交换环的所有零因子和 0 组成的集合是一个理想吗？

是的，交换环的所有零因子和 0 组成的集合，记为 $Z(R) cup {0}$ (更准确地说，是所有零因子与加法单位元 0 组成的集合，我们通常直接称零因子集合包含 0)，是该交换环的一个理想。

要证明这一点，我们需要验证一个子集是否为交换环的理想，需要满足以下三个条件：

1. 非空性： 该集合必须至少包含加法单位元 0。
2. 加法封闭性： 对于集合中的任意两个元素，它们的和也必须在该集合中。
3. 数乘（环乘法）封闭性： 对于集合中的任意一个元素和环中的任意一个元素，它们的乘积也必须在该集合中。

让我们逐一验证这些条件。首先，我们明确一下我们所说的“零因子”的定义。

零因子定义： 在一个环 $R$ 中，一个非零元素 $a in R$ 被称为左零因子，如果存在一个非零元素 $b in R$ 使得 $ab = 0$。类似地，$a$ 被称为右零因子，如果存在一个非零元素 $c in R$ 使得 $ca = 0$。如果 $ab = 0$ 对所有 $b in R$ 都成立，则 $a$ 是零元（即 0）。

在一个交换环 $R$ 中，左零因子和右零因子是同一个概念。所以，我们说一个非零元素 $a in R$ 是零因子，如果存在一个非零元素 $b in R$ 使得 $ab = 0$。

我们考虑集合 $Z(R) = {r in R mid exists s in R, s eq 0, rs = 0}$。这个集合包含了所有非零零因子。我们想要证明的集合是 $Z(R) cup {0}$。通常情况下，我们将零因子集合定义为包含 0 的，即零因子集合本身就包含了 0，所以我们讨论的集合就是零因子集合。为了清晰起见，我们直接称 $N = {r in R mid exists s in R, s eq 0, rs = 0 ext{ or } r=0}$。也就是说， $N$ 是所有零因子（包括 0）的集合。

下面我们来验证 $N$ 是交换环 $R$ 的一个理想：

1. 非空性：
我们知道环的加法单位元 $0$ 总是存在的。根据定义，如果存在一个非零元素 $s$ 使得 $0 cdot s = 0$，那么 $0$ 就是一个零因子。确实，对于环中任何非零元素 $s$，都有 $0 cdot s = 0$。所以，$0$ 属于零因子集合 $N$。因此，$N$ 非空。

2. 加法封闭性：
我们需要证明，如果 $a in N$ 且 $b in N$，那么 $a+b in N$。

情况 1： $a=0$ 或 $b=0$。
如果 $a=0$，那么 $a+b = 0+b = b$。因为 $b in N$，所以 $a+b in N$。同理，如果 $b=0$，则 $a+b = a in N$。

情况 2： $a eq 0$ 且 $b eq 0$。
由于 $a in N$ 且 $a eq 0$，根据零因子的定义，存在一个非零元素 $s_1 in R$ 使得 $as_1 = 0$。
由于 $b in N$ 且 $b eq 0$，根据零因子的定义，存在一个非零元素 $s_2 in R$ 使得 $bs_2 = 0$。

现在我们考虑 $a+b$。我们需要找到一个非零元素 $s$ 使得 $(a+b)s = 0$。

这里我们可能会遇到一个小麻烦：即使我们找到了 $s_1$ 和 $s_2$，我们无法直接将它们作用于 $a+b$ 来证明 $(a+b)s_1=0$ 或 $(a+b)s_2=0$。

让我们重新审视零因子的定义和理想的性质。理想的定义要求的是“任意”元素满足性质，而不是“存在”某个特定的元素。

一个更直接的思路是：如果 $a$ 是零因子，存在非零 $s_1$ 使得 $as_1=0$。如果 $b$ 是零因子，存在非零 $s_2$ 使得 $bs_2=0$。

这里我们需要一个关键的性质：在交换环中，如果 $a$ 是零因子，那么 $a$ 的所有“零化子”（annihilator）的集合构成一个理想。 零化子是指所有与 $a$ 相乘等于 0 的元素的集合。

换个思路，我们知道零因子是那些“可以被某个非零元素‘杀死’”（即乘积为 0）的元素。

让我们考虑集合 $N = {r in R mid exists s in R, s eq 0, rs = 0 ext{ or } r=0}$。

假设 $a, b in N$。
如果 $a=0$ 或 $b=0$，则 $a+b in N$ 是显然的。
如果 $a eq 0$ 且 $b eq 0$。则存在非零 $s_1$ 使得 $as_1=0$，存在非零 $s_2$ 使得 $bs_2=0$。

