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

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

[itex]mX + n + (X^2+a)[/itex] with addition ... and multiplication ...

[itex]Multiplication:[/itex]

[itex](n+mx +(x^{2}+a)(a+bx +(x^{2}+a)[/itex]

[itex]=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[/itex]

[itex]=(na+mbx^{2})+(bn+am)x [/itex]

[itex]=(na-\alpha mb)+ (bn+am)x[/itex]

[itex]As we have x^{2}+a =0 which is the ideal[/itex]

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 [itex]\mathbb{R}[X]/(X^2+a)[/itex] to complex numbers

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# Homework Help: Isomorphic rings

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