# Algebraic closure exists: What's wrong with this proof?

Given a field $K$, let $U = K[X] \times \mathbb{N}$. Identify each $k\in K$ as $(X-k,1) \in U$, so $K \subseteq U$. Consider fields $(S,+,\cdot)$ where $K \subseteq S \subseteq U$, and the inclusion map $K \hookrightarrow S$ is a field homomorphism such that $S:K$ is algebraic. Partially order these fields by $(S_1,+_1,\cdot_1) \leq (S_2,+_2,\cdot_2)$ iff $S_1\subseteq S_2$ and $S_1 \hookrightarrow S_2$ is a field homomorphism. Zorn's lemma is not needed to show this is a partial order, I think.

For any chain $(S_i,+_i,\cdot_i)_{i\in I}$, an upper bound is given by $(S,+,\cdot)$ where $S=\bigcup_{i\in I} S_i$ and $+,\cdot$ are defined per pair of elements in $S_i$, for example define $\alpha + \beta$ to be $\alpha +_i \beta$ if $\alpha,\beta\in S_i$. This is independent of $i$. I don't think Zorn's lemma is needed to show this is indeed an upper bound either...

Finally, apply Zorn's lemma to get a maximal element which is automatically an algebraic closure of $K\subseteq U$.

Question What's wrong with this proof? On the several accounts where I thought Zorn's lemma is not needed, have I been wrong, and why?

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Is the upper bound of the chain you have constructed still algebraic over $K$? –  Prahlad Vaidyanathan Sep 15 '13 at 13:12
I don't grasp what $P(U)$ denotes. Since you have identified "each $k \in K$" as an element of $U$, that lends itself to considering $K \subseteq U$, so interpreting $P(U)$ as a power set of $U$ seems incoherent. –  hardmath Sep 15 '13 at 13:27
@hardmath I'm not sure about $P(U)$ actually. This proof is set as an exercise in 'A Course on Galois Theory' by DJH Garling, and it was hinted that $P(U)$ instead of $U$ should be used. I have edited my post although I don't know if a correct proof will need $\{(X-k,1)\}$ instead of $(X-k,1)$... –  user71815 Sep 15 '13 at 15:27
@PrahladVaidyanathan I think so... Each element in $S$ is algebraic over $K$, isn't it? –  user71815 Sep 15 '13 at 15:28

Although stated as "what's wrong with this proof?", the larger question seems to be how to fix it up, since rather than a fallacious conclusion (as with most "fake-proofs"), the situation here seems to be a bogus proof of a valid result (every field has an algebraic closure).

So I'm going to begin by pointing out a major problem with the proposed proof, and then explain how the difficulty is often circumvented. Although the identification of field elements $k \in K$ the base field with elements $(X-k,1)$ in $U = K[X] \times \mathbb{N}$ seems unobjectionable as a 1-1 mapping, it doesn't explain how some field extension $S:K$, even an algebraic one, might be similarly mapped into $U$ (or in another hinted version, into the power set of $U$).

Indeed the natural ring structure on $K[X] \times \mathbb{N}$ does not seem even to help with realizing the inclusion of $K$ into $U$ as a ring homomorphism. For example, if $k_1,k_2 \in K$ are mapped respectively to $(X-k_1,1),(X-k_2,1) \in U$, there seems to be no obvious rationale for why $(X-k_1,1)+(X-k_2,1) = (X-(k_1+k_2),1) \in U$, much less why $(X-k_1,1)*(X-k_2,1) = (X-(k_1 k_2),1) \in U$.

At this point the role of the second coordinate in $U$ seems largely inexplicable, except that some sort of bookkeeping might be necessary to distinguish roots of higher degree polynomials (as typical "algebraic over $K$" objects).

We do well to reflect on the purpose this machinery presumably serves, namely to put various (algebraic) field extensions of $K$ in some common set where chains of inclusions may be unioned to form a maximum of the chain (then handing off the conclusion to Zorn's lemma, existence of an algebraic closure). Once the existence of said closure is available, the pursuit of Galois theory, treating any two algebraic extensions of $K$ as if they are both subfields of a "universal" extension, becomes easier. But we must resist the temptation to fall into a trap of circular logic, using an algebraic closure to justify operations which lead to existence of an algebraic closure.

Standard treatments of Galois theory, such as Kaplansky's Fields and Rings, will illuminate ways to proceed. In the present context it does seem promising to try associating an irreducible (over $K$) polynomial with each element of an algebraic extension $S$ of $K$, or perhaps the ideal generated by such an irreducible polynomial. After all, included among the beginning steps of Galois theory are mostly constructive demonstrations that the sum of two algebraic elements (likewise their product) is again algebraic.

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Let me discuss your use of Zorn's lemma, since hardmath already discusses the algebra.

You are right on two accounts, but wrong on the third:

1. Zorn's lemma is not needed in order to show that this is a partial order.
2. Zorn's lemma is not needed in order to show that this is an upper bound of a chain.

However you are using Zorn's lemma exactly in the place where it is needed: obtaining a maximal element in the partial order. Zorn's lemma, after all only states: if a certain partial order satisfy certain properties, then there exists a maximal element in that partial order.

Zorn's lemma is not used when we show that the order is a partial order, or when every chain has an upper bound. It is used when we want to assert the existence of a maximal element, whose existence is otherwise unprovable.

Do note, however, that the existence of an algebraic closure requires less than Zorn's lemma in its full power. It has been shown that the existence (and uniqueness, up to isomorphism of course) of an algebraic closure follows from the statement "Every filter can be extended to an ultrafilter". The latter have been shown to be strictly weaker than Zorn's lemma, but still unprovable from the axioms of set theory without choice. These proofs require a lot of nontrivial set theoretic knowledge, and can be found (for example) in Jech's The Axiom of Choice.

(Of course, your proof is missing the proof of the non-obvious statements that every element in your partial order is indeed an algebraic extension, and that a maximal element will indeed be an algebraically closed field.)

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So you're saying the proof is correct? It can't be right: Note how the choice of $U$ does not change the 'proof' at all... –  user71815 Sep 15 '13 at 18:10
Well, it's not correct in the sense that you don't prove that the elements of the posets are algebraic extensions, or that the maximal element is algebraically closed. The points you point out that you are not using Zorn's lemma, you aren't. And as I pointed out in my answer, you do need some fragment of the axiom of choice in order to prove the existence of an algebraic closure for an algebraic closure. –  Asaf Karagila Sep 15 '13 at 18:17
It was hard picking the best answer when each of you two answered a different part of my question, but thanks –  user71815 Sep 15 '13 at 18:34
No problem, hardmath's answer was better with respect to the algebra in the question. I might edit mine to clarify the points discussed in the comments later. –  Asaf Karagila Sep 15 '13 at 18:38