Let $U:=U_3(\Bbb C)$ be the ring of upper triangular matrices over $\Bbb C$ and $I:=\Big\{\begin{pmatrix} 0 & a & b \\ 0& 0& c\\ 0 & 0& 0 \end{pmatrix}: a,b,c\in \Bbb C\Big\}\subset U$ the two-sided ideal of $U$, of strictly upper triangular.
We define the nilradical $N(R)$ of a ring $R$ as the sum of all two sided nilpotent ideals of it.
We want to show that $N(U)=I$.
$“\subseteq": $ Since $I^3=\mathrm{O}_3$, so $I$ is nilpotent, it takes part in the sum. Thus, $I\subseteq N(U)$.
$“\supseteq":$ We consider now the quotient ring
$$U/I:=\{r+I:r\in U\}=\Big\{ \begin{pmatrix} z_1 & 0 & 0 \\ 0 & z_2 & 0 \\ 0 & 0 & z_3 \end{pmatrix} +I: z_1,z_2,z_3\in \Bbb C\Big \}.$$
We observe that there is a ring isomorphism
$$U/I\cong \Bbb C \times \Bbb C \times \Bbb C$$
and that $\Bbb C \times \Bbb C \times \Bbb C$ has not non-zero nilpotent elements, so the same holds for $U/I$.
Now, $N(U)$ is a nil ideal, thus $N(U)/I \trianglelefteq U/I$ is also a nil ideal, which means that all its elements are nilpotent.
But how can we show now that $N(U)\subseteq I$?
Thanks.
Best Answer
It's simply because every nilpotent element of $U$ is contained in $I$, hence every two-sided nilpotent ideal is included in $I$.
To see the first claim, just take any nilpotent element $u$ of $U$ and look at the image of $u$ in the quotient $U/I$. Since $u$ is nilpotent, some power of the image of $u$ in the quotient is $0$.
But that quotient has no non-zero nilpotent element, hence the image of $u$ in the quotient is $0$.