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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?

LaTeXFan
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    One must change the multiplication operation on $\mathbb{R}\times\mathbb{R}$. Then such a homomorphism can exist. – TheNumber23 Nov 16 '13 at 07:28
  • @TheNumber: What is the multiplication operation on $\mathbb R\times \mathbb R$? – user99680 Nov 16 '13 at 07:34
  • @user99680 See the answer below. – TheNumber23 Nov 16 '13 at 07:35
  • The same as the one on $\Bbb C$. The issue here is that $\Bbb C$ is a field, and homomorphic images of fields are again fields: it has no proper, non-trivial ideals (which are the homomorphism kernels). That can be used as an alternative start for Prahlad's answer. – zibadawa timmy Nov 16 '13 at 07:36
  • @zibadawatimmy, how would you use use that? Note that $\mathbb{R} \times \mathbb{R}$ has subfields, for example ${(x,x) : x \in \mathbb{R}}$. – goblin GONE Nov 16 '13 at 07:52

3 Answers3

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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.

goblin GONE
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    That's fine, but you said that you are assuming that it preserves unity, so Step 1 is an assumption, not an exercise. If you don't assume that, then of course it can be the zero homomorphism. – Derek Holt Nov 16 '13 at 09:20
  • ...and also assuming the operations in $;\Bbb R\times\Bbb R;$ (in fact, multiplication) are the "usual" coordinatewise ones. As stated in the comments above, if the multiplication in the cartesian product is changed there exists not only a hom. but in fact an isomorphism. – DonAntonio Nov 16 '13 at 10:01
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    @DerekHolt, yes Step 1 immediately follows from the fact that $(1,1)$ is the identity in $\mathbb{R} \times \mathbb{R}$. But if the OP didn't already know this, then it remains an (admittedly trivial) exercise. As for the operations on $\mathbb{R} \times \mathbb{R}$, I think it goes without saying that we're viewing both $\mathbb{R}$ and $\mathbb{C}$ as rings for this question, in which case the operations on $\mathbb{R} \times \mathbb{R}$ are given by the very definition of ring-theoretic Cartesian product. – goblin GONE Nov 16 '13 at 10:26
  • It seems that the answer to this question depends on the definition of addition and multiplication. Right? – LaTeXFan Nov 18 '13 at 04:14
  • @XuS, yes. But rather than viewing $\mathbb{R}$ as a set, lets view it as a ring. Then $\mathbb{R} \times \mathbb{R}$ has a very specific meaning. Namely, its also a ring. The operations are defined as follows. Given $(x,y)$ and $(x',y')$ in $\mathbb{R} \times \mathbb{R},$ the sum is $(x,y) + (x',y') = (x+x',y+y'),$ and the product is $(x,y)(x',y') = (xx',yy')$. The negation is $-(x,y) = (-x,-y)$, and the zero is given by $(0,0)$. These definitions aren't just arbitrary; they are the unique correct way of defining the product... – goblin GONE Nov 18 '13 at 07:31
  • ...in the category of rings. But this probably won't make sense from the wikipedia page alone. – goblin GONE Nov 18 '13 at 07:33
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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.

TheNumber23
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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?

Prahlad Vaidyanathan
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  • I don't understand the downvote. I do belong to the school thinking that rings have multiplicative neutral elements preserved under homomorphisms, but AFAICT that doesn't invalidate this approach to the problem. – Jyrki Lahtonen Nov 16 '13 at 08:15
  • Your second implication is wrong: a priori you could for instance have $\phi(1) = (1,1)$ and $\phi(i) = (1,-1)$. – Dan Petersen Nov 16 '13 at 08:19