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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 $R\langle x_i\mid f_k\rangle\oplus R\langle y_j\mid g_l\rangle$?

Leo
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    Do you really mean $\oplus$, i.e. the coproduct in Q2? Or do you mean $\times$? See also my answer at http://math.stackexchange.com/questions/345501 – Martin Brandenburg May 23 '13 at 23:18
  • @MartinBrandenburg: Umm, I thought the notation $\oplus$ or $\bigoplus$ always means direct sum, which is equal to the direct product $\times$ or $\prod$ when there are finitely objects, and that the notation for coproduct in a category is $\amalg$ or $\coprod$. This is what I had in mind. – Leo May 24 '13 at 15:05
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    $a \oplus b = a \times b$ is absolutely wrong when you leave the category of modules (think about sets, for example). – Martin Brandenburg May 25 '13 at 10:06

2 Answers2

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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, $e^{\perp}:=1-e$.

Community
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Martin Brandenburg
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  • PS: In the non-unital case Q2 would be much simpler. – Martin Brandenburg May 23 '13 at 23:47
  • $R \langle X,F \rangle \times R \langle Y,G \rangle$ $\cong$ $R \langle X,Y,e : e^2=e, Fe, Ge^{\perp}, x=ex=xe, y=e^{\perp} y=e^{\perp} y\rangle$, I don't understand this. What is $X$ and $x$? Shouldn't we have $Xe,Ye^\perp$ instead of $Fe,Ge^\perp$? Aren't relations $XY,F,G$ missing? Furthermore, are you sure no other relation is missing? – Leo May 25 '13 at 01:44
  • $X$ is the set of chosen generators for the first algebra, $x \in X$ is an element (I was a bit lazy and avoided the set-theoretic mess). Have you understood the universal property of $A \times B$? – Martin Brandenburg May 25 '13 at 10:07
  • Sorry for a late answer. Ok, I think I've vaguely understood the universal property of $\times$. But your formula is still suspicious to me: it doesn't seem to match the formula from Paul VanKoughnett's answer in the commutative case. I still think we should have $Xe^\perp,Ye$ instead of $Fe,Ge^\perp$, and that the relations $XY,F,G$ are missing. – Leo May 26 '13 at 20:40
  • @MartingBrandenburg: For example, if $X!=!=F!=!{x}$ and $Y!=!G!=!{y}$, then by your formula, $R!\times!R\cong R\langle x,y,e,|,e^2!=!e,xe,ye^\perp,x!=!xe!=!ex,y!=!ye^\perp!=!e^\perp y\rangle$ which I don't think is correct. I suspect the following formula shold be more correct, though I'm not sure there are enough relations: $$R\langle x_i|f_k\rangle\times R\langle y_j|g_l\rangle\cong R\langle x_i,y_j,e,|, f_k,g_l,x_iy_j,y_jx_i,x_ie^\perp,e^\perp x_i,y_je,ey_j,ee^\perp\rangle.$$ Is any other relation missing or not needed? – Leo Jun 05 '13 at 18:33
  • I think you meant $xy-yx$ on your displayed equation. – Bruno Stonek Nov 28 '13 at 11:36
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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 $(1, 1) = e_1 + e_2$.

Tunococ
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