我们想要证明 $a+b in N$。也就是说，我们需要证明存在一个非零元素 $s$ 使得 $(a+b)s = 0$。
这里我们可能需要利用环的交换性。

一个更强大的论证是利用零化子的概念，或者直接证明这个集合的结构。

让我们考虑一个特殊的零因子集合：所有零因子的集合，我们称之为 $N$。

如果 $a in N$ 且 $b in N$：
如果 $a=0$ 或 $b=0$，那么 $a+b in N$。
如果 $a eq 0$ 且 $b eq 0$。存在非零 $s_1$ 使得 $as_1 = 0$；存在非零 $s_2$ 使得 $bs_2 = 0$。

我们来考虑一个更强的条件。如果 $a$ 是一个零因子，那么 $a$ 属于零化子 $Ann(S) = {r in R mid forall s in S, rs = 0}$ 的某个非空集合。更精确地说，如果 $a eq 0$ 是零因子，那么 $a in Ann({s})$ 对于某个非零 $s in R$。

Let's reconsider the definition of $N$. $N = {r in R mid exists s in R, s eq 0, rs = 0 ext{ or } r=0}$.

Suppose $a, b in N$.
If $a = 0$ or $b = 0$, then $a+b in N$.
If $a eq 0$ and $b eq 0$.
Then there exists $s_1 eq 0$ such that $as_1 = 0$.
And there exists $s_2 eq 0$ such that $bs_2 = 0$.

We need to show that $a+b$ is a zero divisor (or $a+b=0$).
If $a+b=0$, then $a+b in N$.
If $a+b eq 0$, we need to show there exists $s eq 0$ such that $(a+b)s = 0$.

This is where the proof can become tricky if we don't use the right tool.
Consider the case where $R$ is an integral domain. Then the only zero divisor is 0. The set of zero divisors is ${0}$, which is an ideal.

Let's consider a specific example: $R = mathbb{Z}_6$.
The elements are ${0, 1, 2, 3, 4, 5}$.
Zero divisors (nonzero): $2$ (because $2 cdot 3 = 6 equiv 0 pmod 6$), $3$ (because $3 cdot 2 = 6 equiv 0 pmod 6$), $4$ (because $4 cdot 3 = 12 equiv 0 pmod 6$).
The set of zero divisors and 0 is $N = {0, 2, 3, 4}$.

Let's check the properties for $N = {0, 2, 3, 4}$ in $mathbb{Z}_6$.
Nonempty: Yes, it contains 0.
Additively closed:
$2+0=2 in N$
$2+2=4 in N$
$2+3=5 otin N$.
Ah, wait. $5$ is not a zero divisor because $5 cdot s equiv 0 pmod 6$ only if $s=0$.

This means that my initial understanding or memory might be slightly off, or the statement requires a more precise definition of "zero factor".

Let's clarify the definition of $N$:
$N = {r in R mid r ext{ is a zero divisor, or } r=0}$.
A zero divisor is a nonzero element $r$ such that $exists s eq 0, rs = 0$.

In $mathbb{Z}_6$:
Zero divisors are ${2, 3, 4}$.
$N = {0, 2, 3, 4}$.

Let's recheck addition:
$2+2=4 in N$.
$2+3=5$. Is 5 a zero divisor? No, because $5s equiv 0 pmod 6$ only if $s$ is a multiple of 6, which means $s=0$ in $mathbb{Z}_6$.
So, $N={0, 2, 3, 4}$ is NOT closed under addition.

Therefore, the set of all zero divisors and 0 is NOT always an ideal in a general commutative ring.

What did I miss?

The question asks about "交换环的所有零因子和 0 组成的集合".
Is it possible that the question implies a specific type of zero divisor, or a property that I'm overlooking for the set to be an ideal?

Let's consider the set of nilpotent elements. An element $x$ is nilpotent if $x^n = 0$ for some positive integer $n$. The set of all nilpotent elements in a commutative ring is an ideal, called the nilradical.
Is there a confusion between zero divisors and nilpotent elements?
Zero divisors are elements $x eq 0$ such that $exists y eq 0, xy = 0$.
Nilpotent elements are elements $x$ such that $exists n geq 1, x^n = 0$.

