# Non-zero-divisor in quotient ring: Not a unit over minimal prime?

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.)

## 1 Answer

This is not true. For instance, let $$k$$ be a field, let $$A=k[x,y]$$, and let $$f=xy$$, $$g=x+1$$. Then $$\mathfrak{p}=(x)$$ is a minimal prime containing $$f$$ but $$g$$ is a unit mod $$\mathfrak{p}$$.

It is true that there is always some minimal prime mod which $$g$$ is not a unit, for the reason which you say: every non-unit is in some maximal ideal and is thus not a unit mod any minimal prime contained in that maximal ideal.

A bit more broadly, letting $$X=\operatorname{Spec} B$$, to say that $$g$$ is not a zero divisor in $$B$$ means that it does not vanish identically on any irreducible component of $$X$$, to say $$B/(g)$$ is nonzero means that $$g$$ vanishes at some point of $$X$$, and to say that $$g$$ is a unit mod some minimal prime $$\mathfrak{p}$$ means that $$g$$ does not vanish at any point of the irreducible component corresponding to $$\mathfrak{p}$$. So, you could take a function on $$X$$ which does not vanish at all on one of the irreducible components but does vanish somewhere (but not identically) on another irreducible component, and that will give a counterexample to your question.

• Beat me to it by about one second. : D Was just about to hit “Post Your Answer”. – k.stm Jun 10 at 21:11
• Dear Eric, thank you very much for the counter-example and putting it into geometric perspective. I presume the fact that there is always one irreducible component (where $g$ vanishes at some point) is the intuition behind the statement that "intuitively a regular sequence successively cuts down the ring as much as possible". – johnnycrab Jun 11 at 6:27
• No, that condition says that $g$ doesn't cut down the ring too much (by making it become $0$). Cutting down the ring as much as possible instead corresponds to the condition that $g$ does not vanish identically on any irreducible component. – Eric Wofsey Jun 11 at 6:29
• Sorry, of course that was what I meant. Thank you again. – johnnycrab Jun 11 at 6:31