# Tensor product of a ring with its quotient

Say $A$ is a ring, $I$ an ideal, $A/I$ the usual quotient ring. I am trying to figure what is this tensor product $$A \otimes_{A} A/I$$ My thoughts are these: a generic element of this product is $a \otimes [b]$, where $b$ is a class in the quotient. But since the tensor is $A$-linear, this is also equal to $1 \otimes [a][b]$ which would make me think that $$A \otimes_{A} A/I \simeq A/I$$ Is this correct? If not, can that tensor product be written in another way, or even be described? (suppose $A$ is finitely generated if necessary)

Actually I need to figure out what another tensor product is. I have an algebra $H = P \otimes_{\mathbb{K}} Q$ , $P=\mathbb{K}[X_1, \dots, X_n], Q$ are $\mathbb{K}$-algebras, $\mathbb{K}$ is algebrically closed, $I$ a maximal ideal of $P$ generated by $X_i - a$ where $a$ is a fixed nonzero element of $\mathbb{K}$. Now, I need to know what is $$H \otimes_{P} P/I$$ As before, the generic element is $(p \otimes q) \otimes [r]$. Can I say that is equivalent to $(1 \otimes q) \otimes [pr]$ ?

Thanks to anyone who will help me with this problem. Sadly I am not familiar with tensor products over rings, and I am a bit confused about their behaviour on quotients.

• $A \otimes_A M \cong M$ for any $A$-module $M$ whatsoever. – Zhen Lin Dec 10 '15 at 9:18
• I thought so! Could you take a look at the second part of the question too...? – AnalysisStudent0414 Dec 10 '15 at 9:19
• Well, $M \otimes_A A/I \cong M / I M$. – Zhen Lin Dec 10 '15 at 9:24
• So $H \otimes_P P/I \simeq H/IH$? Can the latter be split into a $P$ and a $Q$ quotient somehow? – AnalysisStudent0414 Dec 10 '15 at 9:39
• Do you know the universal property characterization of tensor products? – Florian Dec 10 '15 at 10:46

You are basically right, with the appropriate interpretation of "the generic element is". You obtain $Q\otimes_{\mathbb K} P/I$, which is a $\mathbb{K}$-vector space that becomes a $P$-module by defining $$p \cdot (q\otimes [\tilde p]):=q\otimes [p\tilde p]$$ Actually such a definition already uses the universal property for tensor products (remember that not all elements of $Q\otimes P/I$ have the form $q\otimes [\tilde p]$): for all $p\in P$ the mapping $Q\times P/I$, $(q,[\tilde p])\mapsto q\otimes [p\tilde p]$ is bilinear, so by the universal property of tensor products there is a linear map with the above property.