every $x≠0$ is a zero divisor or a unit

hbghlyj

“在没有零因子的环中，所有非零元素都是单位”是否正确?
我在ringsandmodulespartsIandII.pdf(这里的环含1)看到

Proposition 1.34
  Suppose that $R$ is a ring with no non-zero zero divisors that is also a
  finite dimensional vector space in such a way that left and right
  multiplication maps are linear. Then $U (R) = R^{\ast}$, and in particular
  if $R$ is an integral domain then $R$ is a field.

Proof.
  For $a \in R$ the map $R \rightarrow R ; x \mapsto xa$ is linear, and since
  $R$ is an integral domain it is injective if $a \in R^{\ast}$. Since $R$ is
  finite dimensional the Rank-Nullity Theorem tells us that the map is
  surjective, and hence there is $x \in R$ such that $x a = 1$. Similarly
  there is $y$ such that $ay = 1$, and finally $x = x 1 = x (ay) = (xa) y = 1 y = y$
  so $a \in U (R)$ as required.

在Zero divisor or unit看到

Let $R$ be a finite ring with unity. Prove that every $x≠0$ is a zero divisor or a unit.

条件“finite ring”包含于Proposition 1.34的条件“a finite dimensional vector space”

Czhang271828

无零因子环的非零元怎么可能是单位? 拿最简单的 $\mathbb Z$, $2$ 既非零因子又非单位.

Proposition 1.34 中都给出有限维向量空间这么强的条件了, 那么 $U(R)=R^\ast$ 是自然的. 下证明 $R^\ast\subseteq U(R)$: $\forall x\in R^\ast$, 则 $x$ 在 $R$ 上的正则作用 $R\to R,r\mapsto xr$ 是单射(零的原像是零, 因为 $x\in R^\ast$). 注意到 $x$ 作为有限维线性空间的自同态, 从而单射等价于满射: 这也是运用有限维线性空间特性证明此题的关键. 一般的, 对任意环 $A$ 与秩相同的有限生成模 $M,N$, 只有 $f:M\twoheadrightarrow N$ 推出 $f:M\simeq N$, 单射推不出满射.

有限维情形就更简单了, 任意非零因子且非零的元素在自身上的作用是单射, 从而是一个置换 $\sigma$. 由于 $\sigma^{|R|!}=e$, 因此有 $\sigma\in U(R)$.

hbghlyj

Czhang271828 发表于 2023-4-19 08:00
有限维线性空间的自同态, 从而单射等价于满射.

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hbghlyj

Czhang271828 发表于 2023-4-19 08:00
有限维情形就更简单了.

我明白了，R 是有限维的条件比要求它是有限环更广泛，R 可以是有限维但不是有限环，例如$\mathbb Z^n$

hbghlyj

Czhang271828 发表于 2023-4-19 08:00
一般的, 对任意环 $A$ 与秩相同的有限生成模 $M,N$, 只有 $f:M\twoheadrightarrow N$ 推出 $f:M\simeq N$, 单射推不出满射.

$\Bbb Z→\Bbb Z$ multiplication by 2 is injective but not surjective.
injective endomorphisms of finite modules need not be surjective

hbghlyj

Rings in which every non-unit is a zero divisor
Any (commutative unitary) ring of Krull dimension 0 has this property. This includes the class of Artinian rings.

Proof: If $A$ has Krull dimension $0$, then any maximal $m$ ideal of $A$ is also a minimal prime ideal by the definition of Krull dimension. Applying Krull's theorem on the intersection of prime ideals to the localization $A_m$, we find that $mA_m$ is the nilradical of $A_m$. So for any $f\in m$, there exists a positive integer $n$ such that $f^n=0$ in $A_m$. So there exists $s\in A\setminus m$ such that $sf^n=0$ in $A$. We can choose $n$ smallest with respect to this property so that $sf^{n-1}\ne 0$. Therefore $f$ is a zero divisor. Now any non-unit element $f$ belong to some maximal ideal, it is a zero divisor.

hbghlyj

Finite Integral Domain is a Field
Proof 1.        Let $r \in R$ be a nonzero element.
We show that $r$ is a unit.
Consider the map $f: R\to R$ sending $x\in R$ to $f(x)=rx$.
We claim that the map $f$ is injective.
Suppose that we have $f(x)=f(y)$ for $x, y \in R$. Then we have
\[rx=ry\]or equivalently, we have
\[r(x-y)=0.\]
Since $R$ is an integral domain and $r\neq 0$, we must have $x-y=0$, and thus $x=y$.
Hence $f$ is injective. Since $R$ is a finite set, the map is also surjective.
Then it follows that there exists $s\in R$ such that $rs=f(s)=1$, and thus $r$ is a unit.
Since any nonzero element of a commutative ring $R$ is a unit, $R$ is a field.
Proof 2.                Let $r\in R$ be a nonzero element.
We show that the inverse element of $r$ exists in $R$ as follows.
Consider the powers of $r$:
\[r, r^2, r^3,\dots.\]Since $R$ is a finite ring, not all of the powers cannot be distinct.
Thus, there exist positive integers $m > n$ such that
\[r^m=r^n.\]
Equivalently we have
\[r^n(r^{m-n}-1)=0.\]Since $R$ is an integral domain, this yields either $r^n=0$ or $r^{m-n}-1=0$.
But the former gives $r=0$, and this is a contradiction since $r\neq 0$.
Hence we have $r^{m-n}=1$, and thus
\[r\cdot r^{m-n-1}=1.\]Since $r-m-1 \geq 0$, we have $r^{m-n-1}\in R$ and it is the inverse element of $r$.
Therefore, any nonzero element of $R$ has the inverse element in $R$, hence $R$ is a field.

Czhang271828

hbghlyj 发表于 2023-6-18 06:44
Finite Integral Domain is a Field
Proof 1.        Let $r \in R$ be a nonzero element.
We show that $r$ is a ...

这只证明是个除环，最后用 Wedderburn 小定理方可说明是域。