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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 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. :)

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. :)

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 $f1,\dots,f_m$ 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 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. :)

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 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. :)

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 taking one element in each minimal prime, their product will produce a function that is vanishing everywhere and hence nilpotent. It is easy to see 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, so the following claim implies that all of these are zero-divisors.

Claim Let $f_1,\dots,f_t$ be such that $f_1\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $f_1$ is a zero-divisor.

Proof Let $(n_1,\dots,n_t)\in\mathbb N^t$ be a smallest $t$-uple (in the lexicographical order) such that $f_1^{n_1}\cdots f_t^{n_t}=0$. Then $f_1\cdot (f_1^{n_1-1}\cdots f_t^{n_t})=0$, but $f_1^{n_1-1}\cdots f_t^{n_t}\neq 0$. $\square$

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. It appears longer, but only because i) setup ii) more details. Also, let me add that the length is the main reason I did not include it as a comment. In any case algebra=geometry. :)

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 $f1,\dots,f_m$ 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 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. :)