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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?

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  • $\begingroup$ 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). $\endgroup$ – Martin Brandenburg Jan 2 '13 at 14:08
  • $\begingroup$ Great, so the noncommutative case also holds, right? Without $^\mathrm{op}$? $\endgroup$ – Leo Jan 2 '13 at 22:18
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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.

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  • $\begingroup$ what do you mean by "One could verify this using the universal property of tensor products, but it's easiest just to do it directly." when you use the universal property in the next sentence (which you cannot avoid anyway)? $\endgroup$ – Martin Brandenburg Jan 4 '13 at 15:31
  • $\begingroup$ By that you mean the homomorphism comes from the universal property? $\endgroup$ – J. Gaddis Jan 4 '13 at 17:21
  • $\begingroup$ 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. $\endgroup$ – Leo Jan 5 '13 at 15:23

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