Let's consider a standard result: The set of all zero divisors in a commutative ring $R$ is denoted by $Z(R)$. This set $Z(R)$ is not necessarily an ideal. For example, in $mathbb{Z}_6$, $Z(R)={2,3,4}$, and $2+3=5 otin Z(R)$.

Could the question be about something else?
Perhaps the phrasing "零因子和 0 组成的集合" has a subtle meaning.

Let's reread the question carefully: "交换环的所有零因子和 0 组成的集合是一个理想吗？"

Maybe the statement is false, and I need to provide a counterexample.
My counterexample $mathbb{Z}_6$ suggests the statement is false.

Let's try to prove it is an ideal to see where the argument fails.
Let $N = {r in R mid exists s eq 0, rs = 0 ext{ or } r=0}$.
We've established nonempty.

Additively Closed:
Let $a, b in N$.
Case 1: $a=0$ or $b=0$. Then $a+b in N$.
Case 2: $a eq 0, b eq 0$.
$exists s_1 eq 0, as_1 = 0$.
$exists s_2 eq 0, bs_2 = 0$.
We need to show $exists s eq 0, (a+b)s = 0$.

Consider $R = mathbb{Z}_n$. Zero divisors are elements that share a common factor with $n$ greater than 1.
If $n = p_1^{a_1} cdots p_k^{a_k}$ is the prime factorization of $n$.
An element $m in mathbb{Z}_n$ is a zero divisor iff $gcd(m, n) > 1$.
Let $n=6$. $m in {2,3,4}$.
$2 in N$, $3 in N$.
$2+3=5$. $gcd(5,6)=1$. So 5 is not a zero divisor.

What if the question meant "nilpotent elements" rather than "zero divisors"?
The set of nilpotent elements is an ideal.
An element $x$ is nilpotent if $x^n = 0$ for some $n geq 1$.
Let $N = {x in R mid exists n geq 1, x^n = 0}$.
1. $0 in N$ since $0^1 = 0$. So $N$ is nonempty.
2. Additively closed: Let $a, b in N$. Then $exists m geq 1, a^m = 0$ and $exists n geq 1, b^n = 0$.
Consider $(a+b)^{m+n1}$. Using the binomial expansion:
$(a+b)^{m+n1} = sum_{k=0}^{m+n1} inom{m+n1}{k} a^k b^{m+n1k}$.
If $k geq m$, then $a^k=0$. If $m+n1k geq n$, then $b^{m+n1k}=0$.
In any term of the sum, either $k geq m$ or $m+n1k geq n$.
If $k geq m$, the term is $inom{m+n1}{k} a^k b^{m+n1k} = inom{m+n1}{k} cdot 0 cdot b^{m+n1k} = 0$.
If $m+n1k geq n$, then $k leq m+n1n = m1$.
So if $k < m$, then $m+n1k > m+n1(m1) = n$. So $b^{m+n1k} = 0$.
The term is $inom{m+n1}{k} a^k b^{m+n1k} = inom{m+n1}{k} a^k cdot 0 = 0$.
Therefore, $(a+b)^{m+n1} = 0$. So $a+b$ is nilpotent, $a+b in N$. $N$ is additively closed.
3. Multiplicatively closed (by ring element): Let $a in N$ and $r in R$.
$exists n geq 1, a^n = 0$.
Consider $(ra)^n = r^n a^n = r^n cdot 0 = 0$.
So $ra$ is nilpotent, $ra in N$. $N$ is closed under multiplication by ring elements.

So, the set of nilpotent elements is indeed an ideal.

However, the question specifically uses the term "零因子" (zero divisors).

Let's look for a source that confirms whether the set of zero divisors is an ideal. Standard algebra texts (like Dummit & Foote, Artinian Rings, etc.) usually state that the set of zero divisors is not an ideal.

Where could the confusion arise?
Perhaps the question comes from a context where "zero factor" has a more restrictive meaning, or there's a typo.

Let's assume the question is as stated and my understanding of "zero divisor" is standard. Then the statement is false.

Formal Counterexample:
Consider the ring $R = mathbb{Z}_6$ (integers modulo 6). This is a commutative ring.
The elements are $R = {0, 1, 2, 3, 4, 5}$.

A zero divisor is a nonzero element $x in R$ such that there exists a nonzero element $y in R$ with $xy = 0$.

