Final answer:
Z[√5] is a subring of R as it is closed under addition, multiplication, and contains the multiplicative identity. It is also an integral domain because it has no zero divisors and is commutative under multiplication. The characteristic of Z[√5], being a subring of R, is 0.
Step-by-step explanation:
Proving Z[√5] is a Subring and an Integral Domain
To prove that Z[√5] = a + b√5 is a subring of the real numbers R, we have to show that it is closed under addition, multiplication, and contains a multiplicative identity. For any two elements u = a + b√5 and v = c + d√5 (where a, b, c, d ∈ Z), their sum is (a + c) + (b + d)√5 which is also in Z[√5], and their product is (ac + 5bd) + (ad + bc)√5 which is likewise in Z[√5]. Moreover, the multiplicative identity 1 (= 1 + 0√5) is clearly in Z[√5].
To show that Z[√5] is an integral domain, we need to prove that it is commutative under multiplication, it has an identity element (which we've shown above), and it has no zero divisors. Assume u*v = 0 for some u, v in Z[√5]. Then (a + b√5)(c + d√5) = 0. Expanding this we get ac + 5bd + (ad + bc)√5 = 0. For this equation to hold, both ac + 5bd and ad + bc must be zero, which is only possible if a = b = c = d = 0, implying u and v are zero, or both are non-zero which implies Z[√5] has no zero divisors, hence it's an integral domain.
The characteristic of an integral domain is the smallest positive integer n such that n⋅1 = 0, where 1 is the multiplicative identity. In any subring of the real numbers, the characteristic is 0 because the only integer multiple of 1 that is equal to 0 is 0 itself (due to the infinite and non-cyclic nature of the real numbers).