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?
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$\begingroup$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.
$\endgroup$ 6 $\begingroup$Let $(\mathbb{R}\times\mathbb{R},+,\cdot)$ be a ring defined by operations $$+:(\mathbb{R}\times\mathbb{R})\times (\mathbb{R}\times\mathbb{R})\rightarrow \mathbb{R}\times\mathbb{R}$$ by $(a,b)+(c,b)=(a+c,b+d)$ and $$\cdot:(\mathbb{R}\times\mathbb{R})\times (\mathbb{R}\times\mathbb{R})\rightarrow \mathbb{R}\times\mathbb{R}$$ where $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$ It's not hard to check the properties for a field hold, distributes, associative, commutes, inverses, etc...
Now define $\varphi:\mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$ by $a+bi\mapsto (a,b)$. This is an isomorphism.
So without defining the multiplication operation on $\mathbb{R}\times\mathbb{R}$ a isomorphism can exist. But given the normal operation on $\mathbb{R}\times\mathbb{R}$. no such homomorphism can exist.
$\endgroup$ $\begingroup$Suppose such a $\phi\neq 0$ exists, then $$ \phi(1)^2 = \phi(1) \Rightarrow \phi(1) \in \{(1,0), (0,1), (1,1)\} $$ Also, $$ \phi(i)^4 = \phi(1) \Rightarrow \phi(i) = \phi(1) $$ Hence, $$ \phi(a+bi) = (a+b)\phi(1) $$ Can you check that such a map cannot be a homomorphism?
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