Ring of Finitely Supported Integer Sequences

Detailed Solution

We are given the set R of all integer sequences having only finitely many nonzero terms. The operations are defined componentwise. Step 1: Closure under addition. Let a = (a1, a2, a3, ...) and b = (b1, b2, b3, ...) belong to R. Then a + b = (a1+b1, a2+b2, a3+b3, ...). Since both sequences have finitely many nonzero terms, their sum also has finitely many nonzero terms. Therefore, a+b belongs to R. Step 2: Closure under multiplication. Under pointwise multiplication, ab = (a1b1, a2b2, a3b3, ...). The product can have nonzero terms only at positions where both sequences have nonzero terms. Hence, ab has finitely many nonzero terms. Therefore, ab belongs to R. Step 3: Ring properties. The set R satisfies the following properties: 1. Addition is associative and commutative. 2. The zero sequence (0,0,0,...) is the additive identity. 3. Every element has an additive inverse. 4. Multiplication is associative. 5. Multiplication distributes over addition. Thus, R is a commutative ring. Step 4: Check the existence of multiplicative unity. Suppose that R has a multiplicative identity e = (e1, e2, e3, ...). For every a in R, we must have ea = a. For each positive integer n, consider the sequence un = (0, ..., 0, 1, 0, ...), where 1 occurs only at the nth position. Since e is an identity, we must have eun = un. Under componentwise multiplication, this implies en = 1. Since n is arbitrary, e = (1, 1, 1, ...). However, this sequence has infinitely many nonzero terms and therefore does not belong to R. This contradicts the assumption that e belongs to R. Consequently, R has no multiplicative identity. Therefore, R is a commutative ring without unity. Hence, option A is correct. Why the other options are incorrect: Option B: Incorrect. The sequence (1,1,1,...) has infinitely many nonzero terms and does not belong to R. Option C: Incorrect. The sum of two finitely supported integer sequences is again finitely supported. Option D: Incorrect. A field must possess a multiplicative identity, whereas R has no multiplicative identity. Final Answer: