Prove R[x]/(x^2)+a Isomorphic to Complex Numbers If a>0, RxR If a<0

[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

 

[lippyka]

, we define the homomorphism f : \mathbb{R}[X]/(X^2+a) -> C by: f(mX + n + (X^2+a)) = mx + n.For a>0, multiplication in \mathbb{R}[X]/(X^2+a) is isomorphic to complex numbers multiplication. For a<0, multiplication in \mathbb{R}[X]/(X^2+a) is isomorphic to multiplication in \mathbb{R}^{2}. To prove this, one can show that the mapping f is an isomorphism by showing that it is bijective and preserves multiplication. To show bijectivity, we note that the kernel of f consists of elements of the form k(X^2 + a) with k \in \mathbb{R}, which implies that every element in \mathbb{R}[X]/(X^2+a) has a unique representation as a coset of (X^2+a). To show that f preserves multiplication, we note that for two elements mX + n + (X^2+a) and aX + b + (X^2+a) in \mathbb{R}[X]/(X^2+a), their product is given by (mn - ab) + (bn + am)x + (X^2+a). Since f maps both elements to (mx + n) and (ax + b) respectively, the product of these two elements under f is given by (mn - ab) + (bn + am)x, which is the same as the product in \mathbb{R}[X]/(X^2+a). Hence, f is an isomorphism, and thus \mathbb{R}[X]/(X^2+a) is isomorphic to the complex numbers if a>0 and \mathbb{R}^2 if a<0.