Subring of continuous function

Question

Is the set of all function $f(x)$ such that $ f(q) =0$ for all $q \in \mathbb{Q} \cap [0,1]$ a subring of $C[0,1]?$

My attempt: I think yes, because both $f- g \in C[0,1]$, and $ fg$ $\in C[0,1]$.

Answer 1

Yes, you are correct, such set is closed under multiplication and subtraction. However, in order to show that it is a unit subring you should also check that this set contains the multiplicative identity, i. e. the constant function $1$. Is that true?

Answer 2

It depends on the definition of ring in this context. If rings are required to have multiplicative identities, then the answer is no since the function $x \mapsto 1$ is not in your set. However, if rings are not required to have multiplicative identities, then your set indeed forms a subring.

For a more concrete answer, note that the set you describe only consists only of the zero function $x \mapsto 0$ because $\mathbb Q$ is dense in $[0,1]$. Thus, I guess the most standard answer is that your set is a ring (with unit), but not a subring (with unit) since the multiplicative unit $x \mapsto 0$ of the subset does not coincide with the multiplicative unit $x \mapsto 1$ of $C[0,1]$.