# Are there any problems about the difference between set theoretic definitions of polynomials?

I am a novice about this question, so if there is a misunderstanding then I apologize for it.

As for Peano axioms, if I choose Zermelo natural numbers, and you choose von Neumann ones, then this doesn't cause any problems because we can prove Peano axioms are categorical i.e. it is guaranteed my model, and your one are isomorphic, using second order logic.

But how is the case of polynomials? There are a few definitions of polynomials. One is to define $$a_n X^n + \cdots + a_0$$ as a function $$S \rightarrow R$$ which are tantamount to $$0$$ except finite number of elements of $$S$$, where let $$S$$ be the subset of an infinite cyclic group generated by an element, say $$X$$, consisting of powers $$X$$(Serge Lang's Algebra on page 97). Another one is to define one as an infinite sequence $$(a_0, \dots, a_n, 0, \dots)$$(Serge Lang's Undergraduate Algebra on page 106). Don't these different definitions lead to problems? I appreciate for answers in advance.

• Is $R$ equal to $\Bbb R$? Apr 17 at 12:08
• For the same reason that we don't particularly care about whether we are talking about the natural number $1$ vs the integer $1$ or real $1$ or quaternion $1$ and so on... we freely switch between these on a whim based on convenience... so too can we freely switch between explicit ways of defining polynomials. So long as it acts like a polynomial, behaves like a polynomial, and sounds like a polynomial... then it may as well be a polynomial to us. In certain contexts one interpretation / explicit definition may be more useful to us than another competing definition. That's fine of course. Apr 17 at 12:09
• @JoséCarlosSantos I'm sorry to abbreviate. It is necessary to define polynomials to take a commutative ring and $R$ is it.(In Lang's book he uses $A$ rather than $R$.) Apr 17 at 12:19
• @JMoravitz I think I will go off on a tangent, but I feel hazy as for it as well. Of course, the integer $1$ is the equivalence class $\{(1, 0), (2, 1), \dots \}$ so this is not equal to the natural number $1$. I can't explain why this is justified. Apr 17 at 12:26
• Because it is useful. Remember... the problems we wish to solve come first. The formalization and definitions come after. Apr 17 at 12:27

Let's just look at the case of a field. There are, as you say, two possible ways to define polynomials over a field $$K$$: you can define a polynomial $$f : K \to K$$ to be a formal expression: $$f(x) = a_0 + a_1x + a_2x^2 + \ldots + a_nx_n$$ for some $$n$$ and constants $$a_1, a_2, \ldots, a_n \in K$$; or you can define it to be the function $$K \to K$$ represented by such a formal expression. These definitions are equivalent if $$K$$ is infinite but not if $$K$$ is finite.
I should explain why it makes a difference whether $$K$$ is finite. If $$K = \Bbb{Z}_2$$, the field with two elements, the the polynomial functions $$g(x) = x^2+1$$ and $$h(x) = x^4+1$$ take on the same values for every $$x \in K$$: $$g(0) = 1 = h(0)$$ and $$g(1) = 0 = h(1)$$. In an infinite $$K$$ this would be impossible: the quartic polynomial $$h(x) - g(x) =x^4 - x^2$$ can have at most $$4$$ roots, so it must be non-zero for some $$x$$ in $$K$$.
• I'm not sure if I can explain well, but let me do it. The definition of natural numbers exists in multiple as well as polynomials: like the von Neumann numbers $0 = \{\}, 1 = \{\{\}\}, 2 = \{\{\}, \{\{\}\}\}, \dots$ and Zermelo numbers $0 = \{\}, 1 = \{\{\}\}, 2 = \{\{\{\}\}\}, \dots$. Of course, the two $2$ is not equal to each other. Then, if I choose Zermelo approach, and you select von Neumann one, is it OK? The answer is yes: Apr 20 at 22:56