Idk how many videos Id been warching in a row but this is very entertaining and enlightning.....I was literally reading category theory a while ago and watched a video yesterday on the fundamental group so all this connections are so OP and beautiful
In the unlikely event that there are any set theorists reading the comments who don't already know this: you don't actually need the full axiom of choice to construct algebraic closures, the ultrafilter axiom is enough. The proof (for which I don't have a reference handy) is a cute application of standard logic tools: "field extensions of K" can be equivalently thought of as models of a particular first order theory (the language is built out of the ring operators and the elements of K, and the laws are the field axiom together with the equations for K). So, consider the set T of finite extensions of K which is inductively defined such that if L in T and q in L[x] is irreducible over L then L[x]/q in T. Define a filter F on T such that S in F if there is some L in S s.t. for all extensions in S which factor through L are also in S. By the ultrafilter axiom this extends to an ultrafilter U, and now we take the ultraproduct M = (Pi_{L in T} L)/U which must be a model of the theory of field extensions of K (and so therefore, a field extension of K). For every non constant polynomial p the splitting field K_p is in T, and the subset [K_p] of T consisting of extensions of K_p is in the filter. Splitting into n factors is captured by a first ordered sentence, which, for p holds for all elements of a set in the filter, and so p splits in M. We can therefore take the subfield of M consisting of algebraic elements over K to be the algebraic closure. (Edit: fixed to index by extensions instead of polynomials)
Thank you, I had actually been looking for a proof on Ultra Filter Lemma implies every field has an algebraic closure. Is it known whether the reverse implication holds or not? It feels like every field has an algebraic closure would be some sort of weak choice principle.
@@bobman8192 I think that question is open. I agree, it seems like a weak choice principle, but to my knowledge exactly where it fits on the hierarchy of such principles is not known. For what it is worth, revisiting my proof I think there is a much easier way of phrasing it: "field extensions of K" are just models of a particular first order theory (the first order theory of fields extended with a constant for each k in K--that is likely uncountable alphabet, but that isn't a problem--and all the associated axioms for K's equations), while "the polynomial p(x) has a root" is a first order statement. So, really, the existence of algebraic closures follows trivially from *compactness* of first order logic: we just need that any finite set of polynomials over K all have roots in some extension of K. Indeed, one might say that existence of algebraic closures is the preeminent example of the compactness theorem in practice. Ultrafilters get you to compactness, and seeing how the ultrafilter construction works in this particular case is nice as it gives a flavor of what the algebraic closure actually looks like and actually is just a really good example of how the compactness theorem works, but for people who accept soundness/completeness/compactness of first order logic, perhaps all the technical stuff isn't needed.
In fact, algebraic closure cannot be made into a functor at all. There is a theorem due to Artin and Schreier which states a finite subgroup of the Galois group of an algebraically closed field has order at most two, so an automorphism of order, e.g, three of a subfield cannot be extended to an automophism of its closure of the same order. This shows that while an automorphism of subfield can always be extended to its algebraic closure, there is no way to make a canonical choice for this extension that would agree with composition and identity in order to be a functor.
If automorphism is not based on sound fundamentals, Cantor's logic falls apart. But finite has a way to include infinite, the fundamental of which remains murky. Attia once commented that after life long pursuit of abstract algebra, he concludes, he understood nothing. What a sad conclusion.
Thank for these videos Professor. Your viewers (myself included) might really appreciate a series on Grothendieck's Galois theory and perhaps some Galois theory of schemes and etale fundamental groups after this one.
I'm confused at 6:30, where you produce an alpha saying it is a root of the polynomial p(x). You are trying to prove that p(x) "has a root", and these are abstract fields, so how can you claim that such an alpha exists? I understand when we're talking about subfields of the complex numbers where the fundamental theorem of algebra guarantees there is some complex alpha, but for abstract fields (in this example we just have K and K bar); what field does alpha live in when you produce it during the proof?
