flat module

hbghlyj

To show $\mathbb{Z}/2\mathbb{Z}$ is not a flat $\mathbb{Z}$-module, we use the definition that a module $M$ is flat if for every injective homomorphism $f: A \to B$, the induced map $f \otimes \text{id}_M: A \otimes M \to B \otimes M$ is also injective.

Consider the injective $\mathbb{Z}$-module homomorphism $f: \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = 2x$.

Now, tensor this map with $\mathbb{Z}/2\mathbb{Z}$:
$$f \otimes \text{id}_{\mathbb{Z}/2\mathbb{Z}}: \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$$
Let $y \in \mathbb{Z}/2\mathbb{Z}$. The map $f \otimes \text{id}_{\mathbb{Z}/2\mathbb{Z}}$ takes an element $x \otimes y$ to$$f(x) \otimes y = 2x \otimes y= x \otimes 2y=0$$
As $f \otimes \text{id}_{\mathbb{Z}/2\mathbb{Z}}=0$ is not injective, $\mathbb{Z}/2\mathbb{Z}$ is not a flat $\mathbb{Z}$-module.

hbghlyj

An $R$-module $P$ is projective if, for every surjective $R$-module homomorphism $f: M \to N$ and every $R$-module homomorphism $g: P \to N$, there exists an $R$-module homomorphism $h: P \to M$ such that $f \circ h = g$.

Equivalently, an $R$-module $P$ is projective if and only if every short exact sequence $0 \to A \to B \to P \to 0$ splits. This means there is a section map $s: P \to B$ such that $g \circ s = id_P$.
Another equivalent definition is that $P$ is projective if and only if it is a direct summand of a free module. That is, there exists an $R$-module $Q$ such that $F \cong P \oplus Q$ for some free module $F$.

An $R$-module $M$ is flat if the tensor product functor $M \otimes_R - : \text{Mod}_R \to \text{Ab}$ (or $\text{Mod}_R \to \text{Mod}_R$ for commutative rings) preserves exact sequences. That is, if $0 \to A \to B \to C \to 0$ is a short exact sequence of $R$-modules, then $0 \to M \otimes_R A \to M \otimes_R B \to M \otimes_R C \to 0$ is also a short exact sequence. Since the tensor product is always right exact, this condition reduces to requiring that for every injective $R$-module homomorphism $\alpha: A \to B$, the induced map $M \otimes_R \alpha: M \otimes_R A \to M \otimes_R B$ is also injective.

hbghlyj

Let $F$ be a free $R$-module. By definition, $F$ is isomorphic to a direct sum of copies of $R$, i.e., $F \cong \bigoplus_{i \in I} R$ for some index set $I$.

To show $F$ is flat, we need to show that for any injective $R$-module homomorphism $\alpha: A \to B$, the induced map $F \otimes_R \alpha: F \otimes_R A \to F \otimes_R B$ is injective.

Consider an elementary tensor $f \otimes a \in F \otimes_R A$. If $F \otimes_R \alpha (f \otimes a) = 0$, this means $f \otimes \alpha(a) = 0$.
Since $F \cong \bigoplus_{i \in I} R$, any element $f \in F$ can be written as a finite sum $f = \sum_{j=1}^n r_j e_j$ where $e_j$ are basis elements and $r_j \in R$.
Then $f \otimes \alpha(a) = (\sum r_j e_j) \otimes \alpha(a) = \sum r_j (e_j \otimes \alpha(a))$.

The tensor product $R \otimes_R A$ is naturally isomorphic to $A$ via $r \otimes a \mapsto ra$.
So, $F \otimes_R A \cong (\bigoplus_{i \in I} R) \otimes_R A \cong \bigoplus_{i \in I} (R \otimes_R A) \cong \bigoplus_{i \in I} A$.

Let $x \in F \otimes_R A$. We can write $x = \sum_{k=1}^m f_k \otimes a_k$ for some $f_k \in F$ and $a_k \in A$.
If $(F \otimes_R \alpha)(x) = 0$, then $\sum_{k=1}^m f_k \otimes \alpha(a_k) = 0$ in $F \otimes_R B$.
Since $F$ is free, we can choose a basis $\{e_i\}_{i \in I}$ for $F$. Each $f_k$ can be written as a finite linear combination of basis elements.
By rearranging terms, any element in $F \otimes_R A$ can be uniquely written as $\sum_{i \in I'} e_i \otimes a_i'$ for some finite subset $I'$ of $I$ and $a_i' \in A$.
So, if $\sum e_i \otimes \alpha(a_i') = 0$, then $\alpha(a_i') = 0$ for all $i \in I'$.
Since $\alpha: A \to B$ is injective, $\alpha(a_i') = 0$ implies $a_i' = 0$ for all $i \in I'$.
Therefore, $\sum e_i \otimes a_i' = 0$ in $F \otimes_R A$. This shows that $F \otimes_R \alpha$ is injective.
Thus, every free module is flat.

hbghlyj

Suppose $M$ is a flat $R$-module, and $P$ is a direct summand of $M$. This means there exists an $R$-module $Q$ such that $M \cong P \oplus Q$.

Let $\alpha: A \to B$ be an injective $R$-module homomorphism. We need to show that $P \otimes_R \alpha: P \otimes_R A \to P \otimes_R B$ is injective.

We have the natural isomorphism $(P \oplus Q) \otimes_R A \cong (P \otimes_R A) \oplus (Q \otimes_R A)$.
Similarly, $(P \oplus Q) \otimes_R B \cong (P \otimes_R B) \oplus (Q \otimes_R B)$.

The map $M \otimes_R \alpha$ can be written as:
$(P \oplus Q) \otimes_R \alpha: (P \otimes_R A) \oplus (Q \otimes_R A) \to (P \otimes_R B) \oplus (Q \otimes_R B)$
which acts as $(P \otimes_R \alpha) \oplus (Q \otimes_R \alpha)$.

Since $M \cong P \oplus Q$ is flat, the map $(P \otimes_R \alpha) \oplus (Q \otimes_R \alpha)$ is injective.
If an element $(x, y) \in (P \otimes_R A) \oplus (Q \otimes_R A)$ is mapped to $0$, i.e., $((P \otimes_R \alpha)(x), (Q \otimes_R \alpha)(y)) = (0,0)$, then it implies that $(P \otimes_R \alpha)(x) = 0$ and $(Q \otimes_R \alpha)(y) = 0$.
Since $(P \otimes_R \alpha) \oplus (Q \otimes_R \alpha)$ is injective, $(x, y) = (0,0)$, which means $x=0$ and $y=0$.
In particular, if $x \in P \otimes_R A$ is such that $(P \otimes_R \alpha)(x) = 0$, then taking $y=0$, we have $(x,0)$ mapped to $(0,0)$, which implies $x=0$.
Therefore, $P \otimes_R \alpha$ is injective.
Thus, any direct summand of a flat module is flat.

Since every projective module is a direct summand of a free module, and we have shown that free modules are flat and flatness is preserved under direct summands, it follows directly that every projective module is flat.