homomorphic image of a sum of infinite elements

Question

I am trying to prove that all ring homomorphisms from $\mathbb{R} \to \mathbb{R}$ is trivial or identity. If I define irrational number as limiting sum of infinite sums of rationals, then can I take $f(q_1+q_2+q_3...)=f(q_1)+f(q_2)+f(q_3)+....$ after I have proved this realtion for rationals and then extend it to irrationals? the usual answers involve density of rationals, continuos map and such things. But if I have defined Irrationals the way I did, will infinite homomorphic image work as well?

Answer 1

It does not follow that a function being a ring homomorphism means infinite sums are preserved. There are multiple ways of thinking about this.

A common approach might be to say something like: "Just because $\sum_{i\in I}x_i$ converges doesn't mean $\sum_{i\in I}f(x_i)$ does." I think this doesn't quite get at the issue.

What makes something a homomorphism generally in algebra, not just for rings, is that it preserves the operations. In more detail, we have the formal theory of rings with constants like $0$, $1$, and operations like $+$, and $\cdot$ from which given a set of "variables" we can build terms like $x+(1+1)\cdot x$ where $x$ is an element of the set of "variables". Given terms, we can talk about axioms like $1\cdot x=x$ which are really just pairs of terms. The significance of axioms comes from interpretations which assigns value to each constant, a function to each operation, and are defined recursively over the terms. Interpretations are required by definition to send any pair of terms forming an axiom to the same element. Given a ring $R$, the interpretation of $x+(1+1)\cdot x$ into $R$ which I'll write $\newcommand{den}[1]{[\![#1]\!]}\den{x+(1+1)\cdot x}_R$ is $\den{+}_R(\den{x}_R,\den{\cdot}_R(\den{+}_R(\den{1}_R,\den{1}_R),\den{x}_R))$. $\den{+}_R$ and $\den{\cdot}_R$ are the functions assigned to $+$ and $\cdot$ respectively, and $\den{0}_R$ and $\den{1}_R$ are the elements assigned to the constant $0$ and $1$ respectively. The easiest way to handle $\den{x}_R$ is to take the set of "variables" as the set of elements of $R$, then state $\den{x}_R=x$. A homomorphism $f:R\to S$ is a function satisfying $f(\den{+}_R(r_1,r_2))=\den{+}_S(\den{f(r_1)}_S,\den{f(r_2)}_S)$ and $f(\den{0}_R)=\den{0}_S$ and similarly for $\cdot$ and $1$. A consequence of this is that for any term $t$, $f(\den{t}_R)$ is entirely determined by $f$'s actions on the "variables" of $t$.

The purpose of reviewing this is the following fact: there are no infinite terms. For the recursive definition of interpretation to be well-defined, i.e. a total function, there can't be any infinitely deep terms. Put a different way, the set of terms is inductively defined and the definition of interpretation on terms is justified by structural induction. Now only infinitely deep terms are problematic, adding infinite arity operations is not a big issue for the induction. We could imagine having an infinite arity operation $\sum$, but first operations are usually required to be total, and second there's no reason for a ring homomorphism to preserve this new operation. For intuition, I do recommend thinking of infinitary operations as like new operations with only limited connection to their finitary counterparts. I think taking this perspective avoids a large variety of common confusions around infinitary operations of various sorts. From this perspective it's very obvious that laws for finitary counterparts of infinitary operations need to be independently reestablished and may well fail.

So to your specific question, if you want to use the property $f(\sum \bar x) = \sum (f \circ \bar x)$, you will need to prove it. It is not an immediate consequence of the fact that $f$ is a ring homomorphism. Of course, a la giannispapav's proof, it does ultimately follow from that, because it turns out the only ring homomorphism we have is the identity and thus the above equation becomes trivial. You could perhaps find a different proof of the above equation, and then use it to show the result via the route you desire, but it will take some work.

Answer 2

So it is easy to see that $f(q)=q\forall q\in\mathbb{Q}$

We know that $kerf$ is either $\{0\}$ or $\mathbb{R}$.If $kerf=\mathbb{R}$ then it is a trivial homomorphism.

Now suppose that $kerf=\{0\}$ so $f$ is $1-1$.

If $r,s\in\mathbb{R}$ and $r>s$ then $f(r)>f(s)$ because:

$r-s=(\sqrt{r-s})^2\Rightarrow f(r-s)=f((\sqrt{r-s})^2)=f(\sqrt{r-s})^2\geq0$.Since $f$ is $1-1$ and $r\not=s$ we have $f(r-s)\not=f(0)=0\Rightarrow f(r-s)>0\Rightarrow f(r)>f(s)$

Let's suppose now that $\exists r\in\mathbb{R}:f(r)\not=r$

• If $r>f(r)$ then from the density of $\mathbb{Q}$ in $\mathbb{R}$,$\exists a\in\mathbb{Q}:r>a>f(r)$,so from above we have $f(r)>f(a)=a$ contradiction
• Similarly if $r<f(r)$ we have a contradiction(I leave it to you!)

So $f(r)=r\forall r\in\mathbb{R}$ which is what you wanted.