= Solution
A <prime ideal> $\mathfrak p$ is minimal over $I$ if $I\subseteq\mathfrak p$ and no strictly smaller prime contains $I$. An <associated prime of a module> is an <annihilator> of an individual nonzero element which happens to be prime. For the quotient <module> this reads
$$
\boxed{\operatorname{Ass}_R(R/I)=\{\operatorname{Ann}_R(\bar a):0\neq\bar a\in R/I,\ \operatorname{Ann}_R(\bar a)\text{ prime}\},}
$$
where $\operatorname{Ann}_R(\bar a)=(I:a)=\{r\in R:ra\in I\}$. This is the <annihilator> of an individual element, rather than necessarily the <annihilator> of the whole quotient.
Minimal primes exist because $I$ is proper. First choose a <maximal ideal> containing $I$. Within the primes contained in it and containing $I$, an intersection of any decreasing chain is again prime. Indeed, if $ab$ is in the intersection and $a$ is absent from one member, then $b$ belongs to that member and to every smaller member; it also belongs to every larger member. The intersection is still proper and contains $I$. <Zorn's lemma>, applied with reverse inclusion, produces a minimal prime. This proves <existence of minimal primes over a proper ideal>, even without the <Noetherian> hypothesis.
Now fix a minimal prime $\mathfrak p$ and put $A=R/I$. The <localization at a prime ideal> $A_{\mathfrak p}$ is nonzero and is a <Noetherian local ring>. By <prime ideal correspondence for localization>, its only prime is $\mathfrak pR_{\mathfrak p}/IR_{\mathfrak p}$. Write this <maximal ideal> as $\mathfrak n$. It is the <nilradical>; since it is finitely generated and each generator is nilpotent, some power of $\mathfrak n$ is zero. Explicitly, if its generators have nilpotence exponents $e_1,\ldots,e_r$, every product of degree $1+\sum_i(e_i-1)$ vanishes.
Choose the smallest $s\geq1$ such that $\mathfrak n^s=0$, and choose a nonzero element in $\mathfrak n^{s-1}$; when $\mathfrak n=0$, choose $1$. Its <annihilator> over $R_{\mathfrak p}$ is exactly $\mathfrak pR_{\mathfrak p}$. Represent it as $a/u$ with $u\notin\mathfrak p$. Multiplication by the unit $u$ shows that $a/1$ has the same <annihilator> and remains nonzero.
Let $p_1,\ldots,p_r$ generate $\mathfrak p$ in $R$. For each $i$, the equality $p_i a/1=0$ supplies $u_i\notin\mathfrak p$ such that $u_i p_i\bar a=0$ in $A$. Put $u=\prod_i u_i$ and $\bar b=u\bar a$. Its localization is nonzero, so $\bar b\neq0$, and every element of $\mathfrak p$ kills it. Conversely, an element outside $\mathfrak p$ becomes a unit and cannot kill the nonzero element $\bar b/1$. Therefore
$$
\operatorname{Ann}_R(\bar b)=\mathfrak p,\qquad\boxed{\varnothing\neq\operatorname{Min}_R(I)\subseteq\operatorname{Ass}_R(R/I).}
$$
The key step in <minimal primes are associated primes> is clearing denominators for a finite generating set of $\mathfrak p$; that is where <Noetherianity> is used.
For an <embedded associated prime>, take $R=k[x,y]$ and $I=(x^2,xy)$. Its radical is $(x)$, so $(x)$ is its unique minimal prime. The nonzero class of $x$ satisfies
$$
\operatorname{Ann}_R(\bar x)=(I:x)=(x,y).
$$
Indeed, $rx\in x(x,y)$ is equivalent to $r\in(x,y)$ by cancellation in the polynomial domain. Thus \b[$(x,y)$ is associated but is not minimal], since $(x)\subsetneq(x,y)$. For comparison, $\operatorname{Ann}_R(\bar y)=(x)$, exhibiting the minimal associated prime as well.
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