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.
My solution goes like this:
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 elemwnt in a ring $R$ with an identity $1$, then, $1+x$ is a unit in $R.$
This means, $1+x$ is a unit in $\Bbb R.$
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.
Is the part of my solution, where I claimed:
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
valid? I want to know precisely whether my reasoning for establishing commutativity an unnecessary property if $y=1$ is justified or not?
Finally, in the portion, where I wrote:
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
does not seem like an answer to the question, "If the elements $x,y$ do not commute, show that the above property does not hold". Thus is because, although a reasoning in a particular lines of a proof may fail, but it does not mean that there does not any other way of proving if the result holds if the elements $x,y$ do not commute. I think I must provide a counter example for this case, but unfortunately, I couldn't find any.
Any help regarding these two apparent issues will be greatly appreciated.
It's not hard to find counterexamples in my favourite noncommutative ring, the $2 \times 2$ matrices over $\mathbb Z$. Take e.g.
$$ x = \pmatrix{0 & 1\cr 0 & 0\cr},\ y = \pmatrix{0 & -1\cr 1 & 0\cr}$$