nov 14 '17 at 15:57
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    @yoyostein: By definition, the tensor product of a right $A$-module with a left $A$-module satisfies the identity $(x \cdot a) \otimes_A y = x \otimes_A (a \cdot y)$. And by construction (or by definition, depending on how you define things), the action of $A$ on $R$ is given by $a \cdot r = \varphi(a) r$. (where the right hand side is the product in $R$) –  Nov 14 '17 at 16:34
  • @Hurkyl Thanks I understand it now. – yoyostein Nov 15 '17 at 01:29
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Define

$$\phi : A/a \times A/b \rightarrow A/(a+b),\; \phi(x+a,y+b) = xy + (a+b).$$

Can you show $\phi$ is a well-defined bilinear map?

Now if you use the universal property of tensor products to get a map $$\psi : A/a \otimes_A A/b \rightarrow A/(a+b),$$ can you show $\psi$ is injective? To show surjectivity, it will probably help you to show that an arbitrary simple tensor in $A/a \otimes_A A/b$ can be written in the form $(1+a)\otimes(x+b)$...

ah11950
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    ok, but I think the difficult point is proving injectivity – WLOG Apr 06 '14 at 18:41
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    It's been a year since I did this question myself, sorry...that characterisation of simple tensors in the final line will also help you with injectivity too I think. – ah11950 Apr 06 '14 at 18:44
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    I recognize that my comment is quite late, but one way that avoids proving injectivity is to view $\phi$ (well, $\phi^-1$, which is also easy to explicitly write down) as a map from $A$ to $A/a \otimes A/b$ and show that it has kernel $(a+b)$. – ASKASK Aug 16 '16 at 06:34