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Let $\text{CAlg}_R$ denote the category of commutative $R$-algebras and $R$-algebra homomorphisms. How can I show that if $A,B \in \text{CAlg}_R$, the tensor product $A \otimes_R B$ can be given the structure of a commutative $R$-algebra so that it is the categorical coproduct of $A$ and $B$ in $\text{CAlg}_R$?

Well, for the first part I dont think there is much of a choice. If $\phi_1: R \to A$ and $\phi_2: R \to B$ is the given homomorphisms, then we might define $\phi: R \to A \otimes_R B$ by $r \mapsto \phi_1(r) \otimes \phi_2(r)$. After this I am stuck, am I supposed to prove that the coproduct of $A$ and $B$ equals $A \otimes_R B$ if I have chosen $\phi$ correctly? I tried, but got stuck on defining the unique function (in the definition of a coproduct). Any hints?

Brad
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    Your function $\phi$ is not even linear! – Mariano Suárez-Álvarez Sep 24 '15 at 18:45
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    For $\phi$, try $r \mapsto \phi_1(r) \otimes 1_B$. For your last question, to show the it is the coproduct, you have to check that $A \otimes_R B$ satisfies the universal property of coproduct - i.e given maps $A \to C$ and $B \to C$, there is a unique map $A \otimes_R B \to C$ satisfying various commutative diagram conditions – ykm Sep 24 '15 at 18:54
  • @MarianoSuárez-Alvarez, I really cant see why it is not linear. If we regard the tensor product as the quotient $C/D$, then $r \mapsto (\phi_1(r), \phi_2(r)) + D$. From here it is easily seen that it is linear, am I missing something? – Brad Sep 24 '15 at 19:38
  • Suppose that $A$ and $B$ are both equal to $R$, so that the map $\psi:a\otimes b\in A\otimes_RB\mapsto ab\in R$ is an isomorphism. Then th composition $\psi\circ\phi$ of your map with mine is $r\in R\mapsto r^2\in R$ which is generally not linear. – Mariano Suárez-Álvarez Sep 24 '15 at 19:41
  • You should really try to actually prove that your map is linear. Saying things like "it is easily seen that it is linear" never contitutes a proof of anything :-) – Mariano Suárez-Álvarez Sep 24 '15 at 19:41
  • @MarianoSuárez-Alvarez. Ok sure. $\phi(a+b)=( \phi_1(a+b), \phi_2(a+b))+ D$ $= ( \phi_1(a)+\phi_1(b), \phi_2(a)+\phi_2(b))+ D = ( (\phi_1(a), \phi_2(a))+ (\phi_1(b),\phi_2(b))+ D$ $=(\phi_1(a), \phi_2(a))+D +(\phi_1(b),\phi_2(b)) + D= \phi(a) + \phi(b)$. I might be wrong, but I really cant see what is wrong with the above proof for linearity. Can you point it out? – Brad Sep 24 '15 at 19:50
  • The third equality you wrote is false. You should review the construction of the tensor product. – Mariano Suárez-Álvarez Sep 24 '15 at 19:52
  • @MarianoSuárez-Alvarez You are right, thanks. – Brad Sep 24 '15 at 20:43
  • @MarianoSuárez-Alvarez the given map is a ring homomorphism. I refer to Atiyah, page 30-31 on tensor product of algebras. Why are you saying it is not linear? –  Sep 25 '15 at 22:13
  • @Sodan, I am saying that it is not linear simply because it is not linear. I even offered a very simple example in my second comment above. You are free to insist, of course. – Mariano Suárez-Álvarez Sep 25 '15 at 22:58

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What you have to prove is this:

There exist $R$-algebra homomorphisms $\;\phi_1\colon A\to A\otimes_RB$, $\;\phi_2\colon B\to A\otimes_RB$, such that for any $R$-algebra $S$ and any pair of $R$-algebra homomorphisms $\;\psi_1\colon A\to S$, $\;\psi_2\colon B\to S$, there exist an $R$-algebra homomorphism $\;f\colon A\otimes_RB\to S$ such that $$f\circ\phi_1=\psi_1,\quad f\circ\phi_2=\psi_2.$$

$\phi_1, \phi_2$ are defined by \begin{align*} \phi_1\colon A& \to A\otimes_RB,&\phi_2\colon B&\to A\otimes_RB.\\ a&\mapsto a\otimes 1_B& b&\mapsto 1_A\otimes b \end{align*} The $R$-algebra homomorphism $\;f\colon A\otimes_RB\to S$ is the composition: \begin{align*} A\otimes_RB&\xrightarrow{\psi_1\otimes\psi_2} S\otimes S\xrightarrow{\phantom{\psi_1\otimes\psi_2}} S,\\ a\otimes b&\mapsto \psi_1(a)\otimes\psi_2(b)\mapsto\psi_1(a)\psi_2(b) \end{align*} (we define it for elementary tensors and extend it by linearity).

Bernard
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