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Polynomials are things of the form:

$$ \sum_{i\geq0} a_i {X}^i= a_0+a_1X^1+\dots $$

Where only finite $a_i$'s are non-zero.

My question is, what kind of object is $X$?

We call it an indeterminate, but how do we define it in terms of a formal system? Can $\bf ZFC$ define indeterminates?

E: I'm not asking for a general definition of a polynomial, but that one of an indeterminate, guys.

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    $\begingroup$ $X$ is a transcendental over the ring that your polynomial takes its coefficients from, simple as that. This insures that two polynomials are unique exactly when they don't have the same coefficients. $\endgroup$
    – Set
    Commented Oct 24, 2015 at 21:07
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    $\begingroup$ Given that you can write polynomials out as finite strings, I wouldn't be too concerned about whether ZFC can handle it. $\endgroup$ Commented Oct 24, 2015 at 21:08
  • $\begingroup$ Do you know how to define infinite sequences in ZFC? The sequence $(a,b,c,\dots)$, I believe, is represented by the set $\{(0,a),(1,b),(2,c),\dots\}$, where the ordered pair $(x,y)$ is short for $\{\{x\},\{x,y\}\}$. (Of course, this isn't the only possible way.). EDIT: $26\approx\infty$, I guess $\endgroup$ Commented Oct 25, 2015 at 0:37

5 Answers 5

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If we’re talking about polynomials over a ring $R$ as formal objects, it’s a notational placeholder. We could just as well define a polynomial over $R$ to be a sequence $\langle a_n:n\in\Bbb N\rangle$ of elements of $R$ such that $\{n\in\Bbb N:a_n\ne 0_R\}$ is finite. (In fact, I’ve taught from a textbook in which this is exactly how polynomials are introduced.)

The advantages of the notation with an indeterminate are first that it’s familiar from every student’s early experience with basic algebra, and secondly that it makes it easy to relate the polynomial as formal object to the polynomial as function from $R$ to $R$.

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The definition I've seen is that a polynomial with coefficients in a ring $ R $ is an infinite sequence $(a_0, a_1,\ldots) $ of elements of $ R $, such that eventually all the remaining terms of the sequence are $0$. And $ X $ is the polynomial $(0,1,0,\ldots) $.

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There is no $X$, actually. ;-) A polynomial is just the sequence of its coefficients.

More formally, let $A$ be a ring (with unity) and consider the set of sequences $n\mapsto a_n$ of elements of $A$ that are eventually zero, which means $a_n=0$ for all $n>\bar{n}$ (where $\bar{n}$ depends on the sequence).

Sum sequences termwise and multiply them according to the rule $$ (a_n)(b_n)=(c_n),\qquad c_n=\sum_{k=0}^n a_kb_{n-k} $$ This is a ring, denoted by $A[X]$ and the mapping $A\to A[X]$ defined by $a\mapsto \varphi(a)=(a,0,0,\dotsc)$ (the sequence whose $0$-th term is $a$ and all others are zero) is a ring monomorphism.

Consider now $X=(0,1,0,0,\dotsc)$, the sequence whose first term is $1$ and all others are zero. Then it's just a question of trivial verifications showing that if $f=(a_0,a_1,\dots,a_n,0,0,\dotsc)\in A[X]$, then $$ f=\varphi(a_0)+\varphi(a_1)X+\dots+\varphi(a_n)X^n $$ The last step is identifying $a$ with $\varphi(a)$ (this is a standard method) so that we can write $$ f=a_0+a_1X+\dots+a_nX^n $$


This is just one of the possible definitions of $A[X]$. The important thing is that the ring $A[X]$ is an overring of $A$ generated by an element $X$ and that it satisfies the following property:

for every ring homomorphism $\gamma\colon A\to B$ and every $b\in B$, there exists one (and only one) ring homomorphism $\gamma_b\colon A[X]\to B$ such that

  1. for all $a\in A$, $\gamma_b(a)=\gamma(a)$
  2. $\gamma_b(X)=b$
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In ZFC it's difficult to define If you don't know anything about Galois theory you can see that you have a ring of polynomial and every element in this ring is like that. Also X is one element of that ring. A ring is obviously a set with some properties that you can formalize logically. I think this is the better way to formalize this types of object. If you know something about Galois theory you can identify transcendetal extensions of degree 1 with the polynomial space in one indeterminate (note that we are assuming that only a finite number of coeficients are non-zero so we have a polinomial).

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I think every answer given here miss the point. It goes this way for a ring $R$ :

  1. Start by defining the set of polynomials $Pol_R$
  2. Define addition and multiplication of polynomials.
  3. Define $X$ to be a polynomial so that you have the following :

$$\forall P \in Pol_R, \exists a_0, \cdots, a_n \in R^n, P = \sum a_i X^i$$

(Note that for this sum to make sense, you must went through points 1. and 2.)

For example using the definition of polynomials as sequences of elements of $R$ with a finite number of non zero elements. One can define $$X = (0,1, 0, 0, \cdots)$$

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