# Thread: Ring Isomorphism Question

1. ## Ring Isomorphism Question

Hello, I am stuck with the following question:

Show that $\displaystyle \mathbb{R}$ is not ring-isomorphic to $\displaystyle \mathbb{Q}$.
Do NOT use a cardinality argument
.
Hint: suppose you had a ring isomorphism $\displaystyle \phi: \mathbb{R} \rightarrow \mathbb{Q}$, look at $\displaystyle \phi(\sqrt{2})$

Thanks in advance

2. any ring homomorphism has to send 1 to one. Because it is a ring Hom, it has to preserve addition, so it must send the integers to the integers. We also have for the rational numbers in the reals, it must send them to the same thing in the rationals because of this.
$\displaystyle \phi(r/s)=\phi(rs^{-1})=\phi(r)\phi(s)^{-1}=\frac{\phi(r)}{\phi(s)}$

But this show we already have the rational numbers inside of the reals mapping onto the rationals, so this $\displaystyle \sqrt{2}$ must go into the rationals somewhere and this will contradict the injectivity of any possible isomorphism, since there are two things going to wherever phi sends $\displaystyle \sqrt{2}$