I think I do need some help with a basic commutative algebra question: Suppose we have a Noetherian integral domain $A$ (we can also assume that $A$ is a $K$-algebra of finite type, though I do not think this is necessary here).
Now let $f,g$ be two nonzero elements in $A$, such that the quotient $A/(f,g) \neq 0$ and such that $g$ is a non-zero-divisor on the quotient $A/(f)$ (i.e. the pair $f,g$ forms a regular sequence).
We now take any minimal prime ideal $\mathfrak p \subseteq B := A/(f)$ and consider the domain $B/\mathfrak p$. I am wondering if we can show that
1.) The element $g$ in $B / \mathfrak p$ is not zero, and
2.) The element $g$ in $B/\mathfrak p$ is not a unit.
I can confirm the first statement, because if $g \in \mathfrak p$, then $g$ is a zero-divisor on $A/(f)$ (as in Noetherian rings, minimal prime ideals consist of zero-divisors), contradicting the assumption.
I am not sure however if the second statement has to be true. It would be very nice if someone could help out! Thank you very much in advance!
EDIT: A quick addition (I have yet to think about). If the second statement is not true in general, is there some minimal prime ideal of $B$ such that $g$ is not a unit? (This should be true, because otherwise $g$ lies in no prime ideal of $A/(f)$, hence $A/(f,g) = 0$, in contradiction to the first assumption.)