**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.6**

Decide which of the following are subrings of the ring of all functions from the closed interval $[0,1]$ to $\mathbb{R}$.

(1) The set of all functions $f(x)$ such that $f(q) = 0$ for all $q \in \mathbb{Q} \cap [0,1]$.

(2) The set of all polynomial functions.

(3) The set of all functions which have only a finite number of zeros, together with the zero function.

(4) The set of all functions which have an infinite number of zeros.

(5) The set of all functions such that $\lim_{x \rightarrow 1^-} f(x) = 0$.

(6) The set of all rational linear combinations of the functions $\sin mx$ and $\cos nx$, where $m,n \in \mathbb{N}$.

Solution:

(1) Let $A \leq [0,1]$, and let $S$ denote the set of all functions $f : [0,1] \rightarrow \mathbb{R}$ such that $f[A] = 0$. Clearly the zero function is in $S$, and if $f,g \in S$, then $(f-g)(x) = 0$ for all $x \in A$, so that $S$ is a subgroup by the subgroup criterion. Now if $f,g \in S$, then $(fg)(x) = f(x)g(x) = 0$ if $x \in A$. So $S$ is closed under multiplication, and is thus a subring.

(2) Since the sum, difference, and product of two polynomials is again a polynomial, this is a subring.

(3) Define $f$ and $g$ as follows: $f(x) = 1$ if $0 \leq x < 1/2$, 0 if $x = 1/2$, and 1 if $1/2 < x \leq 1$, while $g(x) = -1$ if $0 \leq x < 1/2$, 0 if $x = 1/2$, -1 if $1/2 < x < 1$, and 0 if $x = 1$. Both $f$ and $g$ have finitely many zeroes, but $(f+g)(x) = 0$ if $0 \leq x < 1$ and 1 if $x = 1$; so $f+g$ has infinitely many zeros and is not itself the zero function. Since this set is not closed under addition, it is not a subring.

(4) Consider the functions $f(x)$ and $g(x)$ defined as follows: $f(x) = 0$ if $0 \leq x \leq 1/2$ and 1 otherwise, while $g(x) = 1$ if $1/2 \leq x \leq 1$ and 0 otherwise. Both $f$ and $g$ have infinitely many zeroes. However, $(f+g)(x)$ is 2 at 1/2 and 1 otherwise, and thus has no zeroes. Hence this set is not closed under addition, and cannot be a subring.

(5) First we show that this subset is a subgroup. If $f$ and $g$ approach 0 as x approaches 1, then $$\lim_{x \rightarrow 1} (f-g)(x) = \lim_{x \rightarrow 1}f(x) - \lim_{x \rightarrow 1} g(x) = 0.$$ Since the zero function satisfies this property trivially, by the subgroup criterion this set is a subgroup. Similarly, we know from calculus that if $\lim_{x \rightarrow 1} f(x) = 0$ and $\lim_{x \rightarrow 1} g(x) = 0$, then $\lim_{x \rightarrow 1} (fg)(x) = 0$. Thus this set is closed under multiplication, and is a subring.

(6) Recall the product-to-sum formulas for sine and cosine: $$\cos x \cos y = (\cos(x-y) + \cos(x+y))/2,$$ $$\cos x \sin y = (\sin(x+y) - \sin(x-y))/2,$$ $$ \sin x \cos y = (\sin(x+y) + \sin(x-y))/2,$$ and $$\sin x \sin y = (\cos(x-y) - \cos(x+y))/2.$$ Due to these identities, the sum, difference, and product of a rational linear combination of sines and cosines of integer multiples of $x$ are again of this type. Thus this set is a subring.