We de?ne a subring of a ring in the same way we de?ned a subgroup of a group: (S

Question

We de?ne a subring of a ring in the same way we de?ned a subgroup of a group: (S, +, .) is a subring of (R, +, .) if and only if (R, +, .) is a ring, S is a subset of R, and (S, +, .) is a ring with the same operations. For example, the ring of even integers is a subring of the ring of integers, and both are subrings of the ring of rational numbers. (a) Prove that the ring ({0}, +, .) is a subring of any ring (R, +, .) (called the trivial subring). (b) (Subring Test) Prove that if (R, +, .) is a ring. T is a nonempty subset of R, and T is closed under subtraction and multiplication, then (T, +, .) is a subring.

Explanation / Answer

When you start to work on this problem, don't worry about what K is. Just ask yourself, "what makes a subset of R into a subring?" The answer to this question for a nonempty subset K (verify to yourself that K is nonempty) is that K must be (1) closed under subtraction, and (2) closed under multiplication. So now, take two arbitrary elements a and b in K, and prove (1) and (2) in turn. For (1), observe that f(a-b) = f(a) - f(b) because f is a homomorphism. Then f(a-b) = 0 - 0 = 0 in S by the definition of K. Therefore a-b must be an element of K because K contains all elements in R that are mapped to 0. So K is closed under subtraction. For (2), observe similarly that f(ab) = f(a)f(b) = 0 * 0 = 0 Therefore ab is in K as well, and K is closed under multiplication. So K is a subring of R and you're done! Source(s): Hungerford, _Algebra_ Springer GTM 73

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