Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can you give me an example of infinite field of characteristic $p\neq0$?


share|cite|improve this question
up vote 101 down vote accepted

One very important example of an infinite field of characteristic $p$ is $$\mathbb{F}_p(T)=\left\{\,\frac{f}{g}\,\Bigg|\,\,\,f,g\in\mathbb{F}_p[T], g\neq0\right\},$$

the rational functions in the indeterminate $T$ with coefficients in $\mathbb{F}_p$ (the symbol $\mathbb{F}_p$ is just a synonym for $\mathbb{Z}/p\mathbb{Z}$). In other words, these are ratios of polynomials in $\mathbb{F}_p[T]$; this is the same construction as the one we use to make $\mathbb{Q}$ from $\mathbb{Z}$. The field $\mathbb{F}_p(T)$ is infinite because, for example, it contains $1$, $T$, $T^2$, $\ldots$, and it is of characteristic $p$ because it contains $\mathbb{F}_p$ (alternatively, because the kernel of the unique ring homomorphism $\mathbb{Z}\to\mathbb{F}_p(T)$ is $p\mathbb{Z}$.)

Another important example is $\overline{\mathbb{F}_p}$, the algebraic closure of the finite field $\mathbb{F}_p$. 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 $\mathbb{F}_p(T)$, $\overline{\mathbb{F}_p}$ a subfield of the other?
  • Is $\mathbb{F}_p(T)$ 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., $L\supseteq K$), we say that $\alpha\in L$ is an algebraic element over $K$ (or, for short, just that "$\alpha$ is algebraic over $K$") when there exists some (non-zero) $f\in K[x]$ such that $f(\alpha)=0$. If $\alpha\in L$ is not algebraic over $K$, we say it is transcendental over $K$. Wikipedia presents the following (standard) examples:

    • $\sqrt{2}\in\mathbb{R}$ is algebraic over $\mathbb{Q}$, because there is a non-zero polynomial $f\in\mathbb{Q}[x]$ (i.e., $f$ has rational coefficients) such that $f(\sqrt{2})=0$; we could take $f=x^2-2$, or $f=3x^3-6x$, or any other polynomial in $\mathbb{Q}[x]$ with $\sqrt{2}$ as a root.
    • $\pi\in\mathbb{R}$ is transcendental over $\mathbb{Q}$, because there is no non-zero polynomial in $\mathbb{Q}[x]$ with $\pi$ as a root; in other words, $\pi$ satisfies no algebraic relation with the rational numbers.
    • However, $\pi$ is algebraic over $\mathbb{R}$, because there are many easy examples of non-zero polynomials in $\mathbb{R}[x]$ with $\pi$ as a root - first and foremost, $x-\pi$. (Given any field $K$, any $\alpha\in K$ is algebraic over $K$ because $x-\alpha$ is a polynomial in $K[x]$ having $\alpha$ as a root. This demonstrates the importance of specifying algebraic (or transcendental) over what field.)
  • The setup of two fields $K$ and $L$, with $L\supseteq K$, 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 $\alpha\in L$ is algebraic over $K$.

  • A field $K$ is algebraically closed when any non-constant $f\in K[x]$ (i.e., $f$ is a polynomial of degree $\geq 1$) has a root in $K$. The requirement that the root is in $K$ is the key property. For example, any non-constant polynomial $f\in\mathbb{R}[x]$ has a root, but some of them (e.g. $x^2+1$) don't have any roots that are actually in $\mathbb{R}$; thus $\mathbb{R}$ is not algebraically closed. (The fact that $\mathbb{C}$ 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 $L=\overline{K}$, or sometimes $L=K^{\text{alg}}$.

Now we have the concepts necessary to compare and contrast $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$.

First, note that $T\in\mathbb{F}_p(T)$ is transcendental over $\mathbb{F}_p$ - there's no non-zero $f\in\mathbb{F}_p[x]$ 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 $\mathbb{F}_p$, we have only added it in as a formal symbol, so the only way we can get $$a_nT^n+\cdots+a_1T+a_0=0\text{ for }a_i\in\mathbb{F}_p$$ is if every $a_i=0$, so there is no non-zero polynomial with coefficients in $\mathbb{F}_p$ having $T$ as a root. In fact, any element of $\mathbb{F}_p(T)$ that isn't itself an element of $\mathbb{F}_p$ (i.e., anything having a $T$ in it) is transcendental over $\mathbb{F}_p$, by the same argument - so the extension $\mathbb{F}_p(T)/\mathbb{F}_p$ is really super-duper non-algebraic.

However, every element of $\overline{\mathbb{F}_p}$ is algebraic over $\mathbb{F}_p$, because part of the definition of algebraic closure includes that the extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$ is algebraic.

So, if we had $\mathbb{F}_p(T)\subseteq\overline{\mathbb{F}_p}$, then we'd have transcendental elements inside our algebraic extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$, which is a contradiction.

