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.
The set of all \(2 \times 2\) real matrices under usual matrix addition and multiplication is a Unique Factorization Domain.
The set of all real valued functions under usual function addition and multiplication is a Unique Factorization Domain.
The set of all polynomials with coefficients in \(\mathbb{R}\) under usual polynomial addition and multiplication is a Unique Factorization Domain.
The set of Riemann integrable functions in \([0,1]\) under usual function addition and multiplication is a Unique Factorization Domain.
Show Solution
The Correct Option isC
Solution and Explanation
This question is really about one property: a Unique Factorization Domain (UFD) must first be an integral domain, a commutative ring where no product of two non zero elements is zero. If we can find one pair of non zero elements whose product is zero, that ring is finished as a candidate, since it fails to even be a domain. Let's go through the four candidate rings one at a time.
2 by 2 real matrices: the matrices $E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are both non zero, but $E_{11}E_{22}$ is the zero matrix. This ring has zero divisors, and is not even commutative, so it cannot be a UFD.
Real valued functions: pick $f$ equal to $1$ on one part of the domain and $0$ elsewhere, and $g$ equal to $0$ where $f$ is $1$ and $1$ where $f$ is $0$. Then $f$ and $g$ are both non zero but $fg$ is the zero function everywhere. This ring has zero divisors.
Polynomials $\mathbb{R}[x]$: if $p(x)$ and $q(x)$ are non zero real polynomials of degrees $m$ and $n$, the leading coefficient of $p(x)q(x)$ is the product of the two leading, non zero, real coefficients, which is non zero since $\mathbb{R}$ has no zero divisors. So $p(x)q(x)$ is never zero when both factors are non zero. This ring is an integral domain, and since $\mathbb{R}$ is a field, $\mathbb{R}[x]$ is a Euclidean domain, hence a PID, hence a UFD.
Riemann integrable functions on $[0,1]$: the same trick as functions works, take $f$ as the indicator of the first half interval and $g$ as the indicator of the second half; both are Riemann integrable and non zero, but $fg = 0$ on all of $[0,1]$. Zero divisors again.
Only $\mathbb{R}[x]$ survives the very first test of being an integral domain, and being a Euclidean domain over a field pushes it all the way to being a UFD. So the correct statement is option (C).
Let's summarize:
A ring with zero divisors can never be a UFD, since UFD requires being an integral domain first.
Matrix rings, and rings of real (or Riemann integrable) functions on more than one point, always have zero divisors built from split indicator-like elements.
Polynomials over a field form a Euclidean domain, hence a PID, hence a UFD.
So the true statement is option (C): $\mathbb{R}[x]$ under usual addition and multiplication is a Unique Factorization Domain.
\[\boxed{\text{Option (C): } \mathbb{R}[x] \text{ is a UFD}}\]
