Let $I_0 ⊊ I $ be ideals of a ring $R$ with $1$ such that $I/I_0$ is a semiprime ideal of $R/I_0$. Is it a fact that $I$ is semiprime in $R$? I have an argument for this, but I am not sure of its validity.
By Theorem (10.11) of "A First Course in Noncommutative Rings, T.Y. Lam", $$I/I_0=∩_i Q_i/I_0,$$ where $Q_i$ are prime ideals of $R$ containing $I_0$. Now, we have $$x\in I ⇔ x+I_0\in ∩_i Q_i/I_0⇔ x\in ∩_iQ_i $$ for all $i$, which means that $I$ is semiprime, by the same theorem.
Thanks for replies!
Yes.
Said another way, the prime ideals of $R$ containing $I$ correspond exactly to prime ideals of $R/I_0$ containing $I/I_0$. The lattice isomorphism between ideals of $R$ containing $I_0$ and the ideals of $R/I_0$ respects everything, including intersections. Therefore both prime and semiprime ideals correspond to prime and semiprime ideals, respectively.
But even more simply, the third isomorphism theorem says $(R/I_0)/(I/I_0)\cong R/I$, so $I$ is semiprime in $R$ iff $I/I_0$ is semiprime in $R/I_0$.