Proof verification of showing that $M\otimes A/{\frak a} \cong M/{\frak a}M$

abstract-algebramodule-isomorphismsolution-verificationtensor-products

Let $A$ be a (commutative) Ring with $1$. Let $M$ be an $A$-Module and ${\frak a}\subseteq A$ an Ideal.

I was trying to prove that $M\otimes A/{\frak a} \cong M/{\frak a}M$.

I would like to know whether my attempt is reasonable and if not, what i need to take more care of.

I just considered the "obvious choices" via \begin{align*} \varphi\colon M\otimes A/{\frak a} &\to M/{\frak a}M \\ m\otimes (a+\frak a) &\mapsto am+{\frak a}M\end{align*}

and its inverse as

\begin{align*} \psi\colon M/{\frak a}M &\to M\otimes A/{\frak a} \\m+{\frak a}M &\mapsto m\otimes(1+{\frak a})\end{align*}

In order to confirm they are indeed inverses of each other, i argued as follows:

$$(\varphi\circ\psi)(m+{\frak a}M) = \varphi(m\otimes (1+{\frak a})) = 1\cdot m + {\frak a}M = m + {\frak a}M$$

\begin{align*}(\psi\circ\varphi)(m\otimes(a+{\frak a})) = \psi(am+{\frak a}M) &= am \otimes (1+{\frak a}) \\ &= am \otimes 1 + am \otimes \frak a \\ &= {\color{red} {m\otimes a + m\otimes a{\frak a}}} \\ &= m\otimes a + m\otimes \frak a \\ &= m\otimes (a+\frak a).\end{align*}

I didn't include checking for well definedness, but i'm aware that i need to confirm that the maps i've chosen are indeed well defined. This post is more about the way i chose the respective maps.

My questions:

  1. Is my attempt more or less reasonable? Or did i do some technical errors? I'm a bit uncertain about the red part.

  2. Would it be necessary to invoke the homomorphism theorem in order to justify the existence of $\psi$ in the first place?

Thanks for any advise and help!

Best Answer

For well-definedness of the maps you can just use the universal property of tensor products for $\varphi$, and define $\psi$ on $M$ and show that elements of $\mathfrak aM$ go to zero. The maps you have specified are good.

One thing to keep in mind is that general elements of $M\otimes (A/\mathfrak a)$ are sums of pure tensors. But your argument is still fine since the sums will just pull out of the maps since they're homomorphisms.

The calculations you're doing around the red part are essentially correct, although you should keep the "$+\mathfrak a$" part on the right side of the tensors since the right side of the tensor is in $A/\mathfrak a$. You can also just do $$(am)\otimes(1+\mathfrak a)=m\otimes(a(1+\mathfrak a))=m\otimes (a+\mathfrak a).$$

Answers to comments:

The universal property is exactly what's used to justify the map you gave $M\otimes (A/\mathfrak a)\to M/\mathfrak a M$ is well-defined. So you would consider the map $M\times (A/\mathfrak a)\to M/\mathfrak a M$ via $(m,a+\mathfrak a)\mapsto am+\mathfrak a M$ and show that this is $A$-bilinear, and the universal property gives you a unique $A$-linear map $\varphi:M\otimes (A/\mathfrak a)\to M/\mathfrak a M$ which satisfies $\varphi(m\otimes (a+\mathfrak a))=am+\mathfrak a M$. Basically when you want to send a pure tensor $x\otimes y$ to some $z$, you use the universal property and first define a bilinear map which sends $(x,y)$ to $z$.

For showing $\psi$ is well-defined you don't need the first isomorphism theorem (I assume this is what you mean by the isomorphism theorem). You just show that $\mathfrak a M$ is contained in the kernel of the map $M\to M\otimes (A/\mathfrak a)$ via $m\mapsto m\otimes(1+\mathfrak a)$, so that this map factors to a map on $M/\mathfrak a M$. You could also just directly check that your map $\psi$ is well-defined, but this will be equivalent to what I have described here (and what I have described would be a little cleaner).

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