# Thread: ideal,nil,nilpotent ideal in prime ring

1. ## ideal,nil,nilpotent ideal in prime ring

Let R be non-commutative prime ring
let t $\in$ R such that ${t}^{2}=0$
i found statement:
${(tR)}^{ 3}=0$, thus tR is nonzero nil right ideal satisfying ${z}^{ 3}=0$ for all z $\in$ 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

2. Originally Posted by Shurelia
Let R be non-commutative prime ring
let t $\in$ R such that ${t}^{2}=0$
i found statement:
${(tR)}^{ 3}=0$, thus tR is nonzero nil right ideal satisfying ${z}^{ 3}=0$ for all z $\in$ 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 $(tR)^3=0$? i'm saying this because $t^2=0$ does not imply that $(tR)^3=0$, even in prime rings. if you're assuming $(tR)^3=0$, why do you need the condition $t^2=0$? are you trying to prove that a prime ring has no non-zero nilpotent ideal? to prove this, you only need $R$ to be semiprime.

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

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

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

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