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**Shurelia** Let R be non-commutative prime ring

let t $\displaystyle \in$ R such that $\displaystyle {t}^{2}=0 $

i found statement:

$\displaystyle {(tR)}^{ 3}=0 $, thus tR is nonzero nil right ideal satisfying $\displaystyle {z}^{ 3}=0 $ for all z $\displaystyle \in$ tR, then R has nonzero nilpotent ideal.

my question is:

1. tR is nonzero nil right ideal, how can i proof tR is right ideal on R?(proof about closure in substraction) and any guarantee that tR is nonzero?

2. R has nonzero nilpotent ideal. how can i say that? (my mentor say there is lemma proof this in book "I.N. Herstein, Topics in Ring Theory, univ. chicago press,1969" i can't find this book, pretty rare i guess.

thanks

your question is a mess!

are you assuming that $\displaystyle (tR)^3=0$? i'm saying this because $\displaystyle t^2=0$ does not imply that $\displaystyle (tR)^3=0$, even in prime rings. if you're assuming $\displaystyle (tR)^3=0$, why do you need the condition $\displaystyle t^2=0$? are you trying to prove that a prime ring has no non-zero nilpotent ideal? to prove this, you only need $\displaystyle R$ to be semiprime.

anyway, $\displaystyle tR$ is obviously a right ideal. i don't know why you can't prove this! it is non-zero becasue $\displaystyle 1 \in R$ and so $\displaystyle 0 \neq t \in tR.$ i guess you forgot to mention in your question that $\displaystyle 1 \in R$ and $\displaystyle t \neq 0$.

for the second part of your question, you don't need Herstein! suppose that $\displaystyle I$ is a right ideal of a ring $\displaystyle R$ and $\displaystyle I^n=0$ for some integer $\displaystyle n \geq 2$. then $\displaystyle RI$ is clearly an ideal of $\displaystyle R$ and

$\displaystyle (RI)^n=R(IR)^{n-1}I \subseteq RI^{n-1}I = RI^n=0$.

thus $\displaystyle (RI)^n=0$ and so $\displaystyle RI$ is a nilpotent ideal of $\displaystyle R$. note that if $\displaystyle I \neq 0$, then$\displaystyle RI \neq 0$ because $\displaystyle I \subseteq RI.$