Hi I am having some trouble with this question, and I have been looking, at the proof of the first isomorphism theorem several times, but it doesn't seem to click.
The following problem is let R={a+ib$|$a,b $\in \mathbb{Q}$} be a subring of $\mathbb{C}$ Show that ring R is isomorphic to the quotient ring $\mathbb{Q}[x]/(x^2+1)$
I have been trying to construct what I have mentioned, but I really don't know how to do it right. Then I have also considered to show that it is a homomorphism with congruence arithmetic and then finding the inverse. the thing is I have the tools, but I don't know how to use them.. Any help would be appreciated
The First Isomorphism Theorem of Rings is what you want to use in order to prove this. Recall it states the following: If $\phi: R \rightarrow R'$ is a homomorphism of rings, then $R/ker(\phi) \cong image(\phi)$.
In particular the isomorphism from $R/ker(\phi)$ onto its image is induced in a natural way by $\phi$: if $[a]$ is a coset in R/I, then it gets mapped to $\phi(a)$.
If $\phi$ is surjective, then image($\phi)= R'$. In which case $R/ker(\phi) \cong R'$.
In your case you're looking to show $\mathbb{Q}[x]/\langle x^{2} + 1 \rangle$ is isomorphic to $\mathbb{Q}(i)$. Following the first isomorphism theorem it should be enough to find a surjective homomorphism from $\mathbb{Q}[x]$ with the correct kernel i.e. ker$(\phi) = \langle x^{2} + 1 \rangle$. What might this be?
When defining a homomorphism out of $\mathbb{Q}[x]$ (or, any polynomial ring) one can sometimes choose to keep constants fixed by the homomorphism. So, in your case, you can can define a homomorphism of the required form by simplying saying: (i) constants are fixed, and (ii) where $x$ is mapped.
Finally, you must choose where $x$ is sent by this homomorphism; bearing in mind you want a particular kernel.