# Ring homomorphisms from $\mathbb{R}$ to another unital ring $S$.

We know that there is only one non-trivial ring homomorphism from $$\mathbb{Z}$$ or $$\mathbb{Q}$$ to another unital ring $$S$$.

What’s more,when we consider the automorphism of $$\mathbb{R}$$,it is unique determined,too.More explicitly speaking: let $$\varphi \in Aut(\mathbb{R})$$ ,then$$\varphi|_{\mathbb{Q}}=Id_{\mathbb{Q}}$$, and for $$a>b \in \mathbb{R}$$,we have $$\varphi(a)-\varphi(b)=\varphi(a-b)=\varphi(\sqrt{a-b})^2>0$$. So if we have some $$a\in \mathbb{R}$$,and $$\varphi(a)\neq a$$,we can choose a rational number between $$a$$ and $$\varphi(a)$$, which will conclude a contradiction.

Here is my question: Is there a unique ring homomorphism from $$\mathbb{R}$$ to another untial ring $$S$$(send 1 to $$1_S$$)?

Especially,for $$S=M_n(\mathbb{R})$$ the $$n\times n$$ matrix over $$\mathbb{R}$$. (For $$n=1$$, it is the $$Aut(\mathbb{R})$$,and we have determined it.)

Any suggestion will be appreciated.

• Appropriated: take (something) for one's own use, typically without the owner's permission. :-) – Theo Bendit May 20 at 1:00
• Ok,and I feel sorry about that. – user469246 May 20 at 1:01
• You mean homomorphisms sending $1$ to $1$. I guess you know the axiom of choice implies there are plenty of automorphisms of $\Bbb{C}$. Then for $n=2$ you can embed $\Bbb{C}$ as $\Bbb{R} I +\Bbb{R}P \pmatrix{ 0 & -1\\ 1 & 0 }P^{-1}\subset M_2(\Bbb{R})$ for any $P \in GL_2(\Bbb{R})$ which gives plenty of (injective) homomorphisms $\Bbb{R} \to \Bbb{C}\to \Bbb{C} \to M_2(\Bbb{R})$. – reuns May 20 at 1:09
• @reuns Yes, and I edited my question. – user469246 May 20 at 1:10
• Let a ring embedding $\phi : \Bbb{C} \to M_n(\Bbb{R})$, will it be of the form $\phi = \iota \circ \sigma$ with $\sigma \in Aut(\Bbb{C})$ and $\iota(a+ib) = aI+ bJ$ for some $J^2=-I$ or is it possible that $\phi(\Bbb{C})$ will be something complicated involving the axiom of choice ? – reuns May 20 at 1:20