Let $ \mathfrak{p} $ be a prime ideal of $k[x_{1},...,x_{n}]. $ If $ \mathfrak{p} $ has height $1$, is this equivalent to saying that $ V(\mathfrak{p}) $ is of codimension $1$?
Assuming this is true, if $ \mathfrak{p} $ is principal, does it follow from this that the prime divisor $ D:=V(\mathfrak{p}) $ is principal since $ I(D) = (\mathfrak{p}) = (f) $ for some $ f \in k[x_{1},...,x_{n}]? $