Shorter proof that $\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}\cong \mathbb{C}^2$

Question

This Wikipedia page gives an example proof that $\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}\cong \mathbb{C}^2$ by relying on the fact that $\mathbb{C}\cong \mathbb{R}[x]/(x^2+1)$, then using the fact that for $\mathbb{C}$ a commutative $\mathbb{R}$-algebra, $$\mathbb{C}\otimes_{\mathbb{R}} \mathbb{R}[x]/(x^2+1)\cong \mathbb{C}[x]/(x^2+1)\cong \mathbb{C}[x]/(x+i)\times \mathbb{C}[x]/(x-i)\cong \mathbb{C}^2$$ relying on the Chinese remainder theorem in the middle step. However, would it be easier to simply note that $\mathbb{C}$ is free as an $\mathbb{R}$-module with basis $\{1, i\}$, and therefore any element of $\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}$ has a unique representation as $z_1\otimes 1+z_i\otimes i$, with the canonical isomorphism $\varphi : \mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}\to \mathbb{C}^2$ given by $\varphi(z_1\otimes 1+z_i\otimes i) = (z_1, z_i)$? Is there some reason this wouldn't be valid?

Answer 1

Your mapping is not a homomorphism of rings. For example, $$(1\otimes i)\cdot(1\otimes i)=1\otimes (i^2)=-(1\otimes 1),$$ but you declared $ \phi(1\otimes i)=(0,1) $ meaning that $$ \phi(1\otimes i)^2=(0,1)^2=(0,1)\neq(-1,0)=\phi((1\otimes i)^2). $$

Answer 2

Your map $\varphi$ is fine if you just want an isomorphism of $\mathbb{C}$-modules, and is probably the best way to get such an isomorphism. The advantage of the complicated isomorphism given by Wikipedia is that it actually has a stronger property: it's an isomorphism of $\mathbb{C}$-algebras, not just of $\mathbb{C}$-modules. That's why it says "as ring[s]" before presenting the isomorphism (though this is rather poorly explained, since it never mentioned that the previous example gave an isomorphism of rings, nor has the article even mentioned the ring structure on tensor products previously).