# Do common elements of the rings R and S map to one another in a ring homomorphism?

I am taking a course on abstract algebra and I am having trouble understanding some of the concepts of ring theory, one of which is ring homomorphisms. I know the definition of a ring homomorphism (as defined in class) is \mathrm{Given}\ f:R\longrightarrow S\ \mathrm{a\ ring\ homomorphism}\\\begin{align}&1.\ f(1_R)=1_S\\&2.\ f(x+y)=f(x)+f(y)\\&3.\ f(xy)=f(x)f(y)\end{align}\\

However, I am having trouble actually using this to prove anything more than whether or not a given set is a homomorphism. I am wondering if the definition implies that $$\forall r\in R\ \cap\ S,\ f(r)=r \\$$

My reasoning is that given the second requirement, this must be true: \begin{align}&\mathrm{Let}\ r\in\ R\ \cap\ S\ \mathrm{then}\\ &f(r)=f(\sum_{i=1}^r 1_R)=\sum_{i=1}^rf(1_R)=r[f(1_R)]=r(1_S)=r\end{align}\\

Am I missing something here? or is this correct?

Also, I was doing an exercise the other day which asked me to prove that $\require{cancel}\mathrm{Q}[\sqrt2]\mathrm{\ is\ not\ isomorphic\ to\ Q}[\sqrt3].$ I was unable to do the problem and the internet said that $\mathrm{Q}[d] \cong\mathrm{\ Q}[d']\iff d=d'$. Assuming I am correct, is my idea the reason why?

Thanks!

• In your summation what is $r$? Why would $r$ be equal to $r$ sums of $1$? This is true for $\mathbb Z$, but if you consider, for example, ring of matrices, what is a sum from $1$ to some matrix? Commented Nov 17, 2015 at 22:44
• You can't sum from $i=1$ to $r$ because $r$ is not an integer. With minor tweaks you could use this to show that one prime subfield must be mapped to another.
– user208649
Commented Nov 17, 2015 at 22:44

Consider the ring homomorphism $f:\mathbb{C}\to \mathbb{C}$ given by complex conjugation. The element $i$ is in $R\cap S$, yet $f(i)=-i$.
For the other problem, $\mathbb{Q}[\sqrt{2}]$ contains an element $x$ such that $x\cdot x=1+1$. However $\mathbb{Q}[\sqrt{3}]$ contains no such element, so what would a ring homomorphism map $x$ to?