# Isomorphic rings

1. Mar 15, 2010

### zcomputer5

Hello could someone help me out with this question

Show that R[x]/(x^2)+a isomorphic to the complex numbers if a>0 and RxR if a<0.

Here is an attempt! for mulitplication in R[x]/(x^2)+a

Elements of $\mathbb{R}[X]/(X^2+a)$ are cosets:

$f(X) + (X^2+a)$. By the polynomial division algorithm $(X^2+a)g(X) + r(X) = f(X)$ with r(X) degree 2, so we have elements:

$mX + n + (X^2+a)$ with addition ... and multiplication ...

$Multiplication:$

$(n+mx +(x^{2}+a)(a+bx +(x^{2}+a)$

$=na +nbx + n(x^{2}+a) + amx+ bmx^{2} + mx(x^{2}+a) + (x^{2}+a)^{2} + bx(x^{2}+a) +n(x^{2}+a$

$=(na+mbx^{2})+(bn+am)x$

$=(na-\alpha mb)+ (bn+am)x$

$As we have x^{2}+a =0 which is the ideal$

Am I proceeding along the correct lines if I show this is isomorphic to the complex numbers number multiplication????

Clearly under Addition it is a bijection. For the mapping $\mathbb{R}[X]/(X^2+a)$ to complex numbers

Last edited: Mar 15, 2010