Let's find the zero divisors in $mathbb{Z}_6$:
$1$: $1 cdot y = 0 implies y=0$. Not a zero divisor.
$2$: $2 cdot 3 = 6 equiv 0 pmod 6$. Since $3 eq 0$, $2$ is a zero divisor.
$3$: $3 cdot 2 = 6 equiv 0 pmod 6$. Since $2 eq 0$, $3$ is a zero divisor.
$4$: $4 cdot 3 = 12 equiv 0 pmod 6$. Since $3 eq 0$, $4$ is a zero divisor.
$5$: $5 cdot y = 0 pmod 6$. This implies $y$ must be a multiple of 6, so $y=0$. Not a zero divisor.

The set of zero divisors in $mathbb{Z}_6$ is ${2, 3, 4}$.
The set of all zero divisors and 0 is $N = {0, 2, 3, 4}$.

Now, we check if $N$ is an ideal of $mathbb{Z}_6$. We need to check closure under addition and multiplication by ring elements.

Check for additive closure:
We need to check if for any $a, b in N$, their sum $a+b$ is also in $N$.
Let's take $a = 2 in N$ and $b = 3 in N$.
$a+b = 2+3 = 5$.
Is $5 in N$? No, because 5 is not a zero divisor (as shown above).
Since $2+3 = 5 otin N$, the set $N$ is not closed under addition.

Conclusion from the counterexample:
The set of all zero divisors and 0 is not generally an ideal in a commutative ring.

Why might the question be phrased this way, or what could be an intended true statement?

1. Typo: The question might have intended to ask about nilpotent elements. As proven above, the set of nilpotent elements is an ideal.
2. Specific Ring Type: In certain types of rings, the set of zero divisors might form an ideal. For example, in a Noetherian integral domain, the set of zero divisors is just ${0}$, which is an ideal. However, the question specifies a commutative ring in general.
3. Misinterpretation of "Zero Factor": Perhaps there's a nonstandard definition of "zero factor" being used. However, the standard definition of a zero divisor is widely accepted.
4. The statement is intended to be false: The question might be a test of understanding the properties of ideals, and the expected answer is "no" with a counterexample.

Let's consider the properties of the set of zero divisors $Z(R)$ more closely.
If $a in Z(R)$ and $r in R$, is $ra in Z(R)$?
If $a eq 0$ is a zero divisor, then $exists s eq 0, as = 0$.
Consider $ra$. If $r eq 0$, then $(ra)s = r(as) = r cdot 0 = 0$.
If $ra eq 0$, then $ra$ is a zero divisor.
What if $ra = 0$? If $a eq 0$, then $r$ must be a zero divisor or $r=0$. If $r=0$, $ra=0 in Z(R) cup {0}$. If $r$ is a zero divisor, then $ra$ is a zero divisor, so $ra in Z(R)$.
So, for any $a in Z(R) cup {0}$ and $r in R$, we have $ra in Z(R) cup {0}$ (assuming $ra=0$ implies $ra in Z(R) cup {0}$). This property, called absorption or multiplicative closure, holds.

The problem lies in additive closure.

Summary of the Argument:
Let $N$ be the set of all zero divisors and 0 in a commutative ring $R$.
$N = {r in R mid exists s in R, s eq 0, rs = 0 ext{ or } r=0}$.

1. Nonempty: $0 in N$ because $0 cdot s = 0$ for any $s eq 0$.
2. Absorption (multiplicative closure): Let $a in N$ and $r in R$.
If $a=0$, then $ra = 0 in N$.
If $a eq 0$, then $exists s eq 0, as = 0$.
Consider $ra$.
If $ra eq 0$, then $(ra)s = r(as) = r cdot 0 = 0$. Since $s eq 0$, $ra$ is a nonzero zero divisor, so $ra in N$.
If $ra = 0$, then $ra$ is the zero element, so $ra in N$.
Thus, $N$ is closed under multiplication by ring elements.
3. Additive Closure: Let $a, b in N$. We need to show $a+b in N$.
If $a=0$ or $b=0$, then $a+b in N$.
If $a eq 0$ and $b eq 0$.
Then $exists s_1 eq 0, as_1 = 0$.
And $exists s_2 eq 0, bs_2 = 0$.
We need to show that there exists $s eq 0$ such that $(a+b)s = 0$.
This is generally not true.

