Proving the existence of unity in $R$, where $R$ is the ring of polynomials over complex numbers with $f(0)=0$.

Question

My line of thought is this: we want to prove that there exists some $h(x)\in R$ such that $g(x)h(x)=g(x)$. Therefore $h(x)=1$ for all $x$. But if $h(x)$ is in $R$, then is it not equal to zero at zero. So my guess would be $h(0)=0$, and $h(x)=1$ for all other $x$. Is this line of thought correct?

Answer 1

The set of such polynomials is a subring, see here. But it has no unit. Take $g(x)=x$ in your equation $g(x)=g(x)h(x)$, then $x=xh(x)$. Comparison of coefficients implies $h(x)=1$, the constant $1$ polynomial. But this implies $1=h(0)=0$, a contradiction.

Edit: Your suggestion, to define $h$ by $h(0)=0$ and $h(x)=1$ for $x\neq 0$ is not a continous function, hence not a polynomial.