Abstract Algebra: Isomorphic polynomial rings

katarinab316
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Homework Statement



If F is an infinate field, prove that the polynomial ring F[x] is isomorphic to the ring T of all polynomial functions from F to F

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The Attempt at a Solution



T is isomorphic to F[x]
f(a+b) = f(a) + f(b)
f(ab)=f(a)f(b)
It is surjective by the definition of T
Injective: If f is not equal to g in F[x], then f(x) is not equal to g(x) in T.
Let h=f-g and assume h is not 0 in F
Claim: h(x) is not equal to zero in T, then h(a)=0 for every a in F. Thus every a in F is a root of the polynomial h in F[x}. However, there is no nonzero polynomial with infinately many roots, thus h=0 in F, which contradicts the asumption of h is not 0 in F, so injective. So F[x] is isomorphic to the ring T.
 
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Hrm. You haven't said what you want from us. I assume just to review your work?

katarinab316 said:
T is isomorphic to F[x]
f(a+b) = f(a) + f(b)
f(ab)=f(a)f(b)
It is surjective by the definition of T
This doesn't make sense. Why are you starting with the thing you're trying to prove? What is f? a? b? What was the point of those equalities? Why are they true? What is surjective? (I assume some of these latter questions, as well as what follows, would become clearer after you answer the first few)
 
Well basically this is an assignment turned in that got a "redo". As you can probably tell, I'm so confused in this subject I don't even know how to begin to correct it...thanks for the input so far
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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