Counterexample (Revisited):
$R = mathbb{Z}_6 = {0, 1, 2, 3, 4, 5}$
$N = {0, 2, 3, 4}$
$a=2 in N$, $b=3 in N$.
$a+b = 5$.
Is $5$ a zero divisor in $mathbb{Z}_6$? No.
$5x equiv 0 pmod 6$ implies $x equiv 0 pmod 6$.
So $5 otin N$.
Thus, $N$ is not additively closed, and therefore not an ideal.

Final Answer Formulation:
The statement "交换环的所有零因子和 0 组成的集合是一个理想吗？" is false.

The set of zero divisors in a commutative ring is not generally an ideal. This is because the set of zero divisors is not always closed under addition.

To demonstrate this, consider the ring of integers modulo 6, denoted as $mathbb{Z}_6$.
The elements of $mathbb{Z}_6$ are ${0, 1, 2, 3, 4, 5}$.

A zero divisor is a nonzero element $x$ such that there exists another nonzero element $y$ in the ring with $xy = 0$.

In $mathbb{Z}_6$:
$2$ is a zero divisor because $2 cdot 3 = 6 equiv 0 pmod 6$, and $3 eq 0$.
$3$ is a zero divisor because $3 cdot 2 = 6 equiv 0 pmod 6$, and $2 eq 0$.
$4$ is a zero divisor because $4 cdot 3 = 12 equiv 0 pmod 6$, and $3 eq 0$.
The other nonzero elements ($1$ and $5$) are not zero divisors.

The set of all zero divisors and 0 in $mathbb{Z}_6$ is $N = {0, 2, 3, 4}$.

Now, let's check if this set $N$ is closed under addition.
Take $2 in N$ and $3 in N$.
Their sum is $2 + 3 = 5$.
However, $5$ is not a zero divisor in $mathbb{Z}_6$ (as $5x equiv 0 pmod 6$ implies $x=0$). Therefore, $5 otin N$.

Since $2 in N$ and $3 in N$, but $2+3 otin N$, the set $N$ is not closed under addition. For a set to be an ideal, it must be closed under addition.

Thus, the set of all zero divisors and 0 is not an ideal in a general commutative ring.

It is possible that the question intended to refer to the set of nilpotent elements, which is always an ideal in a commutative ring. An element $x$ is nilpotent if $x^n = 0$ for some positive integer $n$.

This detailed explanation addresses the question directly, provides a counterexample, and explains why the property fails.不是，交换环的所有零因子和 0 组成的集合通常不是一个理想。

要理解这一点，我们首先需要明确几个定义：

1. 交换环 (Commutative Ring)：一个环 $R$，如果对于任意的 $a, b in R$，都有 $ab = ba$（乘法可交换），那么称 $R$ 为交换环。

2. 零因子 (Zero Divisor)：在环 $R$ 中，一个非零元素 $a$ 被称为一个零因子，如果存在一个非零元素 $b in R$ 使得 $ab = 0$。在一个交换环中，左零因子和右零因子是同一个概念。我们通常将包含 0 的零因子集合记为 $Z(R) cup {0}$，或者更简洁地，如果我们将“零因子”的定义包含 0，那么就是指所有非零零因子和零元组成的集合。为了避免混淆，我们这里明确指出：设 $N = {r in R mid exists s in R, s eq 0, rs = 0 ext{ or } r=0}$。我们想要考察的集合就是 $N$。

3. 理想 (Ideal)：在环 $R$ 中，一个非空子集 $I$ 是 $R$ 的一个理想，如果它满足以下两个条件：
加法封闭性：对于任意的 $a, b in I$，都有 $a+b in I$。
吸收性 (或数乘封闭性)：对于任意的 $a in I$ 和任意的 $r in R$，都有 $ra in I$ 和 $ar in I$。在一个交换环中，这两个条件是等价的，只需满足 $ra in I$（或 $ar in I$）即可。

现在，我们来分析集合 $N$ 是否满足成为理想的条件。

验证集合 $N$ 的性质：

1. 非空性：
环 $R$ 必须包含加法单位元 0。根据零因子的定义，如果存在一个非零元素 $s in R$ 使得 $0 cdot s = 0$，那么 0 就是一个零因子（或者说，0 本身就属于我们考虑的集合 $N$）。对于环中的任何非零元素 $s$，都有 $0 cdot s = 0$。因此，$0 in N$。所以，$N$ 是非空的。

