Number of ring homomorphism from the ring of real numbers to itself .

Question

I think there maybe two ring homomorphisms as there are two idempotent elements in the ring of real numbers.

Answer 1

The only ring homomorphism $f : \mathbb{R} \to \mathbb{R}$ is the identity (and for me, ring homomorphisms preserve $1$). To see this, start from $f(1) = 1$ and conclude that $f(n) = n, n \in \mathbb{Z}$, then (since $\mathbb{R}$ is uniquely divisible) that $f(q) = q, q \in \mathbb{Q}$.

Now, because $f(r^2) = f(r)^2$, it follows that $f$ sends nonnegative reals to nonnegative reals, and so must be monotonic. Together with the fact that $f(q) = q$, approximating an arbitrary real number $r$ from below and above by rationals shows that $f(r) = r, r \in \mathbb{R}$.

Without the hypothesis that ring homomorphisms preserve $1$ you can also have $f(1) = 0$, I guess, but why would you want to?

Answer 2

Claim: The only ring isomorphism from $\Bbb{R}$ to itself is the identity map.

Proof: It is clear that the identity map is a ring isomorphism from $\Bbb{R}$ to itself. Suppose $\phi:\Bbb{R}\to\Bbb{R}$ is a ring isomorphism that is not the identity map. Then, it must satisfy three conditions: $\forall a,b\in \Bbb{R}$,

1. $\phi(1)=1$,
2. $\phi(ab)=\phi(a)\phi(b)$, and
3. $\phi(a+b)=\phi(a)+\phi(b)$ .

Using Condition [1] and Condition [3], we can prove that $\phi(n)=n$ for all integer $n$ by the Principle of Mathematical Induction. By the uniqueness of multiplicative inverse in a field and Condition [2], we obtain $$\phi(q)=q \quad \forall q \in \Bbb{Q}. \qquad(\$) $$

Any injective ring homomorphism $\psi: G \to H$ must have $ker\, \psi =\{0_G\}$. Hence, we have \begin{equation} \phi(r)=\phi((\sqrt{r})^2)=(\phi(\sqrt{r}))^2 >0 \quad \forall r \in \Bbb{R}_{>0}. \end{equation} Consequently, if $x,y \in \Bbb{R}$ satisfy $x>y\ge0$, then \begin{equation} \phi(x)-\phi(y)=\phi(x-y)>0. \qquad (*) \end{equation} $\phi$ is not the identity map by initial assumption, hence $\exists x \in\Bbb{R}$ such that $\phi(x)\neq x$. It is clear that $x \neq0$. WLOG, assume $|x|<\phi(|x|)$. Since $\Bbb{Q}$ is dense in $\Bbb{R}$, $\exists r \in \Bbb{Q}$ such that $|x|<r<\phi(|x|)$. The inequality $r > |x| \ge 0$ with Eq. $(*)$ imply $\phi(r)-\phi(|x|)>0$. By Eq. ($ \$ $), this means $r-\phi(|x|)>0$. However, this contradicts our initial choice of $r$ such that it satisfies $r<\phi(|x|)$. By assuming $|x|>\phi(|x|)$, we get a contradiction again by the same reasoning.

Therefore, the identity map is the only ring isomorphism from $\Bbb{R}$ to itself. $\square$

Answer 3

This becomes trivial if you have studied orders in a ring.

A ring is ordered if there exists a non-empty set $\mathcal{P}$ (for positive) which enjoys:

1. Closure: $\forall a,b\in\mathcal{P}, a+b,ab\in\mathcal{P}$
2. Trichotomy: $\forall a \in\mathcal{P}, a\in\mathcal{P}\vee a=0\vee-a\in\mathcal{P}$

and we say that $a<b$ iff $b-a\in\mathcal{P}$.

One can prove that the only possible ordering is the "natural" ordering. By the closure property, any square of a non-zero element in $\mathbb{R}$ is positive, so $\mathbb{R}^+\subseteq\mathcal{P}$. On the other hand, suppose $a\in\mathcal{P}-\mathbb{R}^+$, then $a\in\mathbb{R}^-\cap\mathcal{P}$, also $-a\in\mathbb{R}^+\subseteq\mathcal{P}$, contradicting the trichotomy. Hence $\mathcal{P}=\mathbb{R}^+$.

Now, any ring isomorphism $\phi:R\longrightarrow R'$ induces an order $<'$ defined by $\phi(a)<'\phi(b)$ iff $a<b$, with $P'=\phi\mathcal(P)$.

So any isomorphism will preserve the order, and hence the addition and multiplication structure, and the only such isomorphism is the identity.