tensor product and direct product of algebra presentations

Question

Let $R$ be a commutative unital ring and $R\langle x_i\mid f_j\rangle$ denote a unital $R$-algebra presentation.

Q1: What is the presentation of $R\langle x_i\mid f_k\rangle\otimes R\langle y_j\mid g_l\rangle$? It must be $R\langle x_i,y_j\mid f_k,g_l,h_m\rangle$ for some $h_m$.

Q2: What is the presentation of $R\langle x_i\mid f_k\rangle\oplus R\langle y_j\mid g_l\rangle$?

Answer 1

I misunderstood the problem. Martin pointed out that it is for $R$-algebras, not $R$-modules.

So, this is what I think is correct: $$R\langle x_i \mid f_i\rangle \otimes R\langle y_i \mid g_i\rangle = R\langle x_i, y_j \mid f_i, g_i, x_iy_j - y_jx_i\rangle.$$ The direct sum (based on the coproduct of semigroups) is $$R\langle x_i \mid f_i \rangle \oplus R\langle y_i \mid g_i\rangle = R\langle x_i, y_i \mid f_i, g_i \rangle. $$ The direct product (based on the product of semigroups) is $$ R\langle x_i \mid f_i \rangle \times R\langle y_i \mid g_i\rangle = R\langle x_i, y_i, e_1, e_2 \mid f_ie_1, g_ie_2, x_iy_j - y_jx_i, e_1 + e_2 - 1, e_1x_i - x_i, x_i - x_ie_1, e_2y_i - y_i, y_i - y_ie_2, e_1^2 - e_1, e_2^2 - e_2 \rangle. $$ Intuitively, $e_1$ represents $(1, 0)$ and $e_2$ represents $(0, 1)$ in the direct product, while elements originally from $X$ and $Y$ become $(x, 0)$ and $(0, y)$. The unit in the product is $(1, 1) = e_1 + e_2$.

Answer 2

Q1. I assume by $A \otimes_R B$ you mean the $R$-algebra which has as underlying $R$-module the tensor product of the underlying $R$-modules, and whose algebra is structure is given by $1 := 1 \otimes 1$ and $(a \otimes b) (a' \otimes b'):=aa' \otimes bb'$. This $R$-algebra has the following universal property: Homomorphisms $A \otimes_R B \to C$ correspond 1:1 to pairs of homomorphisms $f : A \to C$ and $g : B \to C$ which commute in the following sense: For $a \in A$ and $b \in B$ we have $f(a) g(b)=g(b) f(a)$. Therefore, it might be called the "commutative coproduct". It follows immediately that

$R\langle X,F \rangle \otimes_R R\langle Y,G \rangle = R \langle X \cup Y : F \cup G \cup \{xy-xy : x \in X, y \in Y\}\rangle$

Q2. Do you mean the coproduct in the category of $R$-algebras? In that case, we have $R \langle X,F \rangle \oplus R \langle Y,G \rangle = R\langle X \cup Y ,F \cup G \rangle$. Or do you mean the direct product (which is, unfortunately, often denoted by $\oplus$, see also here)? In that case, we have $R \langle X,F \rangle \times R \langle Y,G \rangle = R \langle X,Y,e : e^2=e, Fe, Ge^{\perp}, x=ex=xe, y=e^{\perp} y=e^{\perp} y\rangle$. By $Fe$ I mean that every occurence of $1$ in a relation has to be replaced by $e$. The reason is the following (co)universal property of the direct product of two $R$-algebras: A homomorphism $A \times B \to C$ corresponds to an idempotent $e \in C$ (the image of $(1,0)$) and two homomorphisms of algebras $A \to Ce$ and $B \to Ce^{\perp}$. Here, $e^{\perp}:=1-e$.