If a polynomial is reducible say $F=fg$ ($f \neq g$), does $V(F)$ still have codimension $1$?
This is a question to clear my conception, what I feel is no because we will have $V(F) \subsetneq V(f)$ so the codimension has to be greater than $1$, but in many books, we directly use that the dimension of a hypersurface is $n-1$.
When a variety is reducible, its spectrum is reducible as a topological space. This makes the codimension a touch more subtle.
The reason is because the codimension is defined to be the Krull dimension of the ring you get when you localize at a prime. In your case, since the equation is reducible, you cannot localize at its equation as a prime ideal because it isn't prime! In this case, the codimesion is defined to be the infimum of all localizations at primes containing this non-prime ideal. In particular, you are taking the infimum over the set containing at least the ideals of the two factors, which have codimension $1$. It is also easy to see that the codimension is at most $1$ because there is no such thing as a codimension $0$ closed subset in affine $n$-space (of course, you can always have things like quasi-projectives, but this is not the issue at hand).