Seeing the real numbers as a $\mathbb{C}$-module

Question

I recently bumped into this question on the community and the accepted answer left me thinking. I always thought that we couldn't view the real numbers as a $\mathbb{C}$-module for the following reason. Any ring is a module over itself. Therefore, in a ring $R$, the left $R$-modules are exactly the left $R$-ideals. Now if we take the ring $(\mathbb{C},+, \cdot)$ with the usual addition and product, $\mathbb{R}$ is not an ideal of this ring. Therefore, if it is not a $\mathbb{C}$-ideal, it cannot be a $\mathbb{C}$-module. Where is my reasoning wrong? Perhaps I need to define different operations in the ring?

Answer 1

$\Bbb R$ is not a $\Bbb C$-submodule of $\Bbb C$ for the reasons you provide.

There is, however, a $\Bbb C$-module structure on the abelian group $\Bbb R$.

These two statements don't contradict each other. The fact that $\Bbb R$ is a subset of $\Bbb C$ may be misleading here, it's irrelevant for the existence of a $\Bbb C$-module structure.

Here's an easier example of the same phenomenon: consider $\Bbb Q[T]/(T^2)$ and the subgroup $\Bbb Q \subset \Bbb Q[T]/(T^2)$. It's not a submodule, because it's not an ideal, but still $\Bbb Q$ can be given a $\Bbb Q[T]/(T^2)$-structure. (Do you see how?)

Answer 2

It seems that you are confusing modules and submodules. $\mathbb R$ can be given structure of a $\mathbb C$-module, but that module can't be submodule of $\mathbb C$. But these are two different things. There are many $\mathbb C$-modules $M$ that are not submodules of $\mathbb C$. For example, take $M$ to be space of functions $\{f\colon X\to\mathbb C\}$, then you can define $(z\cdot f)(x) = zf(x)$, where on the RHS we have multiplication in $\mathbb C$. Clearly, that's not a submodule of $\mathbb C$. The fact that $\mathbb R$ is a subset of $\mathbb C$ is irrelevant, we can define how $\mathbb C$ acts on $\mathbb R$ in other ways than by multiplication in $\mathbb C$.

However, this is not a straightforward thing to do, but we can do it. I'll use the fact that $\mathbb R$ and $\mathbb C$ are isomorphic as abelian groups (you can see why here). Let $\varphi\colon (\mathbb R,+)\to (\mathbb C,+)$ be a group isomorphism and let's define $$z\cdot x = \varphi^{-1}(z\varphi(x)),\ z\in\mathbb C,\ x\in\mathbb R.$$ This defines $\mathbb C$-module structure on $\mathbb R$:

\begin{align}z\cdot(x+y)&=\varphi^{-1}(z\varphi(x+y)) = \varphi^{-1}(z(\varphi(x)+\varphi(y)))=\varphi^{-1}(z\varphi(x)+z\varphi(y))\\ &= \varphi^{-1}(z\varphi(x))+\varphi^{-1}(z\varphi(y)) =z\cdot x + z\cdot y\end{align}

\begin{align} (z+w)\cdot x &=\varphi^{-1}((z+w)\varphi(x)) = \varphi^{-1}(z\varphi(x) +w\varphi(x)) = \varphi^{-1}(z\varphi(x))+ \varphi^{-1}(w\varphi(x))\\ &= z\cdot x + w\cdot x \end{align}

\begin{align}z\cdot(w\cdot x) = \varphi^{-1}(z\varphi(w\cdot x)) = \varphi^{-1}(z\varphi(\varphi^{-1}(w\varphi(x)))) = \varphi^{-1}(zw\varphi(x)) = (zw)\cdot x\end{align}

$$1\cdot x = \varphi^{-1}(1\varphi(x)) = \varphi^{-1}(\varphi(x)) = x.$$