Prove that Z [i] is a commutative ring with unity.

Answer 1

x = a+bi y = c + di where a,b,c,d are integers

x+y = a+bi + c+di = a+c+bi+di <--- commutative of addition

= (a+c)+ (bi+di) <--- associative of addition

= (a+c) + i(b+d) <--- distributive over addition

= (c+a) + i(d+b) <--- commutative

= (c+a) + id + ib <--- distributive

= c + a + id + ib <--- associative

= c + id + a + ib <--- commutative

= (c+id) + (a+ ib) <--- associative

= y + x

Addition operation is closed and commutative

x * y = (a+bi)(c + di) = ac + adi + bci + bd (-1) <--- FOIL

= ac + adi + bci + - bd <-- property of -1

= ac + -bd + adi + bci <--- commutative property

= (ac - bd) + (adi + bci) <--- associative; defintion of subtraction

= (ac - bd) + i(ad + bc) <--- distributive

Note that ac-bd and ad+bc are also integers by closure property of addition and multiplication

over the integers. So the multiplication operator is closed.

y * x = (c + di)(a + bi) = ca + cbi + dai + db(-1) <--- FOIL

= ac + bci + adi + -bd <--- commutative of integer multiplication

= ac + -bd + bci + ad i <-- commutative of addition

= (ac-bd) + (bci + adi) <--- associative

= (ac - bd) + i(bc + ad) <-- distributive

= (ac - bd) + i(ad + bc) <--- commutative

= x * y <--- substitutes same expression above

Multiplication is closed and commutative

Let t = v + iw and constant K , v and w are integers

Kt(x+y) = K(v + iw) [( c + id) + (a + ib)] <--- substitution from expressions above

= K(v + iw)(c + id) + K(v + iw)(a + ib) <--- distributive

= K( vc + vdi + cwi - wd) + K( av + vbi + awi - wb) <---- distributive; steps left for you

= K ( vc - wd + av - wb + vdi + cwi + vbi + awi )

= K ( vc - wd + av - wb) + K(vd+cw+vb+aw)i

closure property of multiplication and addition/subtraction over the integers guarantees

the real and imaginary parts are integers.

Therefore, scalar and group distributive property holds, which satisfies group/ring properties.

The identity is 1 = 1 + 0i

(1+0i)( a + ib) = a + ib + 0i + 0*-1*b

= a + ib