# Maximal Extension Field

• Aug 17th 2006, 09:43 AM
ThePerfectHacker
Maximal Extension Field
Is the algebraic closure of a field the maximal extension field? It happens to be maximal algebraic extension field but is it also maximal extension field. Note: Such as field exists by partial ordering a set of extension fields and applying Zorn's lemma.
• Aug 17th 2006, 12:35 PM
rgep
Take any set I and adjoin a set of transcendentals x_i indexed by I. Given any such extension there is always a strictly larger one obtained by adjoining one more transcendental not in the set { x_i : i in I }.
• Aug 17th 2006, 12:42 PM
ThePerfectHacker
Quote:

Originally Posted by rgep
Take any set I and adjoin a set of transcendentals x_i indexed by I. Given any such extension there is always a strictly larger one obtained by adjoining one more transcendental not in the set { x_i : i in I }.

But how can that be!?! Zorn's lemma gaurentees the existstence of a maximal element.
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While we are on this subject I encountered another simpler problem.

Consider a group $\mathcal{G}$. Assume it have a maximal normal subgroup. Then create a set of all maximal normal subgroups $S$. Applying Zorn's lemma using ordering by inclusion we can show that there is such thing a maximal maximal normal subgroup. Meaning each group has a unique maximal normal subgroup. But that cannot be. For example, consider $\mathbb{Z}_15$ surly, $\mathbb{Z}_3 \mbox{ and }\mathbb{Z}_5$ are maximal normal subgroups and they are different!!!
• Aug 17th 2006, 01:05 PM
BubbleBrain_103
Hey Hacker,

I am pretty sure the answer to your question is no. C(x) for instance, the field of all rational functions with complex coefficients, is a proper extension of C (and R). Note that we had to adjoin the transcendental indeterminate "x" to get beyond C. That is, the only proper extensions of C are necessarily infinite dimensional over C, since finite dimensional extensions are algebraic, and C has no proper algebraic extensions. So I think you can say that the algebraic closure K of a field F is a maximal finite dimensional extension (so long as K/F is finite dimensional, which certainly isn't always true, e.g. C/Q)

This raises the question though... given any field F, can't we just adjoin an indeterminate "x" to yield a proper extension F(x)? Seems like we can, so I was trying to figure out where the reasoning using Zorn's Lemma breaks down. I'm no set theorist, but I think the flaw is in considering "the set of all extension fields of F", which is just a stone's throw from invoking "the set of all sets". In other words, I think that the collection of all extension fields of F is a proper class, not a set, and so Zorn's Lemma doesn't apply. That's my best guess anyhow. Maybe you could prove some things by just considering "the set of all extension fields of F with cardinality less than ____ " or something like that.
• Aug 17th 2006, 01:23 PM
ThePerfectHacker
Quote:

Originally Posted by BubbleBrain_103
Seems like we can, so I was trying to figure out where the reasoning using Zorn's Lemma breaks down.

Okay, given field F, define,
Let, $S=\{ F\leq E\}$ (we note S is non-empty trivially).
Define partial ordering on $S$ as $A\leq B \leftarrow \rightarrow A\subseteq B$
We can easily verify, that the relation $\leq$ on S is: reflexsive, transitive and anti-symettric. Now consider any chain $C\subseteq S$ define a set, $Z=\bigcup_{x\in C}x$. Clearly, $\forall X\in C\leq Z$. Next we turn Z into a field by using the same binary operations as we for the other fields. Thus, $Z\in S$. Therefore, every chain has an upper bound. Thus, there is a maximal element in $S$- "maximal extension field".
But according to you that set is not well-founded? Is that the right term to use? How about the confusion with my second problem that every group that has a maximal normal subgroup has a unique maximal normal subgroup. Same idea?

I find it interesting, my book, when it proved the existence of algebraic closure used Zorn's lemma in the same fashion as I did but for some reason it worked for them.
• Aug 17th 2006, 01:29 PM
BubbleBrain_103

In the conclusion part of Zorn's Lemma, there is no reason to take the maximal element in the poset to be unique. Maximal for a poset just means "there is nothing bigger than it", not necessarily that "everything is smaller than it". In fact every element in the poset of all maximal normal subgroups is maximal, since the chains will consist merely of single element subsets.
• Aug 17th 2006, 01:37 PM
ThePerfectHacker
Quote:

Originally Posted by BubbleBrain_103

In the conclusion part of Zorn's Lemma, there is no reason to take the maximal element in the poset to be unique. Maximal for a poset just means "there is nothing bigger than it", not necessarily that "everything is smaller than it". In fact every element in the poset of all maximal normal subgroups is maximal, since the chains will consist merely of single element subsets.

