Prove or disprove: There exists a ring homomorphism $\varphi: \mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$.
I think it is intuitive to try $\varphi: \mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$ defined by $\varphi: a+bi\mapsto (a, b)$. Then let us check if this works.
- It is trivial that $\varphi[(a+bi)+(c+di)]=(a+b, c+d)=\varphi(a+bi)+\varphi(c+di)$.
- $\varphi(1+0i)=(1, 0)\neq(1,1)$.
- $\varphi[(a+bi)\times(c+di)]=\varphi[(ac-bd)+(ad+bc)i]=(ac-bd, ad+bc)$.
From the above results, it seems that this definition does not satisfy the requirement of a ring homomorphism. Is there a way to modify it, please? Or is it possible at all?
Best Answer
No such homomorphism can exist, so long as ring homomorphisms are assumed to preserve unity.
Main Result. There is no ring homomorphism $\mathbb{C} \rightarrow \mathbb{R} \times \mathbb{R}$.
Proof. Suppose toward a contradiction that $\varphi : \mathbb{C} \rightarrow \mathbb{R} \times \mathbb{R}$ is a ring homomorphism. Then by Lemma 0 below, it follows that $-1$ has a square root in $\mathbb{R} \times \mathbb{R}$. Thus by Lemma 1 below, it follows that $-1$ has a square root in $\mathbb{R}$. But this contradicts Lemma 2 below.
Lemma 0. If a ring homomorphism $\mathbb{C} \rightarrow R$ exists, then $-1$ must have a square root in $R$.
Proof. I claim that $\varphi(i)^2$ is always a square root of $-1$ in $R$. To see this, compute: $$\varphi(i)^2 = \varphi(i^2) = \varphi(-1) = -\varphi(1) = -1.$$
Lemma 1. For all rings $R$ and $S$, if $-1$ has a square root in $R \times S$, then it must have a square root in both $R$ and $S$.
Proof. Suppose $-1$ has a square root in $R \times S$, call it $(r,s)$. Then $(r,s)^2 = -1$. Thus $(r^2,s^2) = (-1,-1)$. Thus $r^2 = -1$. Thus $-1$ has a square root in $R$. A similar argument shows that it must have a square root in $S$.
Lemma 2. The element $-1 \in \mathbb{R}$ does not have a square root in $\mathbb{R}$.
Suppose toward a contradiction that it did, call this value $i_\mathbb{R}.$ Then $(i_\mathbb{R})^2 = -1$. Thus $(i_\mathbb{R})^2 + 1 = 0$. Now we know that $\forall x \in \mathbb{R} : x^2 \geq 0$. Thus $\forall x \in \mathbb{R} : x^2 + 1 > 0.$ Thus $(i_\mathbb{R})^2 + 1 > 0$. Ergo $0 > 0$, a contradiction.