Ring homomorphisms of polynomial rings

[missavvy]

Homework Statement

Let R be a commutative ring and let f_a: R[x] -> R be evaluation at a $$\in$$ R.
If S: R[x] -> R is any ring homomorphism such that S(r) = r for all r$$\in$$ R, show that S = f_a for some a $$\in$$ R.

Homework Equations

The Attempt at a Solution

I don't get this at all.. really.. :S

Is this to show that for any ring homomorphism, you can evaluate it at some a in Z(R)?

Help?

 

[Dick]

Take the polynomial x in R[x]. S(x) is in R. Call it 'a'. Can you show, for example, that S(x^2)=a^2?

 

[missavvy]

So then S(x^2) = x^2, which is an element of R[x] and since fa: R[x] -> R is evaluation at a, we can sub in the a for x and get a^2... ?

 

[Dick]

So then S(x^2) = x^2, which is an element of R[x] and since fa: R[x] -> R is evaluation at a, we can sub in the a for x and get a^2... ?

No. You want to PROVE that S is evaluation at a. Use that S is a ring homomorphism. S(x^2)=S(x*x)=S(x)*S(x), right?

 

[missavvy]

Ok, well then S(x)*S(x) = a*a = a^2 = x^2... ?

 

[Dick]

Ok, well then S(x)*S(x) = a*a = a^2 = x^2... ?

So far so good. Now you've got S(x^2)=f_a(x^2). What about the rest of the polynomials? And saying a^2=x^2 is wrong and sloppy. a^2 is a real number, x^2 is a polynomial. S(x^2)=a^2. x^2 isn't equal to a^2.

 

[missavvy]

So I have to show that S(a_0 + a_1x+ ... + a_nx^n) = f_a(a_0+..+a_nx^n).
I don't really understand how I separate the terms of the polynomial to get to the point where they end up as evaluation at a.. do I have to define each x^k term to be = a^k for k = 0, n?

 

[Dick]

So I have to show that S(a_0 + a_1x+ ... + a_nx^n) = f_a(a_0+..+a_nx^n).
I don't really understand how I separate the terms of the polynomial to get to the point where they end up as evaluation at a.. do I have to define each x^k term to be = a^k for k = 0, n?

You do it the same way you did x^2, missavvy. And use S(a+b)=S(a)+S(b) in addition to S(a*b)=S(a)*S(b). Do you know what 'ring homomorphism' means? If not, could you look it up, please?

 

[missavvy]

Yes sir :) Got it thanks.