0

Let $A$ and $B$ be two finitedimensional $k$-algebras, where $k$ is an algebraically closed field. If $f: A \rightarrow B$ is a homomorphism of $k$-algebras, then $f(radA)\subseteq radB$, why?

flag
3 
 You should ask this on math.stackexchange.com or some other similar site (or consult a textbook :-) ) Here the focus is on research-oriented questions. Good luck! – Mariano Suárez-Alvarez Sep 21 at 3:31
1 
 This is pretty obviously wrong: Take $A$ to be the ring of upper-triangular $2\times 2$-matrices, $B$ to be the ring of all $2\times 2$-matrices, and $f$ to be the canonical inclusion. – darij grinberg Sep 21 at 3:40
4 
 (Actually, let $A$ better be the ring of upper-triangular $2\times 2$-matrices with the diagonal consisting of two equal entries; then it's definitely wrong.) – darij grinberg Sep 21 at 3:41
(darij: This is stated as an exercise in at least one textbook —a misprint, of course— so it may come from that) – Mariano Suárez-Alvarez Sep 21 at 3:58
3 
 «canonical inclusion» is a pet-peeve of mine... While technically correct, there is only one inclusion, so it is canonical only in a very boring sense :-) – Mariano Suárez-Alvarez Sep 21 at 4:07

closed as off topic by Mariano Suárez-Alvarez, unknown (google), Qiaochu Yuan, Yemon Choi, Fernando Muro Sep 21 at 6:24

Browse other questions tagged or ask your own question.