# Is polynomial in general the same as polynomial function?

The algebra text book says, a polynomial in one variable over $$\mathbb{R}$$ is given by,

$$f(x)= a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1x + a_0$$

Where $$x$$ is an unknown quantity which commutes with real numbers, called "indeterminate".

So I have a few question,

1. Is polynomial always a function?if not then what is a polynomial in general?
2. And what's up with the "indeterminate" thingy? Is it wrong to simply call it a variable?
3. What exactly is the $$x$$ in the expression? A number? A matrix? Or...some other object? Why does it have to "commute" with real numbers?
• Can you be more specific in what you mean by "algebra"? At the college level this could be abstract algebra, college algebra, elementary algebra, etc. Can you cite the specific book in use? – Daniel R. Collins Nov 17 '18 at 14:10

No, a polynomial is not a function. However, for each polynomial $$p(x)=a_0+a_1x+\cdots+a_nx^n$$ you may consider the polynomial function$$\begin{array}{rccc}p\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&a_0+a_1x+\cdots+a_nx^n.\end{array}$$And distinct polynomials will be associated with distinct functions. However, although this is true over the reals, it doesn't hold in general. For instance, if you are working over the field $$\mathbb{F}_2$$, then the polynomial $$x^2-x$$ and th null polynomial are distinct polynmials, but the function$$\begin{array}{ccc}\mathbb{F}_2&\longrightarrow&\mathbb{F}_2\\x&\mapsto&x^2-x\end{array}$$is the null function.

So, a polynomial (over the reals) is an expression of the type $$a_0+a_1x+\cdots+a_nx^n$$, where $$x$$ is an entity about which all we assume is that it commutes with each real number. Usually, it is called an “indeterminate” since it is not a specific real number.

• Thank you for writing a simple answer, I don't know what fields or rings are, I have them in the next year. But for the time being, would you be kind enough to explain, or add to your answer, what does the phrase, "over the reals" mean? – William Nov 17 '18 at 8:49
• In this context, if I say that I am working over the reals (or over $\mathbb R$; you used this expression), what I mean is just that all my $a_i$'s are real numbers. I could alse be working over, say, the complex numbers. Or over the integers. – José Carlos Santos Nov 17 '18 at 8:51
• oh, I see, so the coefficients will be from the set of $\mathbb{R}$. Another quick question, in any function $f(x)$, we call the $x$ a variable, but the $x$ in polynomials are called indeterminates? Why is that? – William Nov 17 '18 at 9:00
• In functions the $x$ can take any value in the domain. In a polynomial, the $x$ is an indeterminate thing. It is not supposed to be replaced by a number. – José Carlos Santos Nov 17 '18 at 9:12
• @William : The polynomial as algebraic entity is just the finite sequence of coefficients. Writing it as a sum with powers in some $x$ is just another symbolic way to write the coefficient sequence, where the powers indicate the position in the sequence. Of course everything is arranged so that polynomial multiplication etc. works. – Dr. Lutz Lehmann Nov 17 '18 at 15:07
1. Is polynomial always a function?

No, never.

1. And what's up with the "indeterminate" thingy? Is it wrong to simply call it a variable?

Yes, it is wrong. Formally a polynomial is a sequence $$(a_0, a_1, \ldots)$$ such that $$a_i=0$$ eventually. It's a sequence of coefficients, it's not a function. Each coefficient is taken from a fixed ring, e.g. reals, complex numbers or finite fields. The underlying ring is also called a ring of scalars. With that we can define a specific polynomial addition, polynomial multiplication and scalar multiplication.

1. What exactly is the $$x$$ in the expression? A number? A matrix? Or...some other object? Why does it have to "commute" with real numbers?

$$x$$ is simply a special polynomial of the form $$(0,1,0,0,\ldots)$$. It is a theorem that every polynomial can be uniquely written as a finite sum $$\sum a_ix^i$$ with $$a_i$$ scalars.

It is important to distinguish polynomials from polynomial functions. For example if $$K$$ is a finite field, i.e. $$K=\{a_1,\ldots, a_m\}$$ then take polynomial $$f(X)=(X-a_1)\cdots(X-a_m)$$. For example if $$K=\mathbb{Z}_2=\{0,1\}$$ then that would produce polynomial $$(X-1)X=X^2-X=X^2+X$$ which formally is $$(0,1,1,0,0,\ldots)$$.

Clearly $$f(x)=0$$ for all $$x\in K$$ meaning $$f$$ is a constant $$0$$ as a function $$f:K\to K$$. But as a polynomial $$f(X)$$ is non-zero of positive degree $$m$$.

Also note that over infintie fields these two coincide. Indeed, there is a ring epimorphism $$K[X]\to K\{X\}$$ with polynomials on the left side and polynomial functions on the right. The kernel is non-trivial if and only if $$K$$ is finite. And in that case it is equal to the principal ideal $$\langle (X-a_1)\cdots(X-a_m)\rangle$$.

In particular there are always infinitely many polynomials. But there might be finitely many polynomial functions, depending on the underlying ring of scalars.

• Thanks for answer, but you know what is actually bothering me? The fact that we use the notation of a function i.e $f(x), g(x)$ etc. to represent polynomials.. would you happen to know why do we represent it like that? – William Nov 17 '18 at 11:05
• @William I think this is because of historical reasons. Originally polynomials were indeed understood as polynomial functions. Also it is useful when dealing with polynomials of multiple "variables". – freakish Nov 17 '18 at 11:07