1. ## 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

2. ## Re: Radical is an Ideal

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

3. ## Re: Radical is an Ideal

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

4. ## Re: Radical is an Ideal

What's wrong with this?

5. ## 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?

6. ## 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).

7. ## 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)?