Suppose that is a field and that It may be that is also a field, using the same operations of For example, if then we could have

Definition 15. If is a field and we say that is a subfield of if is a field with the operations of

Let’s examine this definition in some detail. Part of what this is saying is that

If then also i.e., is closed under addition.

If then also i.e., is closed under multiplication.

However, this is not enough. For example, is not a field but it is closed under the addition and multiplication operations of The problem with is that it does not have additive or multiplicative inverses of its elements.

Proposition 16. Suppose that is a field and that

If then

If and then

Proof. Add the additive inverse to both sides of the first equation, and multiply by the multiplicative inverse both sides of the second equation.

The point of Proposition 16 is the following: Suppose that is a subfield of Write for the -th element of and for the -th element of Then in particular, must belong to Similarly, so belongs to as long as contains some element other than But, of course, if is to be a field, then it must have at least two elements, so one of them must be different from

Proposition 17. Suppose is a field and that

If then

If then and

Proposition 17 can be proved by a very similar argument to that of Proposition 16, so I omit the proof. The point of this proposition is that if is a subfield of and then the additive inverse of from the point of view of and its additive inverse from the point of view of must coincide. Similarly, the multiplicative inverse from the point of view of of any nonzero element of is the same as its multiplicative inverse from the point of view of Hence, to properties 1,2 listed above we can add:

3. If then

And:

4. If and then

It turns out that 1–4 characterize subfields:

Theorem 18. Suppose is a field and If satisfies 1–4 and has at least two elements, then is a subfield of

Noticed that we cannot remove the assumption that has two elements. For example, satisfies properties 1–4 but is not a field.

We will prove this theorem next lecture and use it to produce many new examples of fields.

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Friday, February 20th, 2009 at 1:11 pm and is filed under 305: Abstract Algebra I. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

The description below comes from József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038). Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternat […]

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

It is easy to see without choice that if there is a surjection from $A$ onto $B$, then there is an injection from ${\mathcal P}(B)$ into ${\mathcal P}(A)$, and the result follows from Cantor's theorem that $B

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that $$ 1+3+\ldots+(2n-1)=n^2, $$ which can be easily […]

First of all, $f(z)+e^z\ne 0$ by the first inequality. It follows that $e^z/(f(z)+e^z)$ is entire, and bounded above. You should be able to conclude from that.

Yes. The standard way of defining these sequences goes by assigning in an explicit fashion to each limit ordinal $\alpha$, for as long as possible, an increasing sequence $\alpha_n$ that converges to $\alpha$. Once this is done, we can define $f_\alpha$ by diagonalizing, so $f_\alpha(n)=f_{\alpha_n}(n)$ for all $n$. Of course there are many possible choices […]

I disagree with the advice of sending a paper to a journal before searching the relevant literature. It is almost guaranteed that a paper on the fundamental theorem of algebra (a very classical and well-studied topic) will be rejected if you do not include mention on previous proofs, and comparisons, explaining how your proof differs from them, etc. It is no […]

[…] -Fields (5) At the end of last lecture we stated a theorem giving an easy characterization of subfields of a given field We begin by […]