The non-zero elements of the minimal prime ideals of a noetherian commutative ring are zero-divisors.
The proof of this fact I know of uses primary decomposition. Are there alternative (e.g. more direct) proofs ?
[Math] Minimal primes and zero-divisors
ac.commutative-algebra
Related Solutions
Can you show, from whatever you have available to you in the course at this point, that maximal ideals have inverses (as $A$-modules)? It would then follow that if a max. ideal $\mathfrak m$ contains an ideal $\mathfrak a$ then $\mathfrak m$ is a factor of $\mathfrak a$. Then maybe by a Noetherian inductive process show any primary ideal is a prime power by starting with a primary ideal, picking a maximal ideal containing it (we know secretly there's only 1 choice), write your primary ideal as a product of that maximal ideal and another ideal, show that other ideal is primary, and repeat. At the end you could read off that all the maximal ideal factors must be the same. I'm not saying I have worked out the details on that, but it's still not completely clear to me what is known and not known to the students, so this is just a suggestion.
I looked at my own lecture notes to see how I showed a maximal ideal in a Dedekind domain has an inverse, and I used a variant on the determinant trick together with the ring being integrally closed. I vote for using the determinant trick. Then you'd use it now once and you have already told us that you will use it again later. After you use it twice, it becomes a method and not a trick. :) Let's think about it this way: you are willing to use the raw definition of being integrally closed, and you certainly need to use integral closedness somewhere, but if you want to produce a monic polynomial at some point so you can use the fact that your ring $A$ is integrally closed, where in the world are you going to get such polynomials from? The determinant trick is one way. What other way is there in this proof?
I am not sure that any solution you eventually find will be satisfying to the students, to whom this is being seen for the first time. The next time you teach the course, consider covering the material in a different order so you don't get caught in the same way.
The answer is: always, and argument is pretty simple:
Let $R$ be a reduced commutative unital ring. If $a\in R$ is a zero divisor, then $ab=0,$ for some $b\neq 0.$ Hence $b\not \in 0 = \text{nil}(R)= \bigcap \text{Spec}(R) =\bigcap \text{Specmin}(R),$ where $\text{Specmin}(R)$ states for family of all minimal prime ideals of $R.$ So there exists a minimal prime ideal $\mathfrak{p}\triangleleft R$ s.t. $b\not \in \mathfrak{p}.$ But $\mathfrak{p}$ is prime, and $ab\in \mathfrak{p},$ thus $a\in \mathfrak{p}.$ Hence the set of zero divisors is contained in union of minimal prime ideals.
On the other hand it is well-known fact that minimal prime ideals in commutative unital rings consist of zero divisors, so the set of zero divisors in reduced commutative unital ring is exactly the union of minimal prime ideals.
Best Answer
Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$.
If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors).
If $X$ has more than one irreducible component, then do the following: Choose $f_1,\dots,f_m$ such that each $f_i$ is contained in exactly one minimal prime ideal and for each minimal prime there is (exactly) one $f_i$ contained in it. This choice ensures that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent. (The point is, that if their product vanished on an irreducible component, then one of them would have to, but then it would be the one that vanishes on that component). Now take an arbitrary element $g$ in any of the minimal primes and for simplicity assume it is from the one corresponding to $f_1$. (We allow $g$ to be contained in other primes as well, but that has no consequence). Then the following claim implies that $g$ is a zero-divisor.
Claim Let $g,f_2,\dots,f_t$ be such that $g\cdot f_2\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $g$ is a zero-divisor.
Proof Let $n$ be the smallest non-negative integer for which there exists a $t-1$-uple $(n_2,\dots,n_t)\in\mathbb N^{t-1}$ such that $g^n\cdot f_2^{n_2}\cdots f_t^{n_t}=0$. Observe that $n$ exists and $n\geq 1$ by the assumptions. Then $g\cdot (g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t})=0$, but $g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t}\neq 0$. $\square$
Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. Also, it clarifies the unclear step pointed out by Georges. In fact, it seems one needs to do a little yoga to get the result. In any case algebra=geometry. :)