Let R a bSquareroot 5 where a b are rational numbers Let

Let R = {a + bSquareroot 5 where a, . b are rational numbers}. Let be the usual addition operation and be the usual multiplication operation on the- set of real number Prove that (R, ) is a ring. Verify that this ring has n unit. Verify that every non-zero number in R has a multiplicative inverse in R. Prove that if R\' = {a + b Squareroot 2 where a, b are integers} then (R\', ) is a ring hut not a field. Let (R, ) be a given ring and 0 be the identity element under the operation. Prove that 00r = r = r 0 = 0 for all r in R. List all the subrings of Z_25 List all the subrings of Z_18

Solution

1.a) To form a ring we need to satisfy the following things a)addition b)multiplication c)additive zero

    Addition : (a+b*sqrt(2))+(c+d*sqrt(2)) = (a+c)+((b+d)sqrt(2)) Now it becomes p+qsqrt(2) which is also in the set

     multiplication : (a+b*sqrt(2))*(c+d*sqrt(2)) = ac+bd*2+((bc+ad)(sqrt(2))) which is also in the set

   Zero : put a=0 and b=0 we get zero So, it forms a Ring.

1.b) for a ring to be unit it has to follow 1 . a+b*sqrt(2) = a+b*sqrt(2) . 1 put a=1 and b=2 we get that both are equal so the ring has unit

1.c) for multiplicative inverse p.p^-1 = 1 has to be satisfied where p belongs to R

Therefore lets take a=2 b=1 then then p=2+1sqrt(2) now p= ((2+1sqrt(2)) * (2-1sqrt(2))/(2-1sqrt(2)

p= 2/(2-1sqrt(2))     now as it is rational it belngs to R So,for every p there will be a p^-1

1.d) we have to show that a non-zero element inverse is not integer

     lets take 2(a+b*sqrt(2))=1 ==> 2bsqrt(2)=1-2a

if b=0,we get a = 0.5 not an integer

if b not 0,we get sqrt(2)=(1-2a)/2b the rhs belong to Q, a contradiction,since sqrt(r) belongs to R

Proved