# tensoring a polynomial algebra: $R[x_i]/\mathfrak{a}\otimes A \cong A[x_i]/\mathfrak{a}$

Let $R$ be a commutative unital ring, $A$ an associative unital $R$-algebra, $I$ an arbitrary set, and $\mathfrak{a}$ an ideal of $R[x_i; i\!\in\!I]$. If $A$ is commutative, then there is an isomorphism of $R$-algebras

$$R[x_i,i\!\in\!I]/\mathfrak{a} \otimes A \:\cong\: A[x_i,i\!\in\!I]/\mathfrak{a}1_A,$$

where $\mathfrak{a}1_A$ denotes the ideal of $A[x_i,i\!\in\!I]$, generated by $\{1_Af(x); f(x)\!\in\!\mathfrak{a}\}$. For example, there is an isomorphism of $\mathbb{Z}$-algebras $\mathbb{Z}_3\,\otimes\,\mathbb{Z}[x,y]/\langle\langle 1\!+\!3x^2\!-\!5xy\rangle\rangle \:\cong\: \mathbb{Z}_3[x,y]/\langle\langle 1\!-\!2xy\rangle\rangle$.

Question 1: Is everything correct?

Question 2: Does the isomorphism still hold when $A$ is noncommutative? Must $A$ be replaced with $A^\mathrm{op}$ on the right hand side?

• Yes, everything is correct (except that you should not write $\mathbb{Z}_3$ when you actually mean $\mathbb{Z}/3$ ... but let us not start this discussion here). – Martin Brandenburg Jan 2 '13 at 14:08
• Great, so the noncommutative case also holds, right? Without $^\mathrm{op}$? – Leo Jan 2 '13 at 22:18

This actually holds in slightly more generality than you have proposed. Let $S$ be a commutative ring, $\mathfrak{a}$ an ideal in $S$ and $A$ a $S$-algebra, then $S/\mathfrak{a} \otimes_S A \cong A/\mathfrak{a}A$.
One could verify this using the universal property of tensor products, but it's easiest just to do it directly. Define a map $\phi:S/\mathfrak{a} \otimes_S A \rightarrow A/\mathfrak{a}A$ by $\phi(s+\mathfrak{a} \otimes a)=(s+\mathfrak{a})a$ for all $s \in S$, $a \in A$. By the commutivity of $S$, it follows that this defines an algebra homomorphism. It is then left to show that it is bijective. Surjectivity is easy since $\phi(1_S+\mathfrak{a} \otimes a) = a+\mathfrak{a}$ for all $a \in A$. Moreover, if there exists $s \in S$ and $a \in A$ such that $\phi(s+\mathfrak{a},a)=0$ then this implies that $sa \in \mathfrak{a}$. Thus, $\ker\phi=\mathfrak{a}$ and so the map is injective.
• Thank you. But how is this a generalization? Isn't this a special case when there are no variables $x_i$? Now that I think of it, $(A/I) \otimes (B/J) \cong (A\otimes B)/(A\otimes J +I\otimes B)$ seems more useful, though I still can't see if it is a generalization. – Leo Jan 5 '13 at 15:23