2. 吸收性（数乘封闭性）：
我们需要证明：如果 $a in N$ 且 $r in R$，那么 $ra in N$。
情况 1： 如果 $a = 0$。那么 $ra = r cdot 0 = 0$。因为 $0 in N$，所以 $ra in N$。
情况 2： 如果 $a eq 0$ 且 $a in N$。根据 $N$ 的定义，存在一个非零元素 $s in R$ 使得 $as = 0$。
现在考虑 $ra$。
如果 $ra eq 0$：那么我们有 $(ra)s = r(as) = r cdot 0 = 0$。由于 $s eq 0$，根据零因子的定义，$ra$ 是一个非零零因子，所以 $ra in N$。
如果 $ra = 0$：那么 $ra$ 是零元素，根据 $N$ 的定义，$ra in N$。
所以，对于任意的 $a in N$ 和 $r in R$，都有 $ra in N$。吸收性满足。

3. 加法封闭性：
这是最关键且常常不满足的条件。我们需要证明：如果 $a in N$ 且 $b in N$，那么 $a+b in N$。
情况 1： 如果 $a=0$ 或 $b=0$。
例如，如果 $a=0$，那么 $a+b = 0+b = b$。因为 $b in N$，所以 $a+b in N$。同理，如果 $b=0$，则 $a+b = a in N$。
情况 2： 如果 $a eq 0$ 且 $b eq 0$。
根据 $N$ 的定义，存在非零元素 $s_1 in R$ 使得 $as_1 = 0$，并且存在非零元素 $s_2 in R$ 使得 $bs_2 = 0$。
我们需要证明存在一个非零元素 $s in R$ 使得 $(a+b)s = 0$。

এখানেই问题所在。 即使 $a$ 和 $b$ 都是零因子，它们的和 $a+b$ 不一定是零因子。也就是说，$a+b$ 可能不是集合 $N$ 的元素。

反例说明：

考虑交换环 $mathbb{Z}_6$，即模 6 的整数环。其元素为 ${0, 1, 2, 3, 4, 5}$。

1. 找出零因子：
$2 cdot 3 = 6 equiv 0 pmod 6$。因为 $2 eq 0$ 且 $3 eq 0$，所以 2 是一个零因子。
$3 cdot 2 = 6 equiv 0 pmod 6$。因为 $3 eq 0$ 且 $2 eq 0$，所以 3 是一个零因子。
$4 cdot 3 = 12 equiv 0 pmod 6$。因为 $4 eq 0$ 且 $3 eq 0$，所以 4 是一个零因子。
元素 1 和 5 不是零因子，因为如果 $1 cdot x = 0$ 或 $5 cdot x = 0$，那么在 $mathbb{Z}_6$ 中必然有 $x=0$。

2. 构成集合 $N$：
所有零因子和 0 组成的集合是 $N = {0, 2, 3, 4}$。

3. 检查加法封闭性：
我们取 $N$ 中的两个元素 $a=2$ 和 $b=3$。
它们的和是 $a+b = 2+3 = 5$。
现在我们检查 5 是否在 $N$ 中。如上所述，5 不是零因子，也不是 0。所以 $5 otin N$。

由于我们找到了 $a in N$ 和 $b in N$，但 $a+b otin N$，所以集合 $N$ 不满足加法封闭性。

结论：

交换环的所有零因子和 0 组成的集合，在一般情况下，不满足理想的加法封闭性，因此不是一个理想。

可能的混淆：

值得注意的是，幂零元素 (Nilpotent Elements) 的集合在交换环中是构成一个理想的。一个元素 $x$ 是幂零的，如果存在一个正整数 $n$，使得 $x^n = 0$。这个理想被称为幂零根 (Nilradical)。零因子和幂零元素是不同的概念，虽然它们在某些环（如整环）中可能只剩下 0 这一个元素。

反例：

令交换环 ，设平凡与非平凡零因子所构成的集合为

然而

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推而广之， 时（ 为质数），皆为反例. 沿用上面记号

以剩余系为代表元按由小到大排列，而

而在 中介于 到 的元素只有 的某些倍数，但是很显然 ，所以最后只有

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事实上，上面证明条件还可以更宽松一点，只需要 即可，如此一来由裴蜀等式：

而 ，否则存在与相伴的 任意零因子 ，即

而这与零因子非零矛盾.

如此一来，只有 的零因子才可以满足题目，实际上此时有

显然后者是一主理想.