Question

Which one of the following statements is TRUE?

• A. The set of all \(2 \times 2\) real matrices under usual matrix addition and multiplication is a Unique Factorization Domain.
• B. The set of all real valued functions under usual function addition and multiplication is a Unique Factorization Domain.
• C. The set of all polynomials with coefficients in \(\mathbb{R}\) under usual polynomial addition and multiplication is a Unique Factorization Domain.
• D. The set of Riemann integrable functions in \([0,1]\) under usual function addition and multiplication is a Unique Factorization Domain.

Hint:

Check whether each ring is even an integral domain (no zero divisors) first. Matrices and function rings on more than one point always have zero divisors; only one candidate is a domain, and it is a well known type of UFD.

Answer 1

The Correct Option is C

Step 1: Recall what a Unique Factorization Domain needs.
A Unique Factorization Domain (UFD) is first of all an integral domain: a commutative ring with identity that has no zero divisors, meaning \(ab = 0\) forces \(a = 0\) or \(b = 0\). Only then do we ask whether every non zero, non unit element factors into irreducibles in a way that is unique up to order and units. So the first check on every option is whether the ring even has zero divisors.

Step 2: Rule out option (A), the \(2 \times 2\) real matrices.
Matrix multiplication is not commutative, and this ring also has plenty of zero divisors. Take \(A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\). Both are non zero, but \(AB\) is the zero matrix. A ring with zero divisors is not even an integral domain, so it can never be a UFD. Option (A) is false.

Step 3: Rule out options (B) and (D), the function rings.
Let \(f\) be the function that equals \(1\) on the first half of \([0,1]\) and \(0\) on the second half, and let \(g\) equal \(0\) where \(f\) is \(1\) and \(1\) where \(f\) is \(0\). Both \(f\) and \(g\) are Riemann integrable and non zero, yet \(fg\) is the zero function on all of \([0,1]\). The same split works for all real valued functions on any set with more than one point. So both the ring of all real valued functions and the ring of Riemann integrable functions on \([0,1]\) have zero divisors and are not integral domains. Options (B) and (D) are false.

Step 4: Check option (C), the polynomial ring \(\mathbb{R}[x]\).
\(\mathbb{R}\) is a field, and whenever \(F\) is a field, the ring \(F[x]\) is an integral domain, because the degree of a product of two non zero polynomials adds the degrees, so a product of non zero polynomials is never zero. \(F[x]\) is in fact a Euclidean domain under the division algorithm for polynomials, every Euclidean domain is a Principal Ideal Domain, and every Principal Ideal Domain is a Unique Factorization Domain. So \(\mathbb{R}[x]\) is a UFD, where every polynomial factors into irreducible polynomials uniquely up to order and unit (non zero constant) multiples. Option (C) is true.

Final Answer:
Only the polynomial ring \(\mathbb{R}[x]\) is an integral domain among the four choices, and being a Euclidean domain over a field pushes it all the way to being a UFD. \[ \boxed{\text{Option (C)}} \]