The direct sum of two finitely generated algebras is finitely generated

Question

Let $X$ and $Y$ be sets of variables. Let $R$ be a commutative ring. Suppose $S$ and $T$ are finitely generated $R$-algebras. Then $S\cong R[X]/I$ and $T\cong R[Y]/J$ for some $R[X]$-ideal $I$ and $R[Y]$-ideal $J$.

Why is $S\bigoplus T$ a finitely generated $R$-algebra? I wanted to say "just combine the generators of $S$ and $T$" together, but isn't that saying that

$$\frac{R[X]}{I}\oplus\frac{R[Y]}{J}\cong \frac{R[X,Y]}{IR[X,Y]+JR[X,Y]}?$$

But if that's the case, then I'm confused because I thought

$$\frac{R[X]}{I}\otimes\frac{R[Y]}{J}\cong \frac{R[X,Y]}{IR[X,Y]+JR[X,Y]}?$$

But I thought the direct sum and tensor product were very different operations?

Answer 1

I think you’re making it much too hard, perhaps by looking at the situation too abstractly.

In my book, $S\oplus T$ is the set of all pairs $(s,t)$ for which $s\in S$, $t\in T$. In other words, as a set, $S\oplus T$ is equal to $S\times T$. Addition and multiplication take place coordinatewise, so that the zero-element is $(0,0)$ and the multiplicative identity is $(1,1)$. So if $\{x,\cdots,x_m\}$ is a generating set for the algebra $S$ and $\{y_1,\cdots,y_n\}$ does the same for $T$, then a generating set for $S\oplus T$ is $\{(x_1,0),\cdots,(x_m,0),(0,y_1),\cdots,(0,y_n)\}$.

That should be good enough, I think.

Answer 2

What do you mean by $\oplus$ when you are working with $R$-algebras, or more generally with commutative rings? See this wiki section.

If you are writing $\oplus$ to mean the coproduct (in the category of $R$-algebras), then you ought to write $\otimes_R$ instead, and your equation is correct. If you mean the product, then you ought to write $\times$ instead, and your equation is incorrect (e.g., with $I=J=0$, $\mathbb{C}[X]\times \mathbb{C}[Y]$ is not a domain, so it can't be $\mathbb{C}[X,Y]$), but see @Lubin's answer for more in this case. If you mean neither of those, you should clarify what it is that you mean (and your equation is probably incorrect).