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

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- Jul 27th 2012, 01:29 AMCairoRadical 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 AMgirdavRe: 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 $\displaystyle f(rx)=0=f(xr)$.

- Jul 28th 2012, 08:54 AMCairoRe: Radical is an Ideal
- Jul 28th 2012, 08:58 AMgirdavRe: Radical is an Ideal
What's wrong with this?

- Jul 28th 2012, 09:09 AMCairoRe: 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 AMgirdavRe: 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 AMCairoRe: 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)?