4:20 ~ (1)Let L/K be an extension. [[[ LEMMA ]]]: If the characteristic of K is zero, then, its minimal polynomial on K has n-DISTINCT roots, where n = degree(f). (Note that each root is not necessarily contained in L but the algebraic closure of K) [[[ PROOF ]]] If not, then there is some element r of algebraic closure of K such that r is a common zero of minimal polynomial and its derivative, that is f(r) = f’(r) = 0. Because f is a minimal polynomial of r and f’(r) =0, f divides f’ which means that degree of f’ >= degree of f. This contradicts the inequality degree(f’ ) = degree(f) -1 > degree(f) Note that even if L/K is a normal extension, all roots are not in L. Note that if L/K is a splitting field of the minimal polynomial, then all roots are in L. I am confused, Galois theory use many similar terminologies. 5:35~ (2) Let L/K be a normal extension. If characteristic of K is nonzero, then the above LEMMA fails, that is, there is some extension L/K and some element of L such that its irreducible polynomial on K does not have n DISTINCT roots (in the algebraic closure of K), where n = degree(f). PROOF: Because characteristic is nonzero, there is a polynomial f on K such that f’ is identically zero which means that all roots are not distinct. For example, if Char(K) = p and consider an extension L/K = k(t)/k(t^p), where t is not in a field k. Then consider an irreducible polynomial f of K[x] whose derivative is identically zero on L. For example we can define such a polynomial f by f(x) = x^p - t^p, where p = Char(K). Then f’ = px^{p-1} =0 in K[x], so all roots of f are not simple. In fact f = x^p - t^p = (x-t)^p on L. So, f does not have distinct roots. Def: a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K which has a root in L Def: a separable extension L/K means that for any element of L its minimal polynomial has distinct roots. 8:47 ~ f(x) = x^p - t^p is a minimal polynomial of t over the field k(t^p). splitting field of f over k(t^p) is the field k(t). The extension L/K := k(t)/k(t^p) is not separable extension, because there is a element (namely, t ) of L such that its minimal polynomial does not have distinct roots. For nonseparable extension L/K = k(t)/k(t^p), Galois correspondence is broken. In fact, G := Gal( L/K )={ identity } where Char(k) =p. Pick any element g of G = Gal( L/K ). Because t^p is in K, g fix it, that is, g(t^p) = t^p, so g(t) is a solution of the equation X^p -t^p = (X-t)^p = 0. Thus, X = t, that is, g(t) =t. Therefore G does not have any nontrivial subgroup. Now, L/K is not a separable, thus Galoie correspondence cannot be applied for L/K. In fact two subfield of L and K corresponds to same subgroup Gal(L/K) = Gal(L/L) = {identity}.
Galois theory only describes the relation between fields up to their perfect closure, just as homological algebra only describes modules up to their injective envelope. So you might as well take Galois theory over perfect fields, since any other properties have to established seperately anyways. The advantage is that this gets rid of the concept of non-seperable extensions, so normal and Galois extensions are now the same thing, similar to the Galois correspondence in covering maps.
Thank you for these fantastic videos and happy new year!
4:20 ~ (1)Let L/K be an extension.
[[[ LEMMA ]]]: If the characteristic of K is zero, then, its minimal polynomial on K has n-DISTINCT roots, where n = degree(f). (Note that each root is not necessarily contained in L but the algebraic closure of K)
[[[ PROOF ]]] If not, then there is some element r of algebraic closure of K such that r is a common zero of minimal polynomial and its derivative, that is f(r) = f’(r) = 0. Because f is a minimal polynomial of r and f’(r) =0, f divides f’ which means that degree of f’ >= degree of f. This contradicts the inequality degree(f’ ) = degree(f) -1 > degree(f)
Note that even if L/K is a normal extension, all roots are not in L.
Note that if L/K is a splitting field of the minimal polynomial, then all roots are in L.
I am confused, Galois theory use many similar terminologies.
5:35~ (2) Let L/K be a normal extension. If characteristic of K is nonzero, then the above LEMMA fails, that is, there is some extension L/K and some element of L such that its irreducible polynomial on K does not have n DISTINCT roots (in the algebraic closure of K), where n = degree(f). PROOF: Because characteristic is nonzero, there is a polynomial f on K such that f’ is identically zero which means that all roots are not distinct. For example, if Char(K) = p and consider an extension L/K = k(t)/k(t^p), where t is not in a field k. Then consider an irreducible polynomial f of K[x] whose derivative is identically zero on L. For example we can define such a polynomial f by f(x) = x^p - t^p, where p = Char(K). Then f’ = px^{p-1} =0 in K[x], so all roots of f are not simple. In fact f = x^p - t^p = (x-t)^p on L. So, f does not have distinct roots.
Def: a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K which has a root in L
Def: a separable extension L/K means that for any element of L its minimal polynomial has distinct roots.
8:47 ~
f(x) = x^p - t^p is a minimal polynomial of t over the field k(t^p).
splitting field of f over k(t^p) is the field k(t). The extension L/K := k(t)/k(t^p) is not separable extension, because there is a element (namely, t ) of L such that its minimal polynomial does not have distinct roots.
For nonseparable extension L/K = k(t)/k(t^p), Galois correspondence is broken. In fact,
G := Gal( L/K )={ identity } where Char(k) =p.
Pick any element g of G = Gal( L/K ). Because t^p is in K, g fix it, that is, g(t^p) = t^p, so g(t) is a solution of the equation X^p -t^p = (X-t)^p = 0. Thus, X = t, that is, g(t) =t. Therefore G does not have any nontrivial subgroup. Now, L/K is not a separable, thus Galoie correspondence cannot be applied for L/K. In fact two subfield of L and K corresponds to same subgroup Gal(L/K) = Gal(L/L) = {identity}.
lol ok nerd
Galois theory only describes the relation between fields up to their perfect closure, just as homological algebra only describes modules up to their injective envelope.
So you might as well take Galois theory over perfect fields, since any other properties have to established seperately anyways.
The advantage is that this gets rid of the concept of non-seperable extensions, so normal and Galois extensions are now the same thing, similar to the Galois correspondence in covering maps.
Any reading recommendations to learn about Galois groups of inseparable extensions?
ye
Anyone has an example of an inseparable extension that is not purely inseparable?
math.stackexchange.com/questions/1275070/does-an-inseparable-extension-have-a-purely-inseparable-element
@@TheAcer4666 awesome, thank you