When does restriction of scalars induce flat Modules?

Question

Let $f:Y\rightarrow X$ be a ring homomorphism, and let us consider $X$ to be a $Y$-module via restriction of scalars, i.e. given some $y\in Y$, this element acts on $X$ via $f(y)\cdot x$, for $x\in X$

I'm trying to determine what conditions we must put on $f$ in order for $X$ to be a flat $Y$-module. By definition, this means that given an injective morphism $m:N\rightarrow M$ of $Y$-modules, we have that $1\times m: X\otimes_Y N \rightarrow X\otimes M$ is also injective.

I did a bit of computation that seemed a bit off to me, and I just wanted a bit of a sanity check. To prove what conditions on $f$ would be necessary in order for $1\times m$ to be monic, suppose we had $x_1\otimes m(n_1)= x_2\otimes m(n_2)$. We want to see how we might obtain the equation $x_1\otimes n_1 = x_2\otimes n_2$.

I would like to prove this in a setting more general than vector spaces (really, I'd like to work in the category of sup-lattices, as found in Joyal-Tierney), but for now let's just follow the characterisation provided in the answer here: Necessary and sufficient condition for equality of two tensor products

Then, there are three cases in which $x_1\otimes m(n_1)= x_2\otimes m(n_2)$:

1. Suppose $m(n_1)\neq 0$, and $\exists y\in Y$ such that $y\cdot m(n_1)=m(n_2)$. Then this forces the relation that $f(y)\cdot x_2 = x_1$. Since $Y$-module morphisms respect the action, we have that $m(n_2)=y\cdot m(n_1)= m(y\cdot n_1)$, which implies that $n_2=y\cdot n_1$ since $m$ is injective. Since $f(y)\cdot x_2=x_1$, this means that $x_2\otimes n_2 = x_2\otimes y\cdot n_1 = f(y)\cdot x_2\otimes n_1 = x_1\otimes n_1$.

2. Suppose $m(n_1),m(n_2)$ are linearly independent. Then, $x_1=x_2=0$, which implies that $x_1\otimes n_1 = x_2\otimes n_2$.

3. Finally, suppose $m(n_1)=0$. Then, we have that either $m(n_2)=0$ or $x_2=0$. Since $m$ is monic, we have that $n_1=0$, and thus $x_1\otimes n_1=0$. If $m(n_2)=0$, then $n_2=0$ as well, and thus $x_2\otimes n_2=0$; and if $x_2=0$, then $x_2\otimes n_2=0$ as well.

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This little computation seems to imply that restriction of scalars always induces a flat module in this way - is that right? But that seems strange. For instance, take the quotient map $f:\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$. It is well-known that for $n>1$, $\mathbb{Z}/n\mathbb{Z}$ is not flat over $\mathbb{Z}$ since while $n:\mathbb{Z}\rightarrow\mathbb{Z}$, $x\mapsto nx$ is injective, the map $1\times n:\mathbb{Z}/ n\mathbb{Z}\otimes \mathbb{Z}\rightarrow \mathbb{Z}/ n\mathbb{Z}\otimes \mathbb{Z}$ is not injective.

Have I done something silly in the process?

Answer 1

The problem is that linearly dependency is subtler when you work with rings.

You could have $m(n_1)\neq 0$ and $m(n_1),m(n_2)$ linearly dependent, but having $m(n_2)\neq y\cdot m(n_1)$ for all $y\in Y$.

For example, in $\mathbb{Z}$, $2,3$ are linearly dependent since $2.3-3.2=0$, but $2\neq m\cdot 3$ for all $m\in\mathbb{Z}$.

(Also, I hope you know that even if your three cases work (for example in the case of vector spaces), it is not enough to prove injectivity of of $1\times m$, since an element of the tensor product is a sum of elementary tensors.)

What you need to prove is for all $\sum_i x_i\otimes n_i$, $\sum_i x_i\otimes m(n_i)=0$ implies $\sum_i x_i\otimes n_i=0$.

You hae compelte characterization of flat modules, but the easiest case to handle is when your module is finitely presented (ie isomorphic to $F/F'$, where $F$ is free of finite rank and $F'$ is finitely generated). In this case, flat is equivalent to finitely generated and projective (which is equivalent to : isomorphic to a direct factor a free module of finite rank)