First isomorphism theorem for rings

  • Thread starter Kate2010
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Homework Statement



I have to show that [tex]\sum[/tex] ai xi -> (a0 [tex]\sum[/tex] 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(([tex]\sum[/tex] ai xi)([tex]\sum[/tex] bj xj) = f ([tex]\sum[/tex] (aibj) xk) {summing over i+j=k} = (a0,[tex]\sum[/tex]ai)(bo,[tex]\sum[/tex]bj) = f([tex]\sum[/tex] ai xi) f([tex]\sum[/tex] bj xj)

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 + ... + anxn-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 :)
 

Answers and Replies

  • #2
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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 + ... + anxn-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).
 

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