I presume alpha lives in "a splitting field for p(x)" - and you go on to show that any such splitting field must be equal to k bar. Leaving this comment in case others get confused
3:08 Union of fields is not necessarily a field, so probably it is a field generated by the union...in case of increasing sequence, union is a field, Inductive limit? 12:17~18:00 A proof that the set of complex numbers is algebraically closed. THEOREM Any polynomial has a root on complex plane. PROOF Fix a polynomial P(z) and let n be its degree.. 1. Suppose contrarily that P dose not have zeros on complex plane. 2. On large radius circle, P(z) = z^n + (relatively small terms). So, P(z) moves n times around origin when z moves this large circle. 3. If z moves on small radius circle, because P(0) is not zero by assumption, P(z) moves 0 times around origin. This is because, if z is on sufficiently small circle, P(z) is close to P(0). 4. For large and small circles, winding number of P(z) changes, and this is possible if and only if there is an intermediate circle and if z moves on it, then P(z) pass the origin. This contradicts the assumption that polynomial P(z) has no zeros. 21:01 Puiseux series is algebraically closed?!….so, what is a Galois group of the extension of these fields, e.g., puiserx series is an extension of Leurent series and puiseux series is algebraic closure of Leurent series, so Galois group can be calculated..? 22:03 Any two algebraic closures of fixed field F, is gaois. 24:43 Correspondence of topological space X and it first fundamental group is not an isomorphism. But the pair of topolocical space X together with some base point of X are functor …. I cannot follow the story…….. Similarity of Fundamental group and absolute Galois group. Fix a topological space X corresponds fix a field K. Groupoid is a category whose morphism is invertible. Objects are base points of X as the base points of the first fundamental group. Each base point corresponds to algebraic closures of K. Morphisms of two base points x1,x2 is a loop which induces a isomorphism between fundamental groups of base points x1,x2. Morphisms of two algebraic closure of K induces isomorphisms for absolute Gaolis group of algebraic closures….hmm? I do not understand…
There are situations where this is justified, and some people are working on systems where it is natural to do so, under the names, "homotopy type theory," and, "univalent foundations." (Possibly others.) The "groupoid" structure he mentioned is a natural setting for assembling together objects (like fields) and isomorphisms between them. If an object is identified uniquely up to _unique_ isomorphism in a groupoid, then that is in some sense just as good as specifying something uniquely in a set. On the other hand, if an object in a groupoid is isomorphic to _itself_ in multiple distinct ways, that can require similar bookkeeping as the object not being unique at all (and in fact, if you have two objects in a groupoid with two distinct isomorphisms between them, then they are each isomorphic to themselves in distinct ways). The reason is the same as the explanation with the complex numbers in the last video. Just as the electrical engineers may be using 'j' to mean what you are used to calling '-i', they may be using 'i' to mean what you are used to calling '-i', even if you agree on the letter to use (perhaps you convinced them to use the letter 'i', but they just replaced the letters in their backwards results using the 'i = j' isomorphism). So it may be necessary to keep track of which isomorphism is used to map the details of the results between your two conventions, even on 'the same' object.
I wait every day for these gems. Thank you for making insightful lectures.
Idk how many videos Id been warching in a row but this is very entertaining and enlightning.....I was literally reading category theory a while ago and watched a video yesterday on the fundamental group so all this connections are so OP and beautiful
In the unlikely event that there are any set theorists reading the comments who don't already know this: you don't actually need the full axiom of choice to construct algebraic closures, the ultrafilter axiom is enough.
The proof (for which I don't have a reference handy) is a cute application of standard logic tools: "field extensions of K" can be equivalently thought of as models of a particular first order theory (the language is built out of the ring operators and the elements of K, and the laws are the field axiom together with the equations for K). So, consider the set T of finite extensions of K which is inductively defined such that if L in T and q in L[x] is irreducible over L then L[x]/q in T. Define a filter F on T such that S in F if there is some L in S s.t. for all extensions in S which factor through L are also in S. By the ultrafilter axiom this extends to an ultrafilter U, and now we take the ultraproduct M = (Pi_{L in T} L)/U which must be a model of the theory of field extensions of K (and so therefore, a field extension of K). For every non constant polynomial p the splitting field K_p is in T, and the subset [K_p] of T consisting of extensions of K_p is in the filter. Splitting into n factors is captured by a first ordered sentence, which, for p holds for all elements of a set in the filter, and so p splits in M. We can therefore take the subfield of M consisting of algebraic elements over K to be the algebraic closure.
(Edit: fixed to index by extensions instead of polynomials)
Thank you, I had actually been looking for a proof on Ultra Filter Lemma implies every field has an algebraic closure. Is it known whether the reverse implication holds or not? It feels like every field has an algebraic closure would be some sort of weak choice principle.
@@bobman8192 I think that question is open. I agree, it seems like a weak choice principle, but to my knowledge exactly where it fits on the hierarchy of such principles is not known.
For what it is worth, revisiting my proof I think there is a much easier way of phrasing it: "field extensions of K" are just models of a particular first order theory (the first order theory of fields extended with a constant for each k in K--that is likely uncountable alphabet, but that isn't a problem--and all the associated axioms for K's equations), while "the polynomial p(x) has a root" is a first order statement. So, really, the existence of algebraic closures follows trivially from *compactness* of first order logic: we just need that any finite set of polynomials over K all have roots in some extension of K. Indeed, one might say that existence of algebraic closures is the preeminent example of the compactness theorem in practice. Ultrafilters get you to compactness, and seeing how the ultrafilter construction works in this particular case is nice as it gives a flavor of what the algebraic closure actually looks like and actually is just a really good example of how the compactness theorem works, but for people who accept soundness/completeness/compactness of first order logic, perhaps all the technical stuff isn't needed.
In fact, algebraic closure cannot be made into a functor at all. There is a theorem due to Artin and Schreier which states a finite subgroup of the Galois group of an algebraically closed field has order at most two, so an automorphism of order, e.g, three of a subfield cannot be extended to an automophism of its closure of the same order.
