The tensor product of two flat right modules is flat

Let R and S be rings with 1. Let M be a right R-module and let N be an (R,S)-bimodule. Prove that if M is flat over R and N flat over S, then M \otimes_R N is flat over S.

Let 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 be a short exact sequence of left S-modules. Since N is flat over S, the natural sequence 0 \rightarrow N \otimes_S A \rightarrow N \otimes_S B \rightarrow N \otimes_S C \rightarrow 0 is short exact. Moreover, this is a sequence of left R-modules since N is an (R,S)-bimodule. Since M is flat over R, 0 \rightarrow M \otimes_R (N \otimes_S A) \rightarrow M \otimes_R (N \otimes_S B) \rightarrow M \otimes_R (N \otimes_S C) \rightarrow 0 is short exact. Using the natural isomorphisms, 0 \rightarrow (M \otimes_R N) \otimes_S A \rightarrow (M \otimes_R N) \otimes_S B \rightarrow (M \otimes_R N) \otimes_S B \rightarrow 0 is short exact.

Thus M \otimes_R N is flat as a right S-module.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: