Question about the muliplicative identity, inverse elements and nonzero elements of a ring?

Question

Let $R$ be the set of continuous real-valued functions on the interval $[0, 1]$. Show that $R$ is a ring with respect to the operations $(f + g)(x) = f(x) + g(x)$ and $(fg)(x) = f(x)g(x)$.

So my homework asked me to show that R has a multiplicative identity, which I believe is the function with constant value $1$ (hopefully that's right?)

It also asked us to show what elements of $R$ will have multiplicative inverses. I understand that some functions (like $f(x)=x^2+1$) will have one while others (like $g(x)=x-x^2$) will not. But I can't recognize a consistent pattern as to determining which elements will and which won't.

Also would R be an integral domain? I can't think of any nonzero elements that can be manipulated to equal zero, but I just want to be sure on that front as well.

Thank you for any help! :)

Answer 1

As stated in the comments, you can't compute $\frac{1}{f(x)}$ for all $x$ if $f(x)=0$ for some $x$. For a hint on the zero divisors, I will tell you that they do exist. $f(x)$ and $g(x)$ satisfy $f(x)g(x)=0$ if whenever $f(x)\neq 0$ we have that $g(x)=0$. Try to construct a continuous function $f$ that is nonzero if $x<\frac{1}{2}$ and zero if $x\geq \frac{1}{2}$, and also construct a continuous function $g$ such that $g(x)=0$ if $x\leq \frac{1}{2}$ and $g(x)\neq 0$ if $x>\frac{1}{2}$.

Answer 2

It is correct to say that the constant function $1$ is the multiplicative identity of $R$.

Observe that, since $[0,1]$ is compact, a continuous function $f:[0, 1] \to \Bbb R$ which is nowhere zero on $[0, 1]$ is bounded below, in absolute value, by some real $\delta > 0$; that is

$\vert f(x) \vert > \delta, \; \; \forall x \in [0, 1]; \tag{1}$

(1) implies that

$\vert \dfrac{1}{f(x)} \vert = \dfrac{1}{\vert f(x) \vert} < \delta^{-1} \; \; \forall x \in [0, 1], \tag{2}$

so that $1 / f(x)$ is bounded on $[0, 1]$. It is thus a bounded continuous function on $[0, 1]$, hence an element of $R$; this shows that the units of $R$ are precisely those $f(x) \in R$ which never take the value $0$. I believe this is the "consistent pattern" our OP ConfusedSoul seeks.

Though as we have seen $R$ has no shortage of units, it also has plenty of zero divisors. For example, picking $a$ such that $0 < a < 1$, let $f_a(x)$ be the continuous function

$f_a(x) = 1 - \dfrac{x}{a}, \; \; 0 \le x \le a, \tag{3}$

$f_a(x) = 0, \; \; a \le x \le 1, \tag{4}$

and let $g_a(x)$ be defined by

$g_a(x) = 0, \; \; 0 \le x \le a, \tag{5}$

$g_a(x) = \dfrac{1}{1 - a} (x - a), \; \; a \le x \le 1. \tag{6}$

It is easy to see that both $f_a(x) \in R$ and $g_a(x) \in R$, neither are the constant function $0$, and

$(f_a(x))(g_a(x)) = 0, \; \; \forall x\in [0, 1]; \tag{7}$

thus $f_a(x)$ and $g_a(x)$ are both zero divisors in $R$, so $R$ is mos' def' not an integral domain. Many more zero divisors may be constructed by taking functions of the form $h(x) f_a(x)$, $k(x) (x)g_a(x)$ where $h(x), k(x) \in R$ and $h(y) \ne 0$ for some $y \in (0, a)$ whilst $k(y) \ne 0$ for some $y \in (a, 1)$. There are indeed others as well.

Hope this helps. Cheers,

and as ever,

Fiat Lux!!!