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I've been asked to prove that given $f$ and $g$ are polynomials in $R[x]$, where $R$ is a commutative ring with identity,

$(f+g)(k) = f(k) + g(k)$, and $(f \cdot g)(k) = f(k) \cdot g(k)$. However, I always took these things as definition. What exactly is there to prove? How exactly are $f+g$ and $f \cdot g$ defined here?

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up vote 3 down vote accepted

For general rings there is a huge difference between a polynomial in $R[X]$ and the function it associates. In more detail, a polynomial $f\in R[X]$ is a formal expression of the form $f(X)= \sum_{k=0}^na_kX^k$ where the coefficients $a_0,\cdots ,a_n$ come from the ring $R$. With such a polynomials $f(X)$ one can associate a function $f(x):R\to R$ by substitution: $f(t)=\sum_{k=0}^na_kt^k$.

When the ring $R$ is the ring of real numbers there is little difference between the polynomial and the function. If two polynomials $f(X),g(X)$ have associated functions that are equal, $f(x)=g(x)$, then the polynomials are identical (in the sense that their coefficients are identical).

However, consider the polynomials $f(X)=0$ and $f(X)=X^2+X$ considered as polynomials in $\mathbb Z_2[X]$. The associated functions are identical (both are constantly $0$) but the polynomials are different.

Now, polynomials (as formal entities) can be added and multiplied (whatever definition of the polynomial ring you use will have such a definition). The equations you wrote in your question are functional equations. You are asked to show that the combinatorial definition of addition and multiplication of polynomials agree with the functional definitions of addition and multiplication of the associated functions.

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Is there a one-to-one correspondence in the association of polynomials to functions? – chubbycantorset Jan 11 '13 at 4:00
above I gave two distinct polynomials with identical associated functions, so the answer is no. – Ittay Weiss Jan 11 '13 at 4:01
ok, I now understand the question and know exactly what to do! Thanks! – chubbycantorset Jan 11 '13 at 4:08

One construction (I do not like using constructions as definitions) of $R[x]$ is as the set of formal sums $\sum r_i x^i$ of the symbols $1, x, x^2, x^3, ...$ with coefficients in $R$. Addition here is defined pointwise and multiplication is defined using the Cauchy product rule. For every $k \in R$ there is an evaluation map $\text{eval}_k : R[x] \to R$ which sends $\sum r_i x^i$ to $\sum r_i k^i$, and the question is asking you to show that this map is a ring homomorphism.

To emphasize that this doesn't follow immediately from the construction, note that this statement is false if $R$ is noncommutative.

The reason I don't want to use the word "definition" above is that "polynomial ring" should more or less mean "the ring such that the above thing is true, and if it isn't, we messed up and need a different construction."

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If I could upvote this answer twice, I would. Just for your commitment to rigor! Thanks. This helped me understand what's going on in a much better manner. – chubbycantorset Jan 11 '13 at 4:10

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