关于交换与非交换环中幂零元与单位元之和的单位性的问题咨询

原问题 #

Let $x$ be a nilpotent element and $y$ be a unit in a commutative ring $R$ with identity. Show that $x+y$ is also a unit in $R.$ If $y=1,$ is commutativity of $R$ necessary? Justify: If the elements $x,y$ do not commute, show that the above property does not hold.

我的解法 #

Let $R$ be a commutative ring with identity.

We note that, $\exists k\in\Bbb Z^+$ such that $x^k=0$ as $x$ is a nilpotent element in $R.$

Let $x'$ be any nilpotent element in $R$ such that $(x')^n=0,$ where $n\in\Bbb Z^+.$

We note that,
$$(1-x')(1+x'+(x')^2+(x')3+...+(x')^{{n-1})=1-(x')}n.$$

But this means, that
$$(1-(-x'))(1-x'+(x')^2-(x')3+...+(-1)^{n-1}(x'){n-1})=1-(-1)^n(x')n\implies (1+x')S=1,$$
where $S=1-x'+(x')^2-(x')3+...+(-1)^{n-1}(x'){n-1}.$

As $R$ is commutative so we have, $S(1+x')=(1+x')S=1.$

So, if $x'$ is a nilpotent element in a ring $R$ with an identity $1$, then, $1+x'$ is a unit in $R.$

This means, $1+x$ is a unit in $R.$

Now, $x+y=y+x=y(1+xy^{-1}).$

We note that $(xy^{-1})k=x^ky{-k}=0$ and so, $xy^{-1}$ is a nilpotent element in $R.$

Thus, $1+xy^{-1}$ is a unit element in $R.$

So, $x+y=y(1+xy^{-1})$ is a unit in $R$ as $y$ is a unit element and product of two unit elements in a ring is also a unit element.

If $y=1,$ then
$$(1-(-x))(1-x+(x)^2-(x)3+...+(-1)^{n-1}(x){n-1})=1-(-1)^n(x)n\implies (1+x)S=1,$$
where $S=1-x+(x)^2-(x)3+...+(-1)^{n-1}(x){n-1}.$

But the identity, $(1+x'+(x')^2+(x')3+...+(x')^{{n-1})(1-x')=1-(x')}n$ is also valid. So, we may also have that $(1+x)S=1,$ due to which we again, have $1+x$ as a unit of $R.$

So, if $y=1,$ then commutativity of $R$ not necessary, as the identity
$$(1+x'+(x')^2+(x')3+...+(x')^{{n-1})(1-x')=1-(x')}n=(1-x')(1+x'+(x')^2+(x')3+...+(x')^{{n-1})=1-(x')}n$$
is valid in any ring $R$ (having a $1\neq 0$ ) commutative or non-commutative.

Finally, we note that, if $x,y$ did not commute then examining our lines of proof, we find that we could not have written the line,
$(xy^{-1})k=x^ky{-k}=0$ and so, $xy^{-1}$ is a nilpotent element in $R.$

Hence, the proof above would fail for a ring where $x,y$ wouldn't have commuted.

我的疑问 #

1. 我关于“当$y=1$时，$R$的交换性不是必要条件”的推理是否合理？具体来说，我引用的那个乘法恒等式在任意含幺环（$1\neq0$）中都成立的说法是否正确？
2. 对于“若$x,y$不交换，则上述性质不成立”这部分，我只说明了原证明失效，但这不能排除存在其他证明方法的可能。我需要一个具体的反例，但找不到合适的，希望得到帮助。

解答 #

问题1：当$y=1$时交换性是否必要？ #

你的推理完全正确！那个关键的乘法恒等式在任意含幺环中都是成立的，不管交换与否。我们可以直接展开验证：

• 对于$(1 - x')(1 + x' + x'^2 + ... + x'^{n-1})$，把左边乘开：
  $1*(1 + x' + ... + x'^{n-1}) - x'*(1 + x' + ... + x'^{n-1}) = 1 + x' + x'^2 + ... + x'^{n-1} - x' - x'^2 - ... - x'^n = 1 - x'^n$
• 反过来$(1 + x' + ... + x'^{n-1})(1 - x')$的展开结果也是一样的，因为每一项的抵消不需要交换性——只是同类项的加减，乘法分配律在非交换环中依然成立（左右分配都有效）。

既然$x$是幂零元，取$n$为$x$的幂零指数，那么$x^n=0$，所以$(1 + x)(1 - x + x^2 - ... + (-1)^{n-1}x{n-1}) = 1 - (-1)^n x^n = 1$，同时反过来的乘积也是1。这就直接证明了$1+x$是单位，完全不需要交换性。所以你的结论没问题，这里交换性确实是多余的。

问题2：非交换情形下的反例 #

你说得对，只说原证明失效不够，得给具体反例。这里有个经典的例子，用实数域上的2阶方阵环（非交换）来构造：

考虑环$R = M_2(\mathbb{R})$，也就是实数域上的2阶方阵环，它是含幺环（单位元是单位矩阵$I$），但非交换。

• 取幂零元$x = \begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$，显然$x^2 = \begin{pmatrix}0 & 0 \ 0 & 0\end{pmatrix}$，满足幂零性。
• 取单位元$y = \begin{pmatrix}1 & 0 \ 1 & 1\end{pmatrix}$，它的逆矩阵是$\begin{pmatrix}1 & 0 \ -1 & 1\end{pmatrix}$，所以$y$是单位。
• 验证$x$和$y$不交换：计算$xy = \begin{pmatrix}0 &1\0&0\end{pmatrix}\begin{pmatrix}1&0\1&1\end{pmatrix} = \begin{pmatrix}1&1\0&0\end{pmatrix}$，而$yx = \begin{pmatrix}1&0\1&1\end{pmatrix}\begin{pmatrix}0&1\0&0\end{pmatrix} = \begin{pmatrix}0&1\0&1\end{pmatrix}$，显然$xy \neq yx$。
• 计算$x+y = \begin{pmatrix}1&1\1&1\end{pmatrix}$，这个矩阵的行列式是$11 - 11 = 0$，所以它不可逆，不是单位。

这个例子完美符合要求：幂零元加不交换的单位元，结果不是单位，直接推翻了原性质在非交换不交换情形下的成立性。

备注：内容来源于stack exchange，提问作者Thomas Finley