R X R is a field

Consider R X R with the usual coordinatewise addition and a new multiplication given by

(a, b)(c, d) = (ac - bd, ad + bc).

Show that with these operations, R X R is a field.

For clarity, X represents the Cartesian product, R is the set of real numbers, and coordinatewise addition is defined as such:

For a,b,c,d in R, (a, b) + (c, d) = (a + c, b + d)

In order to show that R X R is a field, we must first show that it is a commutative ring with identity.

Proof:

Let a,b,c,d,e,f belong to R.

1) Let (a, b), (c, d) belong to R X R. Then, (a, b) + (c, d) = (a + c, b + d), which also belongs to R X R, so R X R is closed under addition.

2) For (a, b), (c, d), (e, f) belonging to R X R, (a, b) + [ (c, d) + (e, f) ] = (a, b) + (c + e, d + f) = (a + c + e, b + d + f) = (a + c, b + d) + (e, f) = [ (a, b) + (c, d) ] + (e, f), so associative addition holds in R X R.

3) (a, b) + (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) + (a, b), so commutative addition holds in R X R.

4) Since (a, b) + (0, 0) = (a + 0, b + 0) = (0 + a, 0 + b) = (0, 0) + (a, b) = (a, b), (0, 0) is the zero element for R X R.

5) The equation (a, b) + x = (0, 0) has the solution x = (-a, -b), which since -a, -b belong to R, also belongs to R X R.

6) For (a, b), (c, d) belonging to R X R, (a, b)(c, d) = (ac - bd, ad + bc), which also belongs to R X R since ac - bd, ad + bc belong to R, so R X R is closed under multiplication.

7) For (a, b), (c, d), (e, f) belonging to R X R, (a, b)[ (c, d)(e, f) ] = (a, b)(ce - df, cf + de) = (a*(ce - df) - b*(cf + de), a*(cf + de) + b*(ce - df)) = (ace - adf - bcf - bde, acf + ade + bce - bdf) = ((ac - bd)*e - (ad + bc)*f, (ac - bd)*f + (ad + bc)*e) = (ac - bd, ad + bc)(e, f) = [ (a, b)(c, d) ](e, f), so associative multiplication holds in R X R.

8) (a, b)[ (c, d) + (e, f) ] = (a, b)(c + e, d + f) = (a*(c + e) - b*(d + f), a*(d + f) + b*(c + e)) = (ac + ae - bd - bf, ad + af + bc + be) = ((ac - bd) + (ae - bf), (ad + bc) + (af + be)) = (ac - bd, ad + bc) + (ae - bf, af + be) = (a, b)(c, d) + (a, b)(e, f). Also, [ (a, b) + (c, d) ](e, f) = (a + c, b + d)(e, f) = ((a + c)*e - (b + d)*f, (a + c)*f + (b + d)*e) = (ae + ce - bf - df, af + cf + be + de) = ((ae - bf) + (ce - df), (af + be) + (cf + de)) = (ae - bf, af + be) + (ce - df, cf + de) = (a, b)(e, f) + (c, d)(e, f). Thus, the distributive laws hold in R X R.

By these 8 axioms, R X R is a ring under normal coordinatewise addition and new multiplication.

Next, we show that R X R is a commutative ring with identity.

9) For (a, b), (c, d) belonging to R X R, (a, b)(c, d) = (ac - bd, ad + bc) = (ca - db, cb + da) = (c, d)(a, b), so commutative multiplication holds in R X R.

10) Since (a, b)(1, 0) = (a*1 - b*0, a*0 + b*1) = (a, b) = (1*a - 0*b, 1*b + 0*a) = (1, 0)(a, b), (1, 0) is the multiplicative identity for R X R.

By axioms 9 and 10, R X R is a commutative ring with identity.

All that remains is to show that R X R is a field.

11) Let (a, b) belong to R X R, (a, b) does not equal (0, 0). Then, the equation (a, b)x = (1, 0) has the solution x = (a/(a^2 + b^2), -b/(a^2 + b^2)), since (a, b)(a/(a^2 + b^2), -b/(a^2 + b^2)) = (a*[a/(a^2 + b^2)] - b*[-b/(a^2 + b^2)], a*[-b/(a^2 + b^2)] + b*[a/(a^2 + b^2)]) = (a^2/(a^2 + b^2) + b^2/(a^2 + b^2), -ab/(a^2 + b^2) + ab/(a^2 + b^2)) = ((a^2 + b^2)/(a^2 + b^2), (-ab + ab)/(a^2 + b^2)) = (1, 0). Also, (a/(a^2 + b^2), -b/(a^2 + b^2)) belongs to R X R since a/(a^2 + b^2), -b/(a^2 + b^2) belong to R.

By axiom 11, R X R is a field under normal coordinatewise addition and new multiplication, and the proof is complete. []