Let $\alpha \in K$ for $F$ is a field and $K$ is an extension of $F$. Then $F(\alpha)$ is a simple extension of $F$ i.e. $F(\alpha)$ is the smallest subfield containing $F$ and $\alpha$. Is it possible to write $F(\alpha) = F + (\alpha)$ where $(\alpha)$ is the ideal generated by $\alpha$? ($``+"$, meaning the usual addition of ideals in rings) Well,
$$(\alpha)=\{r \alpha: r \in F\}$$
And so altogether
$$F+(\alpha) = \{x+ r \alpha: x,r \in F\}$$
But this is the same as $F(\alpha)$. $\big($or rather, the above is how $F[\alpha]$ is defined, but $F[\alpha] = F(\alpha) \big)$
Have I made any grave error here? This seems intuitively true; I can't find a counterexample. If this reasoning is flawed, please let me know!
What you call "ideal generated by $\alpha$" is actually the $F$-vector space generated by $\alpha$ inside $F(\alpha)$.
Since $\alpha\notin F$ we must have (using your notation) $$ \dim_F(F+(\alpha))=2. $$ Thus the equality $F(\alpha)=F+(\alpha)$ holds for quadratic extensions, i.e. when $\alpha$ is a root of an irreducible polynomial of degree $2$ with coefficients in $F$ but fails for extensions of higher degree.
As an example $\mathbb{C}=\mathbb{R}+\mathbb{R}i$ but the set of numbers $$ L=\{a+b\alpha\,|\,\alpha^3=2, a,b\in\mathbb{Q}\} $$ is not a field since $\alpha^2\notin L$.