Ahh! I see my mistake. When I used Zorn's lemma and got a maximal maximal normal subgroup I assumed it meant that every Normal subgroup is a subset of it. But that was the mistake. Because the initial ordering was $\leq$ not $\subseteq$. Stupid me... I hate it when you such a mistake in algebra. If you are really curious I will show you the stupidest mistake I ever made.

(I think the reason why I arrived at a faulty conclusion is because I was using a famous theorem that every ideal is contained in some maximal ideal. But the problem is that that set is ordered by the sub set relation).
• Aug 17th 2006, 01:44 PM
BubbleBrain_103
Your use of Zorn's Lemma for fields is absolutely correct as far as I can tell, assuming that the set S exists, which is not guaranteed the way you've defined it. You wrote:

S = {F <= E}

But from what universe is F being taken from? I'm really getting out of my league here, but a principle that seems to come up often in axiomatic set theory is that you can't just invoke "the set of all 'things' that have property X". If I remember correctly, this is essentially what Frege tried to do, and Russel put the hammer down on him with his famous paradox. You have to, in one way or another, start with a previously shown (or assumed) to exist set, a universe, and snatch your elements from there.

In my book, Hungerford's "Algebra", his argument for algebraic closures is indeed very reminiscent of your argument. However, as he states before he goes into the proof,

"The chief difficulty in proving that every field K has an algebraic closure is set-theoretic rather than algebraic. The basic idea is to apply zrn's Lemma to a suitably chosen set (his italices) of algebraic extension fields of K"

He then goes on to spend a great deal of energy showing that algebraic extensions of a field K only get so big... that their cardinality is bounded in fact by |K|*aleph[0]. This allows him to later construct an actual bonafied set in which the K-algebraic extensions can be embedded.
• Aug 17th 2006, 01:57 PM
ThePerfectHacker
Quote:

Originally Posted by BubbleBrain_103
"the set of all 'things' that have property X".

This demonstrates the problem of not having a formal definition of a set. I once on this forum asked a question which I found interesting. The set of all finite sets. I was able to demonstrate that this set is not contained by any cardinal number. My attempt was to hopefully introduce a new concept in set theory. Just like there are some sets which cannot be contained by finite number, so too there are sets which cannot be contained by cardinal numbers. The discussion ended with CaptainBlank explaining why I cannot do that (I did not really understand because I never studied axiomatic set theory).

Quote:

In my book, Hungerford's "Algebra", his argument for algebraic closures is indeed very reminiscent of your argument. However, as he states before he goes into the proof,
In my algebra book my author does the same thing. Half of the discussion is on showing the set can be contained in some other set. But back then I did not see the purpose in doing that and just ignored that (in fact I once posed a question on this forum of my version of the proof of algebraic closure asking whether it is correct. Now I realize that it was not because I did not prove my set was countained).

Quote:

that their cardinality is bounded in fact by |K|*aleph[0].
The only sets that can be used are those bounded by cardinal numbers?
• Aug 17th 2006, 02:28 PM
rgep
Zorn's Lemma: Every non-empty partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.

The extensions of a field do not form a set in which every chain has an upper bound. In fact it's doubtful that they form a set at all, as opposed to a proper class.
• Aug 17th 2006, 03:19 PM
BubbleBrain_103
I disagree rgep... my suspiscion, like yours, is that the collection S of ALL extension fields of F is a proper class, and so ZL doesn't apply. But if we could take it be an actual set, we would have that:

1) S is partially ordered by inclusion
2) for any chain of extension fields in S, their union is again an extension field of F, and is an inclusion-upper bound for that chain (not that the union of ANY collection of extension fields is an extension field, but the union over a CHAIN certainly is)

which are the hypotheses for ZL.
• Aug 17th 2006, 05:29 PM
ThePerfectHacker
Quote:

Originally Posted by rgep

The extensions of a field do not form a set in which every chain has an upper bound. In fact it's doubtful that they form a set at all, as opposed to a proper class.

But if you take the union of all the fields in this chain they do form a field!!
And since the union of sets always contains the sets this union set is the upper bounf. Thus, I believe you made a mistake.
• Aug 17th 2006, 10:49 PM
BubbleBrain_103
Hacker,

You had some good questions earlier, wish I could say I could answer all of them off the top of my head. In any event I'd recommend "Classic Set Theory: for guided independent study" by Derek Goldrei. Not terribly dense, good for a first look at the subject, and definitely something you can get through on your own. And when you're done with it it's still good reference (for awkward little math help forum moments like these).