• July 27th 2012, 02:29 AM
Cairo
Let S be the set of all scalar homomorphisms of an algebra X. Then the radical rad{X} of X is defined to be rad{X}=intersection_{f in {S}} {Ker{f}}.

Can anyone help me prove that a radical of an algbera is an ideal?

Thanks
• July 27th 2012, 05:11 AM
girdav
Take $r$ in the radical , $x$ an element of the algebra and $f$ a scalar homomorphism. Show that $f(rx)=0=f(xr)$.
• July 28th 2012, 09:54 AM
Cairo
Quote:

Originally Posted by girdav
Take $r$ in the radical , $x$ an element of the algebra and $f$ a scalar homomorphism. Show that $f(rx)=0=f(xr)$.

Well f(xr)=f(x)f(r)=0.f(r)=0. A similar result holds for f(rx). But this surely can't be a proof.
• July 28th 2012, 09:58 AM
girdav
What's wrong with this?
• July 28th 2012, 10:09 AM
Cairo
Do we not need to prove that it is a subspace of X too?

I'm really confused and having be drowning in functional analysis problems all day. Are you able to offer a full proof?
• July 28th 2012, 10:14 AM
girdav
Yes, we also need to check that the radical is stable by sums: the sum of two elements of the ideal still is in the ideal. After that, I think we get all the condition that an ideal needs to satisfy (ideal in sense of ring theory, I mean).
• July 28th 2012, 10:18 AM
Cairo