Isomorphic

**Coda202** · October 13th 2009, 10:05 AM

Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)

**eeyore** · October 13th 2009, 10:47 AM

I'll give it a shot. Assume $\phi: \mathbb{R} \rightarrow \mathbb{Q}$ is such an isomorphism. Say $\phi(1)=k$ where $k \in \mathbb{Q}$ . $\phi$ must satisfy
$\phi(ab) = \phi(a)\phi(b)$
so
$kab = k^2 ab \Rightarrow k = 1$

Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist.

**Taluivren** · October 13th 2009, 10:51 AM

Hi,
here's my try:
sentence
$\forall x \exists y \; x=y\cdot y\cdot y$
holds in $\mathbb{R}$ but not in $\mathbb{Q}$ . So they are not elementarily equivalent, therefore they can't be isomorphic.

**clic-clac** · October 13th 2009, 12:08 PM

Another one: assume $f:\mathbb{R}\rightarrow\mathbb{Q}$ is a ring isomorphism, then $f(\sqrt{2})^2=f(\sqrt{2}^2)=f(2)=f(1+1)=f(1)+f(1)= 1+1=2$ .

But no rational is a squared root of 2, so $f$ cannot exist.

(assume $p,q$ belong to $\mathbb{N}^*, gcd(p,q)=1$ and $(\frac{p}{q})^2=2$ then $p^2=2q^2$ .
But this is impossible since $\nu_2(p^2)$ is even and $\nu_2(2q^2)$ is odd (where $\nu_2(n)$ is the greatest $k$ such that $2^k$ divides $n$ ))

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