First isomorphism theorem for rings

[Kate2010]

Homework Statement

I have to show that \sum ai xⁱ -> (a0 \sum ai) is a ring homomorphism from C[x] to C x C

I then have to use the first isomorphism theorem to show that there is an isomorphism from C[x]/ (x(x-1)) to C x C where (x(x-1)) is the principal ideal (p) generated by the element p=x(x-1) of C[x]

It then asks is C[x]/(x(x-1)) an integral domain.

Homework Equations

First isomorphism theorem

The Attempt at a Solution

I think I may have done the first part but I'm a little unsure if my notation/understanding is fully correct, considering multiplication can I say f((\sum ai xⁱ)(\sum bj x^j) = f (\sum (aibj) x^k) {summing over i+j=k} = (a0,\sumai)(bo,\sumbj) = f(\sum ai xⁱ) f(\sum bj x^j)

I'm more stuck on the second part. I think I am aiming to show that ker f = (x(x-1))

ker f = {a0 + a1x + ... + anxⁿ : a0 = 0, a1 + ... +an = 0}
= {x (a1 + ... + anx^n-1) : a1+ ... +an =0}

Now I thought about trying to pull out a factor of x-1 but I don't think that would work?
I can see that x(x-1) does satisfy that there is no constant term and the sum of ai is 0 but I don't see how to get that all polynomials multiplied by it do.

Thanks :)

 

[eok20]

I'm more stuck on the second part. I think I am aiming to show that ker f = (x(x-1))

ker f = {a0 + a1x + ... + anxⁿ : a0 = 0, a1 + ... +an = 0}
= {x (a1 + ... + anx^n-1) : a1+ ... +an =0}

Now I thought about trying to pull out a factor of x-1 but I don't think that would work?
I can see that x(x-1) does satisfy that there is no constant term and the sum of ai is 0 but I don't see how to get that all polynomials multiplied by it do.

Since you know that 1 is a root of the polynomial a1x + ... + an x^(n-1), x-1 will divide it.

Alternatively, you know that x(x-1) must generate the entire ideal because C[x] is a PID and so ker f is generated by any element of lowest degree (and clearly there cannot be an element in ker f of degree less than 2 since 0 and 1 are both roots).