What is an indeterminate in a polynomial ring? I am currently studying polynomial ring. I have a basic doubt. What is $x$ in a polynomial?
A polynomial is an expression $\sum_{k = 0}^n a_k x^k$, where $n,k \in \mathbb{N}$ and $a_k \in R$ where $R$ is a ring. Here $x$ is considered as an indeterminate. Can you please explain what an indeterminate is with some examples? What is its definitions and properties ? 
I am studying the book of Dummit and Foote. I am not getting the explanation there. Similarly Herstein has kept silent also. In the chapter Polynomial Ring of the book Linear Algebra of Hoffman Kunze it is explained a little. $x$ is considered as a vector $( 0, 1, 0, \dots, 0$ $(n$ times $) ) $ and so not an element of the ring or the field on which the polynomial ring is constructed. $x^k$ can also be expressed similarly as a vector having $1$ at the $k$-th position and all other terms are $0$. 
In analysis when we consider polynomial function we just use $x$ as a dependent variable may be real or a complex. It may be different in Algebra. Please explain the difference and similarity of the two concepts.
Thanking you very much. 
 A: There are two differents aspects of polynoms : in algebra, polynoms can be defined with sequences $a_0,a_1,...$ such as $a_n=0$ below a given rank. With this definition, you can define laws on your sequences in order to have an algebra. That is in this case that $x$ is a "vector" that represents the polynom $0 + X + 0\cdot X^2 + ...$ so this is the vector $0,1,0,0,...$.
In analysis, polynoms are only usuals fonctions, that have a given kind of formula.In this case the indeterminate is only the variable of the function.
 The similarity of the concepts is that if $R$ is principal and infinite (i don't know if it is optimal),the application from $R[X]$ to $R[x]$ (the second set is seen as a subset of $\mathcal F(R,R)$) is an isomorphism between algebras (and rings a fortiori).
Indeed, this application is obviously a surjective morphism. It is injective because the kernel is trivial. In order to prove it, you need to use ideals and show that you can factorize your polynom an infinity of times if he is in the kernel.
A: An indeterminate $x$ over $R$ is simply an element that is trascendental over $R$, that is, such that
$$
a_{0} + a_{1} x + \dots + a_{n} x^{n} = 0
$$
only when $a_{i} = 0$ for all $i$.
The construction with sequences you are mentioning is a safe method to construct an indeterminate over any (commutative) ring.
You are surely aware that over a finite field there are pairs of distinct polynomials that induce the same function. For instance, if $F$ is a field with $q$ elements, the polynomial $x^{q} - x$ evaluates to zero for all elements of $F$, exactly like the zero polynomial. That's why the approach via functions is not adequate in these situations.
