Show that \(A\injects B \implies A\tensor \QQ \injects B\tensor \QQ\)
Solution
1. The \(\mathbb Z\)-module \(\mathbb Q\) is flat.
It is the localization \(S^{-1}\mathbb Z\) for \(S=\mathbb Z\setminus\{0\}\). Localization is a flat module over the original ring.
2. Therefore tensoring an injection \(A\hookrightarrow B\) with \(\mathbb Q\) preserves injectivity:\[ \boxed{A\otimes_{\mathbb Z}\mathbb Q\hookrightarrow B\otimes_{\mathbb Z}\mathbb Q.} \]
Start with the exact sequence \[ 0\longrightarrow A\longrightarrow B. \] Since \(\mathbb Q\) is flat, applying \(-\otimes_{\mathbb Z}\mathbb Q\) preserves exactness at the left, yielding \[ 0\longrightarrow A\otimes\mathbb Q\longrightarrow B\otimes\mathbb Q. \]