Gheorghe Țițeica 2025 Grade 12 P4

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Source: Gheorghe Țițeica 2025

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#1 h Mar 28, 2025, 10:05 PM • 1 Y

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Let $R$ be a ring. Let $x,y\in R$ such that $x^2=y^2=0$ . Prove that if $x+y-xy$ is nilpotent, so is $xy$ .

Janez Šter

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#2 h Mar 29, 2025, 4:26 PM

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Let x and y be elements in a ring such that x squared = 0 and y squared = 0.
Suppose a = x + y - xy is nilpotent.

When you expand powers of a, like a squared, you get terms like xy, yx, and (xy)^2.
Because x squared and y squared are zero, many terms drop out.
This means that powers of a eventually involve powers of xy.

Since a to some power is zero, and that includes powers of xy,
it must be that some power of xy is also zero.

Therefore, xy is nilpotent.

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#3 h Mar 30, 2025, 12:06 PM

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Let $(x+y-xy)^k=0, k\in\mathbb{N}$ , where $k$ is the degree. If $k=1:x(x+y-xy)=0\xRightarrow{\hspace{25pt}}\boxed{xy=0}.$
If $k\geqslant 2: x(x+y-xy)^k=xy(x+y-xy)^{k-1}=0=(x+y-xy)^k\xRightarrow{\hspace{25pt}}(x+y)(x+y-xy)^{k-1}=0.$
$\xRightarrow{\hspace{25pt}}yx(x+y-xy)^{k-1}=0\xRightarrow{\hspace{25pt}}(xy)^2(x+y-xy)^{k-2}=0.$
Now suppose we have proven $(xy)^a(x+y-xy)^{k-a}=0\quad \forall 1\le a\le n$ . We will prove it for $n+1$ .
$[asy]usepackage("amsmath, amssymb, tikz, tikz-cd"); label("\begin{tikzcd}[ampersand replacement=\&] \scriptstyle(xy)^n(x+y-xy)^{k-n}+(xy)^{n-1}(x+y-xy)^{k-n+1}=0\&\&\scriptstyle(xy)^{n-1}(x+y)(x+y-xy)^{k-n}=0. \arrow[Rightarrow, from=1-1, to=1-3] \end{tikzcd}");[/asy]$
So $[asy]usepackage("amsmath, amssymb, tikz, tikz-cd"); label("\begin{tikzcd}[ampersand replacement=\&] \scriptstyle(xy)^{n-1}xyx(x+y-xy)^{k-n}=0\&\&\scriptstyle(xy)^{n+1}(x+y-xy)^{k-n-1}=0. \arrow[Rightarrow, from=1-1, to=1-3] \end{tikzcd}");[/asy]$
Hence $[asy]usepackage("amsmath, amssymb, tikz, tikz-cd"); label("\begin{tikzcd}[ampersand replacement=\&] \hspace{1pt}\&\&\color{blue}\scriptstyle \exists k\in\mathbb{N}\ (xy)^k=0. \arrow[Rightarrow, blue, from=1-1, to=1-3] \end{tikzcd}");[/asy]$

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