Answer to Question #180289 in Abstract Algebra for Hriday nath

Question

Let S= {(x,y)|x,y € R}. How do we show that S is a ring with identity with the operations defined by (x,y) +(u,v) = (x+u, y+v) and (x,y)(u,v) = (xu-yv,xv+yu)?

Answer 1

(0,0) $\in$ S

Since 0 $\isin$ R

Additive identity is (0,0)

Considering (x,y) $\isin$ S

(x,y)+(0,0)=(x+0,y+0)=(x,y)

$\therefore$ Additive identity exists

If e=(r,s) is the multiplicative identity then,

(x,y)(r,s)=(x,y)

=(xr-ys,xs+yr)

Solving this we get r=1 and s=0

$\therefore$ the multiplicative identity s (1,0)

hence the identity element exists