Why is the set $A=\{f\in F:f(0)=0\}$ an ideal where $F=\{f|f:[-1,1]\to \mathbb{R}\}$?

Question

Why is the set $A=\{f\in F:f(0)=0\}$ an ideal where $F=\{f\mid f:[-1,1]\to \mathbb{R}\}$ and $(F,+,\times)$ is the usual commutative ring?

I first checked whether $A$ is a subring. For this observe that $A$ is non-empty and for any $f,g\in A$ $$f-g\in A \text{ and }f\cdot g\in A.$$ So finally we want to check whether the multiplicative identity $f(x)=1$ is in $A.$ Clearly not as $f(0)\not = 0.$ So $A$ cannot be a subring. Hence $A$ cannot be an ideal also.

But the solution says that $A$ is an ideal. Why?

For reference here is the problem $2.1$ part $(4)$

Answer 1

Hint:

An ideal $I$ is a subset of a ring $R$, for which the elements are closed under addition and $ra\in I$ for all $r\in R$ and $a\in I$.

Answer 2

Yes it is an Ideal : $A$ is a subgroup of $(F,+)$ and if $f\in F$, then clearly $f.g\in A$ for all $g\in A$. ($A$ does not have to be a subring of $F$)

Answer 3

Generally for a subring $S$ or a ring $R$ it is not true that $1_S=1_R$.

For example $R=\mathbb{Z}\times \mathbb{Z}$,$S=\{(a,0)\in \mathbb{Z}\times \mathbb{Z} \}$,$1_R=(1,1),1_S=(1,0)$