Question

Let R →f S be a ring homomorphism and T≤S. Prove that K={r∈R:f(r)=0 S } and P= {r∈R:f(r)∈T} are both subrings of R. Explain why the result for K is a special case of the result for P

Answer 1

To prove that K={r∈R:f(r)=0S} and P={r∈R:f(r)∈T} are subrings of R, we need to show that they satisfy the subring criteria. K is a subring of R where T is the trivial subgroup {0S}, whereas P is a subring of R where T is a non-trivial subgroup of S.

Step-by-step explanation:

To prove that K={r∈R:f(r)=0S} and P={r∈R:f(r)∈T} are subrings of R, we need to show that they satisfy the subring criteria:

1. They are non-empty subsets of R.
2. They are closed by subtraction.
3. They are closed in multiplication.

For K, since f(0R) = 0S, we know that 0R is in K. And if r, s are in K, then f(r - s) = f(r) - f(s) = 0S - 0S = 0S, so r - s is in K. Similarly, if r, s are in K, then f(rs) = f(r)f(s) = 0S0S = 0S, so rs is in K. Therefore, K is a subring of R.

For P, since T is a subset of S, if f(0R) = 0S is in T, then 0R is in P. And if r, s are in P, then f(r - s) = f(r) - f(s) is in T since T is a subgroup under subtraction. Therefore, r - s is in P. Similarly, if r and s are in P, then f(rs) = f(r)f(s) is in T since T is closed under multiplication. Therefore, RS is in P. Thus, P is a subring of R.

Now, the result for K is a special case of the result for P because K is a subring of R where T is the trivial subgroup {0S}, whereas P is a subring of R where T is a non-trivial subgroup of S.