Notes

Description:

Form 233 is true in \(\cal N1\)

Content:

The following proof that Form 233 is true in \(\cal N1\) is due to David Pincus. Form 233 is "Algebraic closures of fields, if they exist, are unique." If \(K\) is an algebraically closed field, if \(\pi\)is a non-trivial automorphism of \(K\) satisfying \(\pi^2 = 1_K\) (the identity on \(K\)), and if \(i\in K\) is a square root of \(-1\), then\(\pi(i) = -i \ne i\). (H. Läuchli has given a proof of this lemma with the additional hypothesis that the characteristic of \(K\) is not equal to 2.  See Jech [1973b], p. 147.) Let \(L = \{x\in K:\pi(x) = x\}\).The order of the group of automorphisms generated by \(\pi\) is 2. Therefore by Artin's theorem (theorem 1.8 on page 264 of Lang [1993]) the index of \(L\) in \(K\) is 2.  By corollary 9.3, page 299 of Lang [1993], (If \(L\) is a field and \(K\) is the algebraic closure of \(L\) and is of finite degree over \(L\), then \(K = L(i)\) where \(i^2=-1\) and L has characteristic 0.) \(K=L(i)\), where \(i^2 = -1\) and \(L\) has characteristic 0, and hence \(K\) has characteristic 0. Choose \(x\in K\) such that \(\pi(x) \ne x\).  Let \(y = \pi(x) - x\) so that \(y\ne 0\) and \(\pi(y) = -y\). Let \(z\) be a square root of \(y\).Then \(\pi(z)\) is a square root of \(\pi(y) = -y\) hence \(\pi(z)=iz\) for some square root \(i\) of \(-1\).  Since \(y\ne 0\), \(z\ne 0\).  We can therefore compute \[\pi(i) =\pi\left(\frac{\pi(z)}{z}\right) = \frac{z}{\pi(z)} =\frac{1}{i}= -i \ne i.\] The last inequality is a consequence of the fact that the characteristic of \(K\) is not 2. Since \(\pi\) has order 2, and \(\pi(-i)= i\).   This completes the proof of the lemma.

Now let \((K,+,\cdot,0,1)\) be a field in \(\cal N1\) with finite support \(E \subseteq A\) where \(A\) is the set of atoms. Assume that \(K\) is algebraically closed. We will show that every element of \(K\) has support \(E\). This implies that all algebraically closed fields are well orderable in \(\cal N1\) and therefore the standard proof of the uniqueness of algebraic closures (using \(AC\)) is valid in \(\cal N1\).

The proof is by contradiction. Assume that \(x\in K\) does not have support \(E\). We will arrive at a contradiction in two steps. The first is to show that for some permutation \(\pi\) of \(A\) which fixes \(E\) pointwise, \(\pi(x)\ne x\) and \(\pi^2 \) is the identity. The permutation \(\pi\) induces an automorphism of \((K,+,\cdot,0,1)\) and we can therefore apply the lemma to conclude that \(\pi(i) = -i \ne i\) for some square root \(i\) of -1 in \(K\).  The second step is to show that for every permutation \(\pi\) of \(A\) that fixes \(E\) pointwise, \(\pi(i) = i\) for every square root \(i\) of -1 in \(K\).

For step one, suppose that \(x\) has support \(E\cup F\) where \(E\cap F= \emptyset\).  Let \(F'\) be a subset of \(A\) for which \(|F'| = |F|\) and \(F'\cap (E\cup F) =\emptyset\).  Let \(\pi\) be a permutation of \(A\) which interchanges the elements of \(F\) and \(F'\), that is, \(\pi\) is a product of cycles of length 2, \(\pi = (a_1,b_1)(a_2,b_2)\cdots(a_n,b_n)\) where \(F = \{a_1,\ldots,a_n\}\) and \(F' =\{b_1,\ldots,b_n\}\).   Clearly \(\pi^2\) is the identity and \(\pi\) fixes \(E\) pointwise. The proof that \(\pi(x)\ne x\) is a standard permutation model argument and depends on the fact that \(x\) does not have support \(E\).

For step two, we note that if \(i\) is a square root of \(-1\) in K, then \(i\) is a solution to the equation \(x^2 + 1 = 0\) all of whose coefficients are fixed by any permutation that fixes \(E\) pointwise.  Therefore if\(\eta\) is a permutation of \(A\) that fixes \(E\) pointwise, \(\eta(i)\) is also a solution to \(x^2 + 1 = 0\).  Suppose that \(i\) does not have support \(E\).  Let \(E\cup F\) be a support of \(i\) where \(E\cap F = \emptyset\). By a standard permutation model argument, we can show that if \(\pi\) and \(\pi'\) are any two permutations of \(A\) that fix \(E\) pointwise and \(\pi(F) \cap \pi'(F) = \emptyset\) then \(\pi(x) \ne \pi'(x)\).  It follows that the equation \(x^2 + 1 = 0\) has infinitely many solutions in \(K\) which is a contradiction and the proof is complete.

Howard-Rubin number: 41

Type: Proof

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