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 1


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.

  • $\begingroup$ Beat me to it by about one second. : D Was just about to hit “Post Your Answer”. $\endgroup$
    – k.stm
    Commented Jun 10, 2019 at 21:11
  • $\begingroup$ 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". $\endgroup$
    – johnnycrab
    Commented Jun 11, 2019 at 6:27
  • $\begingroup$ 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. $\endgroup$ Commented Jun 11, 2019 at 6:29
  • $\begingroup$ Sorry, of course that was what I meant. Thank you again. $\endgroup$
    – johnnycrab
    Commented Jun 11, 2019 at 6:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .