Abstract Algebra problem (related to Rings of Polynomails

[JamesF]

We'll I made it through another semester, but it seems that I am completely stuck on the last problem of the last homework assignment. I've made a little progress, but I'm really having trouble understanding the question. Perhaps someone on these forums will have some insight

Homework Statement

Let F be a field. An element $$\phi \in F^F$$ is a polynomial function on F if there exists an $$f(x) \in F[x] \text{ s.t. } \phi(a) = f(a) \; \forall a \in F$$

a) Show that the set $$P_F$$ of all polynomial functions on F form a subring of $$F^F$$

b) Show that the ring $$P_F$$ is not necessarily isomorphic to F[x] (Hint: show that if F is a finite field, then $$P_F$$ and F[x] don't have the same number of elements)

Homework Equations

In this problem, $$F^F$$ refers to the set of all functions mapping F to F, and F[x] refers to the ring of polynomials with coefficients in F.

The Attempt at a Solution

Part b) doesn't seem so difficult. If we let $$F = \mathbb{Z}_2$$, then there are only 4 functions in all since there are only two possible values in the domain and range. But $$\mathbb{Z}_2 [x]$$ has infinitely many elements, so they cannot be isomorphic.

Part a) though has me stumped. We have that $$P_F \subseteq F^F$$, so for any $$\phi \in P_F, \longrightarrow \phi: F \rightarrow F$$. If we let $$\phi, \psi \in P_F \; f(x), g(x) \in F[x]$$, then we have that $$\phi(a) = f(a), \; \psi(a) = \g(a)$$.

Am I understanding this correctly? How can I demonstrate that it satisfies the ring axioms? Just directly apply them and see if it works? Or am I missing something?

 

[mighty2000]

I can understand your frustration with this problem, but don't worry, I am here to help. First, let me clarify that your understanding of the problem is correct. The set P_F is a subset of F^F, and for any element \phi in P_F, it is a function that maps elements of F to elements of F. Now, to show that P_F is a subring of F^F, we need to show that it satisfies the ring axioms.

First, let's consider the closure under addition. Let \phi, \psi \in P_F. Then, we have \phi(a) = f(a) and \psi(a) = g(a) for some polynomials f(x) and g(x) in F[x]. Now, we know that f(x) + g(x) is also a polynomial in F[x], so we have \phi(a) + \psi(a) = f(a) + g(a) = (f+g)(a). Therefore, \phi + \psi is also a polynomial function on F, and thus \phi + \psi \in P_F.

Next, we need to show closure under multiplication. Let \phi, \psi \in P_F. Then, we have \phi(a) = f(a) and \psi(a) = g(a) for some polynomials f(x) and g(x) in F[x]. Now, we know that f(x) * g(x) is also a polynomial in F[x], so we have \phi(a) * \psi(a) = f(a) * g(a) = (f*g)(a). Therefore, \phi * \psi is also a polynomial function on F, and thus \phi * \psi \in P_F.

Finally, we need to show that P_F is closed under additive and multiplicative inverses. Let \phi \in P_F, then we have \phi(a) = f(a) for some polynomial f(x) in F[x]. Since F[x] is a ring, we know that f(x) has an additive inverse (-f(x)) and a multiplicative inverse (f(x)^{-1}) in F[x]. Therefore, we have (-f(x))(a) = -f(a) and (f(x)^{-1})(a) = f(a)^{-1}. This means that \phi(a) + (-f(a)) = 0 and