Proof re: fields of char 0 and Q

calvino

I need to prove that if K is a field of characteristic (I'll call is "char") 0, then k contains a subfield isomorphic to Q (rationals).

The only way I can think of, is like so (it doesn't seem "good enough").

Let K be a field with char 0. By a certain theorm (in the text), any Integral domain with char 0 contains a subring isomorphic to Z (the integers).

Thus K must also contain a field isomorphic to F_Z (the quotient field of Z)= Q. //



[The last line is a result from another theorem which states that "if K is any field containing an Integral domain isomorphic to D, then K contains a field isomorphic to F_D" - i) Do I need to "show" that the subring isomorphic to Z is an integral domain? ii) Have i lost rigor anywhere?]

thanks

Hurkyl

The property of being an integral domain is preserved by isomorphism, is it not?


It looks fine. The steps of your proof are:

(1) K is an integral domain with char 0.
(2) Thus, K has a subring S isomorphic to Z.
(3) Thus, K has a subfield isomorphic to Q.


But I don't think the extra rigor wouldn't hurt at all. Sometimes stating the obvious makes things clearer for those who aren't yet experts. (which does include me, of course)

In other words, it wouldn't hurt to say that S is an integral domain, and why.

calvino

thanks a lot. I know what needs to be done now.

Hurkyl

I meant to also say that I wouldn't think it necessary either, unless you really did want to be very rigorous.



Anyways, this is a useful technique, I think. When you think you have a proof, but feel unsure about it, it helps to try and rewrite what you've done... or even just describe what you've done.

Sometimes, I find that I can make exactly the same proof seem much clearer the second time I write it.

matt grime

The way you quote those theorems you use it is almost as if you don't understand them.

Let F be any field, it contains 1. It also contains n:=1 added up n times. Characerstic zero tells you these must all be different. As it's a field it must contain the additive inverse to n, call it -n, thus it contains Z. Now, as it's still a field it must contain 1/n for all n, and hence m/n for all m,n, ie ti contains Q. The point is that for any characteristic zero field there is a map from Q to that field that must send 1 to 1 and this determines where all of Q goes. It must be an isomorphism onto the image.

calvino

It's true. I didn't understand the theorems completely when I reasoned out the proof. I decided to formulate the proof, and then try to better understand it. It didn't work. You're explanation helped a great deal, however (no lie). Thank you. [note: One of my main problems was that i didnt understand the definition of characteristic. In my text book, and I quote " iF there is a positive integer n such that na=0 for each a in R (R a ring), then the least such integer is called the characteristic of R. If there is no such integer, then it has characteristic 0". I guess I didn't try hard enough to understand it well. ]