What does $R^{(E)} = \mathbb{Q}$ mean?

Question

Okay, I think I understand- using the section you have linked to as a guide, letting $E$ be the generating set for $\mathbb{Q}$ and $R^{(E)}$ $= \mathbb{Q}$, in particular we can use the fact that $\mathbb{Z}$ is a $\mathbb{Z}$-module over itself, so let $M = \mathbb{Z}$ and then given an arbitrary map $\phi : E \rightarrow \mathbb{Z}$, there exists a unique homomorphism $\psi : \mathbb{Q} \rightarrow \mathbb{Z}$ such that $\phi = \psi \circ i$, where $i : E \rightarrow \mathbb{Q}$ is the inclusion map. However, we have just shown that the only $\psi$ that exists is $0$, and $0 \circ i = 0 \neq \phi$.

score 1 · Accepted Answer · answered Jul 17 '18 at 18:07

In this context, $R^{E}=\mathbb Q$ is the statement that if $\mathbb Q$ were a free $\mathbb Z$-module with basis $E$, then this structure would define an isomorphism $\oplus_{E} \mathbb Z\to \mathbb Q$, where here we think about $\mathbb Z$ as being the ring $R$.

What does $R^{(E)} = \mathbb{Q}$ mean?

1 Answers1