# Radical is an Ideal

• Jul 27th 2012, 01:29 AM
Cairo
Radical is an Ideal
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
• Jul 27th 2012, 04:11 AM
girdav
Re: Radical is an Ideal
Take $r$ in the radical , $x$ an element of the algebra and $f$ a scalar homomorphism. Show that $f(rx)=0=f(xr)$.
• Jul 28th 2012, 08:54 AM
Cairo
Re: Radical is an Ideal
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.
• Jul 28th 2012, 08:58 AM
girdav
Re: Radical is an Ideal
What's wrong with this?
• Jul 28th 2012, 09:09 AM
Cairo
Re: Radical is an Ideal
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?
• Jul 28th 2012, 09:14 AM
girdav
Re: Radical is an Ideal
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).
• Jul 28th 2012, 09:18 AM
Cairo
Re: Radical is an Ideal
So if x, y in Ker(f) and lambda, mu in C, then

f(lambda(x) + mu(y)) = f(lambda(x)) + f(mu(y))= lambda f(x) + mu f(y) = lambda.0 + mu.0 = 0, so that lambda(x)+mu(y) in Ker(f)?