Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $F$ is a field of characteristic $p$ prime, how can one create a field $K$ such that $K$ is created from $F$ (either by modding out or by taking a product which includes $F$ or by some other method which involves $F$) such that $K$ has a different characteristic $j \ne p$? Also, does this question depend on $p$? I would also like to consider creations which are field extensions of $F$. Also, how many of each kind of extension are possible?

share|improve this question
5  
How about Witt vectors? –  Tom Goodwillie Mar 14 '12 at 0:43
2  
Since you allow products of fields, consider $A= \prod_{p} {\mathbf Z}/p{\mathbf Z}$. This ring has characteristic 0: for no positive integer $n$ does $n=0$ in A. Let I be the ideal of elements of A which have finitely many nonzero coordinates. By Zorn's lemma, I is contained in a maximal ideal M of A. What's the characteristic of the field A/M? If it were a prime $p$ then $p=0$ in A/M, so $p$ is in M. But as a (diagonal) sequence in A, $p$ has all but one nonzero coordinate. Use this and the fact that I is in M to show 1 is in M, which is a contradiction. So char(A/M) = 0. –  KConrad Mar 14 '12 at 1:19
7  
You really should read about Witt vectors, as Tom suggests. Although the construction in my previous comment leads to a (nonconstructive) field of char. 0 out of infinitely many fields of different positive characteristics, Witt vectors are a mathematically more significant way to create fields (well, domains) of characteristic 0 out of finite fields in a good (= functorial) manner. –  KConrad Mar 14 '12 at 1:21
2  
I think this question is quite vague. –  Martin Brandenburg Mar 14 '12 at 10:11
add comment

3 Answers

up vote 4 down vote accepted

It should be noted that the characteristic of a field is either prime or zero. If it is zero, then it contains the rational numbers. These are two statements you can probably prove even if algebra isn't your cup of tea.

You can study valuation rings of mixed characteristic. A classic example is $\mathbb{Z}_p$ the p-adic integers. This is a ring of characteristic 0 and its fraction field $\mathbb{Q}_p$ is a field of characteristic 0. However, $\mathbb{Z}_p$ has a (unique) maximal ideal generated by $(p)\mathbb{Z}_p$ such that $$\mathbb{F}_p \cong \mathbb{Z}_p/(p)\mathbb{Z}_p$$ which is a finite field of characteristic $p$.

There is the famous Ax-Kochen theorem, which is the following

$$\Pi_\mathcal{F} \mathbb{Q}\_p \cong \Pi_\mathcal{F} \mathbb{F}_p((t))$$

where $\mathcal{F}$ is a non-principal ultra-filter on $\mathbb{N}$. This result depends on the continuum hypothesis. (J. Ax and S. Kochen, Diophantine problems over local fields I, American Journal of Mathematics,87 (1965), 605–630.)

However, there is also an isomorphism

$$\Pi_\mathcal{F} \mathbb{Z}\_p \cong \Pi_\mathcal{F} \mathbb{F}_p[[t]]$$

where $\mathcal{F}$ is a non-principal ultrafilter on $\mathbb{N}$, which does not depend on the Continuum Hypothesis, in "Use of Ultrapoducts in Commutative Algebra" by Hans Schoutens, found here http://www.springer.com/mathematics/algebra/book/978-3-642-13367-1

Witt Vectors allow you to move from pure characteristic (either zero or positive) to mixed characteristic.

Another way of studying different characteristics is through Algebraic Geometry, by thinking of fields of different characteristics as fibers living over primes in $\mathbb{Z}$.

Edit: It occurred to me that an example here might be useful in order to illustrate this point. Given a curve $C$ over a field of characteristic p>0, we know (Winter's theorem) that there exists a discrete valuation ring (for example, $\mathbb{Z}_p$ of mixed characteristic), call it B, and a family of curves over B. Over the generic fiber, we have a GAGA principle, which means that things we can define over $\mathbb{C}$ (e.g., fundamental group), we can define over a field of $char(k)=0$. Now using Winter's theorem we can 'transfer' our definitions from analytic geometry to algebraic geometry over a field of positive characteristic, which gives us new tools when studying number theory.

Finally, there is even an "physical" way to interpret moving between fields. I highly recommend this article of A. Connes on the topic: "Characteristic one, entropy and the absolute point," Connes & Consani http://arxiv.org/abs/0911.3537

share|improve this answer
1  
I've fixed it, I hope I haven't destroyed any of the mathematical content. –  David Roberts Mar 14 '12 at 2:00
    
Thanks. No, you haven't. –  Andrew Stout Mar 14 '12 at 2:17
add comment

"By some other method" is quite vague. If, by modding out, you are thinking of taking a subring and then modding that ring by a maximal ideal (i.e., you obtain $\mathbb{Z}/p\mathbb{Z}$ from the rationals in this way), then it cannot be done.

Call the field $F$, the subring $R$, and the maximal ideal $I$, and assume $F$ has finite characteristic $p$. By definition of subring, the unit of the $R$ is also the unit of $F$. If the field has characteristic $p$, then $p \cdot 1 = 0$ is also true in the subring, and hence, in the quotient field $R/I$. Thus, the quotient field must also have characteristic $p$ if the original field does.

By definition, an extension field $K$ of $F$ would then contain $F$ as a subring, so again, they would have to have the same characteristic.

share|improve this answer
    
I see from your answer that you cannot change the characteristic of a field by an inner field, by modding out. –  Erin K Carmody Mar 14 '12 at 0:47
    
How many fields can one construct in this way? –  Erin K Carmody Mar 14 '12 at 1:10
add comment

Even if you are interested in fields $K$ that are extensions of $F$, you left the door open for something else.

If $K$ is an extension of $F$, then $K$ has the same characteristic $p$ as $F$, because $p$ is the number of terms in the shortest non-trivial identity $1+\cdots+1=0$.

But you can also start from $F=\mathbb Z/p\mathbb Z$ and build $\mathbb Q_p$, the field of $p$-adic numbers. This one is has characteristic zero. See Dwork's book.

share|improve this answer
add comment

Your Answer

 
discard

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.