If $F$ is a ring of all real-valued functions defined on $\mathbb{R}$, is $S = \{f ∈ F | f(0) = 1\}$ an ideal?
What I'm thinking is $(f+g)(0) = f(0)+g(0) = 1+1 = 2$
and hence $f + g$ is in $S$? Is that a real-valued function? Also, the additive inverse of $f$ is $−f$ and we have
$(−f)(0) = −f(0) = −1$ and hence $−f$ is in $S$? Again not sure if it's a real-valued function. Finally, suppose that $h ∈ F$. Then we have
$(hf)(0) = h(0)·f(0) = h(0)·1 = 1$ and hence $hf$ is in the set $S$. I think $F$ is a commutative ring, so we have $fh = hf$ and hence $fh$
is in $S$? I'm not sure if any of these are actually in $S$ however, I feel like I need to establish a constant function or something in the beginning for this to make sense though
Also what if $\{f ∈ F | f$ is continuous$\}$
Best Answer
S is not an ideal. It isn't closed under addition.
To boot, take $h\in F$ such that $h(0)\neq1$. Then $(hf)(0)=h(0)\cdot f(0)=h(0)\cdot 1=h(0)\neq1$. Thus $hf\not\in S$, violating the condition to be an ideal.
$\{f\in F\mid f \text{ is continuous}\}$ is a subring, but not closed under multiplication on the left, hence is also not an ideal. That is, consider the constant function, $f(x)=1\,,\forall x$. And any discontinuous function $g$. Then $gf=g$ is discontinuous. Hence again, the condition for an ideal is not met.