On the other hand, if we had $\overline{\mathbb{F}_p}\subseteq\mathbb{F}_p(T)$, then we would have that there were some $\frac{f}{g}\in \mathbb{F}_p(T)$ such that $\frac{f}{g}\notin\mathbb{F}_p$ and $\frac{f}{g}\in\overline{\mathbb{F}_p}$ (because $\overline{\mathbb{F}_p}$ is infinite and $\mathbb{F}_p$ is finite), and they would have to be algebraic over $\mathbb{F}_p$, which we've also seen is a contradiction.

Thus, neither $\mathbb{F}_p(T)\subseteq\overline{\mathbb{F}_p}$ nor $\overline{\mathbb{F}_p}\subseteq\mathbb{F}_p(T)$.

It is also easy to see that $\mathbb{F}_p(T)$ is not algebraically closed; for the sake of simplicity, let's set $K=\mathbb{F}_p(T)$. If $K$ were algebraically closed, then every non-constant $f\in K[x]$ would have to have a root in $K$; but there are many such $f$'s that do not, for example $f=x^2-T$ (remember, $T\in K$, so don't be thrown by the existence of two indeterminates; this is just like $x^2-2$ in $\mathbb{Q}[x]$). If the polynomial $x^2-T$ had a root $\frac{f}{g}\in K=\mathbb{F}_p(T)$, then $$\left(\frac{f}{g}\right)^2-T=0$$ $$f^2=Tg^2$$ $$2\cdot\deg(f)=\deg(f^2)=\deg(Tg^2)=1+2\cdot\deg(g)$$ which is a contradiction (can't have even = odd). This mirrors the usual proof that there is no $\frac{a}{b}\in\mathbb{Q}$ such that $\left(\frac{a}{b}\right)^2=2$, i.e. that $\sqrt{2}$ is irrational; the $\deg$ function is analogous to the 2-adic order function. (As Pierre-Yves 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 $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$? Other than the fact that they are both extensions of $\mathbb{F}_p$, not much. The extension $\mathbb{F}_p(T)/\mathbb{F}_p$ is not algebraic, while the extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$ is; in fact the only elements $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$ have in common are $\mathbb{F}_p$ itself. They are both extremely important in algebra, number theory, algebraic geometry, and they both fundamental examples of characteristic $p$ fields.

share|cite|improve this answer
What is the relationship between these two fields? Is one a subfield of the other? Is the first one algebraically closed? – Michael Hardy Aug 19 '11 at 17:27
@Michael: I got carried away answering your questions :) – Zev Chonoles Aug 20 '11 at 7:35
@Zev: You can use \quad as a long space; and \Bigg| as a long pipe. No need for excessive TeX code :) – Asaf Karagila Aug 20 '11 at 8:00
@Asaf: Thanks for the tips :) I'd forgotten about \Bigg, although the spacing was intentional; sometimes I want to fine tune the spacing and \quad's and \qquad's are too big (but after reconsidering, it looks fine with a single \, on each side, anyway). – Zev Chonoles Aug 20 '11 at 8:09
Dear Zev: Your argument shows that K(T) is not algebraically closed for any field $K$, doesn't it? Wonderful answer! +1 – Pierre-Yves Gaillard Aug 20 '11 at 8:25

Another construction, using a tool from the formal logic: the ultraproduct.

The cartesian product of fields $$P = {\Bbb F}_p\times{\Bbb F}_{p^2}\times{\Bbb F}_{p^3}\times\cdots$$ isn't a field ("isn't a model of..." ) because has zero divisors: $$(0,1,0,1,\cdots)(1,0,1,0\cdots)=(0,0,0,0,\cdots).$$ The solution is taking a quotient: let be $\mathcal U$ a nonprincipal ultrafilter on $\Bbb N$. Define $$(a_1,a_2,\cdots)\sim(b_1,b_2,\cdots)$$ when $$\{n\in\Bbb N\,\vert\, a_n=b_n\}\in\mathcal U.$$ The quotient $F=P/\sim$ will be a infinite field of characteristic $p$.

share|cite|improve this answer
Maybe I'm missing something, but won't $\mathbb{F}_p^\mathbb{N} / \mathcal{U}$ be isomorphic to $\mathbb{F}_p$, because every element of $\mathbb{F}_p^\mathbb{N}$ is constant on a set in $\mathcal{U}$? – Trevor Wilson Apr 23 '15 at 19:01
@TrevorWilson, I see the problem (is obvious for $p=2$). I will think if the example is repairable. – Martín-Blas Pérez Pinilla Apr 24 '15 at 7:12
@TrevorWilson, the solution is taking the product of all the $F_{p^n}$. I will edit the answer later. – Martín-Blas Pérez Pinilla Apr 24 '15 at 7:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.