Can you give me an example of infinite field of characteristic p≠0?
Thanks.
Answer
One very important example of an infinite field of characteristic p is
Fp(T)={fgf,g∈Fp[T],g≠0},
the rational functions in the indeterminate T with coefficients in Fp (the symbol Fp is just a synonym for Z/pZ). In other words, these are ratios of polynomials in Fp[T]; this is the same construction as the one we use to make Q from Z. The field Fp(T) is infinite because, for example, it contains 1, T, T2, …, and it is of characteristic p because it contains Fp (alternatively, because the kernel of the unique ring homomorphism Z→Fp(T) is pZ.)
Another important example is ¯Fp, the algebraic closure of the finite field Fp. If you accept, for the moment, that every field has an algebraic closure (which is certainly not an obvious statement), then the fact that there are no finite algebraically closed fields means that the algebraic closure of a field of characteristic p will have to be an infinite field of characteristic p.
Michael Hardy raises some good questions below.
 Is one of Fp(T), ¯Fp a subfield of the other?
 Is Fp(T) algebraically closed?
 What is the relationship between these two fields?
These questions are all related. First, we should look at the definitions of algebraic element, algebraic extension, algebraically closed field, and algebraic closure.

Given a field K, and another field L containing K (i.e., L⊇K), we say that α∈L is an algebraic element over K (or, for short, just that “α is algebraic over K“) when there exists some (nonzero) f∈K[x] such that f(α)=0. If α∈L is not algebraic over K, we say it is transcendental over K. Wikipedia presents the following (standard) examples:
 √2∈R is algebraic over Q, because there is a nonzero polynomial f∈Q[x] (i.e., f has rational coefficients) such that f(√2)=0; we could take f=x2−2, or f=3x3−6x, or any other polynomial in Q[x] with √2 as a root.
 π∈R is transcendental over Q, because there is no nonzero polynomial in Q[x] with π as a root; in other words, π satisfies no algebraic relation with the rational numbers.
 However, π is algebraic over R, because there are many easy examples of nonzero polynomials in R[x] with π as a root – first and foremost, x−π. (Given any field K, any α∈K is algebraic over K because x−α is a polynomial in K[x] having α as a root. This demonstrates the importance of specifying algebraic (or transcendental) over what field.)

The setup of two fields K and L, with L⊇K, is referred to as a field extension. We refer to the extension as a single entity by the expression L/K (this is not a quotient like this or this, though). An extension L/K is an algebraic extension when every α∈L is algebraic over K.

A field K is algebraically closed when any nonconstant f∈K[x] (i.e., f is a polynomial of degree ≥1) has a root in K.
The requirement that the root is in K is the key property. For example, any nonconstant polynomial f∈R[x] has a root, but some of them (e.g. x2+1) don’t have any roots that are actually in R; thus R is not algebraically closed. (The fact that C is algebraically closed is often referred to as the Fundamental Theorem of Algebra). 
Given any field K, there exists an algebraic extension L/K such that L is algebraically closed; such an L is called an algebraic closure of K. It is unique up to isomorphism (so we often talk about “the” algebraic closure of K), and we write L=¯K, or sometimes L=Kalg.
Now we have the concepts necessary to compare and contrast Fp(T) and ¯Fp.
First, note that T∈Fp(T) is transcendental over Fp – there’s no nonzero f∈Fp[x] such that f(T)=0. This is really what we originally meant when we said T is an “indeterminate” – it stands in no relation to Fp, we have only added it in as a formal symbol, so the only way we can get
anTn+⋯+a1T+a0=0 for ai∈Fp
is if every ai=0, so there is no nonzero polynomial with coefficients in Fp having T as a root. In fact, any element of Fp(T) that isn’t itself an element of Fp (i.e., anything having a T in it) is transcendental over Fp, by the same argument – so the extension Fp(T)/Fp is really superduper nonalgebraic.
However, every element of ¯Fp is algebraic over Fp, because part of the definition of algebraic closure includes that the extension ¯Fp/Fp is algebraic.
So, if we had Fp(T)⊆¯Fp, then we’d have transcendental elements inside our algebraic extension ¯Fp/Fp, which is a contradiction.
On the other hand, if we had ¯Fp⊆Fp(T), then we would have that there were some fg∈Fp(T) such that fg∉Fp and fg∈¯Fp (because ¯Fp is infinite and Fp is finite), and they would have to be algebraic over Fp, which we’ve also seen is a contradiction.
Thus, neither Fp(T)⊆¯Fp nor ¯Fp⊆Fp(T).
It is also easy to see that Fp(T) is not algebraically closed; for the sake of simplicity, let’s set K=Fp(T). If K were algebraically closed, then every nonconstant f∈K[x] would have to have a root in K; but there are many such f‘s that do not, for example f=x2−T (remember, T∈K, so don’t be thrown by the existence of two indeterminates; this is just like x2−2 in Q[x]). If the polynomial x2−T had a root fg∈K=Fp(T), then
(fg)2−T=0
f2=Tg2
2⋅deg(f)=deg(f2)=deg(Tg2)=1+2⋅deg(g)
which is a contradiction (can’t have even = odd). This mirrors the usual proof that there is no ab∈Q such that (ab)2=2, i.e. that √2 is irrational; the deg function is analogous to the 2adic order function. (As PierreYves Gaillard points out in the comments, this actually shows that K(T) is not algebraically closed, for any field K.)
So, in summary, what is the relationship between the fields Fp(T) and ¯Fp? Other than the fact that they are both extensions of Fp, not much. The extension Fp(T)/Fp is not algebraic, while the extension ¯Fp/Fp is; in fact the only elements Fp(T) and ¯Fp have in common are Fp itself. They are both extremely important in algebra, number theory, algebraic geometry, and they both fundamental examples of characteristic p fields.
Attribution
Source : Link , Question Author : Aspirin , Answer Author : Community