Relation between a filter and its complement being an ideal.

Exercise. Show that a true filter $\mathcal{F}$ in $X$ is an ultrafilter iff the complement of $\mathcal{F}$ is an ideal, i.e., $Y,Z \notin \mathcal{F} \Leftrightarrow Y \cup Z \notin \mathcal{F}$, for $Y,Z \subset X$.

My resolution.

$(\Rightarrow)$ Suppose $\mathcal{F}$ is an ultrafilter in $X$. Then, $\mathcal{F}$ is also a prime filter on $X$, i.e., \begin{equation*} \forall \hspace{.15cm}Y,Z \subset X, \hspace{.25cm} Y \cup Z \in \mathcal{F} \Rightarrow Y \in \mathcal{F} \text{ or }Z \in \mathcal{F}. \end{equation*} Directly from the property above (i.e., the definition of a prime filter) we have that $Y,Z \notin \mathcal{F} \Rightarrow Y \cup Z \notin \mathcal{F}$. Thus all we want to prove now is that $Y \cup Z \notin \mathcal{F} \Rightarrow Y,Z \notin \mathcal{F}$. Let us assume now that $Y \cup Z \notin \mathcal{F}$ but $Y \in \mathcal{F}$ or $Z \in \mathcal{F}$. WLOG, assume that $Y \in \mathcal{F}$. So, $Y \subseteq Y \cup Z \Rightarrow Y \cup Z \in \mathcal{F}$, which is a contracdition, proving what we wanted.

$(\Leftarrow)$ Suppose now that the complement of $\mathcal{F}$ is an ideal, i.e., $Y,Z \notin \mathcal{F} \Leftrightarrow Y \cup Z \notin \mathcal{F}$, for $Y,Z \subset X$. So it's trivial that $\mathcal{F}$ is a prime filter just by applying $(p \Rightarrow q) \Leftrightarrow (\lnot q \Rightarrow \lnot p)$, and if $\mathcal{F}$ is a prime filter in $X$, then it is also an ultrafilter.

Well, I just wanted to check if what I made is correct. To me, its makes plenty of sense but at the same time I feel like I made it way too simple. Thanks for all the help in advance.

Yes, that argument works provided you know the equivalence

$$\mathcal{F} \text{ is an ultrafilter on } X \iff \forall F,G \subseteq X: (F \cup G \in \mathcal{F}) \to (F \in \mathcal{F} \lor G \in \mathcal{F})\tag{1}$$

(where ultrafilter is defined as inclusion-maximal filter on $X$).

If $(1)$ is an established fact (in both directions; you only claim the direction left to right to be known to you) then the right hand statement is indeed the same as

$$\forall F,G \subseteq X: (F \notin \mathcal{F} \land G \notin \mathcal{F}) \to (F \cup G \notin \mathcal{F})\tag{2}$$ by modus tollens and de Morgan. And logical equivalence is transitive..