Abstract Algebra: Isomorphic polynomial rings

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SUMMARY

The discussion centers on proving that the polynomial ring F[x] is isomorphic to the ring T of all polynomial functions from an infinite field F to F. The proof involves demonstrating both surjectivity and injectivity of the mapping between F[x] and T. Key arguments include the properties of polynomial functions, specifically that if a polynomial h is not zero in F[x], it cannot have infinitely many roots in F, leading to a contradiction. The conclusion drawn is that F[x] is indeed isomorphic to T.

PREREQUISITES
  • Understanding of polynomial rings, specifically F[x]
  • Knowledge of isomorphism in algebra
  • Familiarity with the concepts of surjectivity and injectivity
  • Basic properties of infinite fields
NEXT STEPS
  • Study the properties of isomorphic structures in abstract algebra
  • Learn about polynomial function behavior in infinite fields
  • Explore examples of surjective and injective functions
  • Review the fundamentals of polynomial roots and their implications
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying polynomial rings and isomorphisms, as well as educators seeking to clarify these concepts for their students.

katarinab316
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Homework Statement



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

Homework Equations





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
 

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