This shows that while an automorphism of subfield can always be extended to its algebraic closure, there is no way to make a canonical choice for this extension that would agree with composition and identity in order to be a functor.
If automorphism is not based on sound fundamentals, Cantor's logic falls apart. But finite has a way to include infinite, the fundamental of which remains murky. Attia once commented that after life long pursuit of abstract algebra, he concludes, he understood nothing. What a sad conclusion.
Excellent Lecture as usual
Thank for these videos Professor. Your viewers (myself included) might really appreciate a series on Grothendieck's Galois theory and perhaps some Galois theory of schemes and etale fundamental groups after this one.
The comments at 7:40 really improved my personal experience with this.
Ah yes, the lesser known "proof by rope" of the fundamental theorem of algebra
I'm confused at 6:30, where you produce an alpha saying it is a root of the polynomial p(x). You are trying to prove that p(x) "has a root", and these are abstract fields, so how can you claim that such an alpha exists? I understand when we're talking about subfields of the complex numbers where the fundamental theorem of algebra guarantees there is some complex alpha, but for abstract fields (in this example we just have K and K bar); what field does alpha live in when you produce it during the proof?
I presume alpha lives in "a splitting field for p(x)" - and you go on to show that any such splitting field must be equal to k bar. Leaving this comment in case others get confused
3:08 Union of fields is not necessarily a field, so probably it is a field generated by the union...in case of increasing sequence, union is a field, Inductive limit?
12:17~18:00 A proof that the set of complex numbers is algebraically closed.
THEOREM Any polynomial has a root on complex plane.
PROOF Fix a polynomial P(z) and let n be its degree..
1. Suppose contrarily that P dose not have zeros on complex plane.
2. On large radius circle, P(z) = z^n + (relatively small terms). So, P(z) moves n times around origin when z moves this large circle.
3. If z moves on small radius circle, because P(0) is not zero by assumption, P(z) moves 0 times around origin. This is because, if z is on sufficiently small circle, P(z) is close to P(0).
4. For large and small circles, winding number of P(z) changes, and this is possible if and only if there is an intermediate circle and if z moves on it, then P(z) pass the origin. This contradicts the assumption that polynomial P(z) has no zeros.
21:01 Puiseux series is algebraically closed?!….so, what is a Galois group of the extension of these fields, e.g., puiserx series is an extension of Leurent series and puiseux series is algebraic closure of Leurent series, so Galois group can be calculated..?
22:03 Any two algebraic closures of fixed field F, is gaois.
24:43 Correspondence of topological space X and it first fundamental group is not an isomorphism. But the pair of topolocical space X together with some base point of X are functor …. I cannot follow the story……..
Similarity of Fundamental group and absolute Galois group.
Fix a topological space X corresponds fix a field K.
Groupoid is a category whose morphism is invertible.
Objects are base points of X as the base points of the first fundamental group. Each base point corresponds to algebraic closures of K. Morphisms of two base points x1,x2 is a loop which induces a isomorphism between fundamental groups of base points x1,x2. Morphisms of two algebraic closure of K induces isomorphisms for absolute Gaolis group of algebraic closures….hmm? I do not understand…
14:45
No, P(x) won't wiggle at all, it will actually be really smooth.
12:49 *Rouche's theorem* proves fundamental theorem of algebra
3:57 Anyone can explain more about the proof for K is uncountable using Aoc?
12:13 Proof of the Fundamental Theorem of Algebra (Topological)
If anyone knows: at 10:56, how do we know L even exists by only doing a countable number of iterations?
also the proof at 16:00 is really beautiful!
7:01 Does K bar contain a power series? A root of a power series (with rational coefficients) is not algebraic, isn't it?
Never mind. Confused closure with completion.
İs algebraic numbers are algebraically closed?
Can you be more specific and mention the fields you are talking about? Q bar is indeed algebraically closed. As is C.
I think I've been treating "unique" and "unique up to isomorphism" as synonymous in math. Oh no.
There are situations where this is justified, and some people are working on systems where it is natural to do so, under the names, "homotopy type theory," and, "univalent foundations." (Possibly others.)
The "groupoid" structure he mentioned is a natural setting for assembling together objects (like fields) and isomorphisms between them. If an object is identified uniquely up to _unique_ isomorphism in a groupoid, then that is in some sense just as good as specifying something uniquely in a set. On the other hand, if an object in a groupoid is isomorphic to _itself_ in multiple distinct ways, that can require similar bookkeeping as the object not being unique at all (and in fact, if you have two objects in a groupoid with two distinct isomorphisms between them, then they are each isomorphic to themselves in distinct ways).
The reason is the same as the explanation with the complex numbers in the last video. Just as the electrical engineers may be using 'j' to mean what you are used to calling '-i', they may be using 'i' to mean what you are used to calling '-i', even if you agree on the letter to use (perhaps you convinced them to use the letter 'i', but they just replaced the letters in their backwards results using the 'i = j' isomorphism). So it may be necessary to keep track of which isomorphism is used to map the details of the results between your two conventions, even on 'the same' object.
ye