Is that Ring a field? Given a commutative Ring $R$ of ordered pairs $(x,y)$ of reals $x,y$ with addition and multiplication defined in the following way.
$$(x,y) + (u,v) = (x+u,y+v)$$ 
$$(x,y).(u,v) = (xu-yv,xv + yu)$$
I already showed that $R$ is an integral domain , now i need to show to prove $R$ is a field or not .
If $R$ is a field then every nonzero element $(x,y) \in R$  have a multiplicative inverse , which is $(x,y)(m,n) = (xm-yn,xn + ym) =  (1,0)$   
$(1,0)$ is the ring unity and $(m,n)$ is the multiplicative inverse and $xm - yn =1$, $xn + ym = 0$ how do we show that such $(m,n)$ exists or not ?
 A: Solve for $m$ and $n$ to get $n=\frac {-y}{x^2+y^2}$ and similarly for $m$ .Since $x,y$ are not both zero inverse exists
A: Alternatively, you can recognize that $(x,y)$ is kind of like $x + yi \in \mathbb{C}$ with the given operations.
From your definition, $$(u, v) + (x,y) = (u + x, v + y)$$ and $$(u, v)\cdot (x, y) = (ux - vy, uy + vx).$$
But this is the same behavior as $$(u + vi) + (x + yi) = (u + x) + (v + y)i$$ and $$(u + vi)(x + yi) = ux + uyi + vxi + vyi^{2} = ux + uyi + vxi - vy = (ux - vy) + (uy + vx)i$$
in the complex numbers.
But if we have a non-zero complex number $x + yi$, what is its multiplicative inverse?  Well, the multiplicative inverse should be $\dfrac{1}{x + yi}$.  But how do we express this as $u + vi$?  Well, $\dfrac{1}{x + yi} = \dfrac{1}{x + yi} \cdot \dfrac{x - yi}{x - yi} = \dfrac{x - yi}{x^{2} + y^{2}} = \dfrac{x}{x^{2} + y^{2}} + \dfrac{-y}{x^{2} + y^{2}}i$.
Since the space you are dealing with "acts" like the complex numbers, that means the multiplicative inverse of $(x,y)$ should be $(\dfrac{x}{x^{2} + y^{2}}, \dfrac{-y}{x^{2} + y^{2}})$.  You should check that this multiplied by $(x,y)$ gives you $1$.
