Galois theory: Splitting fields

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Galois theory: Splitting fields

Рет қаралды 24,999

Күн бұрын

This lecture is part of an online course on Galois theory.
We define the splitting field of a polynomial p over a field K (a field that is generated by roots of p and such that p splits into linear factors). We give a few examples, and show that it exists and is unique up to isomorphism.

Пікірлер: 46

@maxwibert 3 жыл бұрын

I'm loving this series for its own sake but it's just icing on the cake watching Dr. Borcherds dunk on electrical engineers

@hyperduality2838 3 жыл бұрын

Poles (eigenvalues) are dual to zeroes -- optimized control theory. Signals are dual to noise -- signal to noise ratio in electronics. Alternating currents AC are dual to direct currents DC. Antinomy (duality) is two truths that contradict (ambiguity) each other -- Immanuel Kant. "Always two there are" -- Yoda.

@lisadominiqueh.4644 3 жыл бұрын

Normally I don't like lectures, but this channel is wonderful and exciting. I keep coming back for more, and watching just one of them is never enough. Thank you so much, Prof. Borcherds. I especially love all those examples. :)

@hyperduality2838 3 жыл бұрын

Galois groups are dual to modular forms. Modular forms are dual to elliptic curves. Poles (eigenvalues) are dual to zeroes -- optimized control theory. Signals are dual to noise -- signal to noise ratio in electronics. Alternating currents AC are dual to direct currents DC. Antinomy (duality) is two truths that contradict (ambiguity) each other -- Immanuel Kant. "Always two there are" -- Yoda. Positive charge is dual to negative charge -- electric charge. North poles are dual to south poles -- magnetic fields. Electro is dual to magnetic -- electro-magnetic energy is dual.

@gunyolk55 2 жыл бұрын

2:41 It's more accurate to call L the splitting field of p over K since it's dependent on the polynomial.

@ayshashahul7133 3 жыл бұрын

At 13:55 I think you meant K1 instead of K0

@m.g.3618 2 жыл бұрын

the single dislike is by russian electrical engineer

@Suav58 3 жыл бұрын

20:54 No need to apologize. It doesn't take much effort for a Russian to distinguish between "какой либо" and "определённый". (I do feel embarrassed by some, often very good, Russian mathematicians speaking English and it has had always puzzled me, if it is by lack of motivation, or by mirror symmetry). Use of articles adds efficiency to English language, but I can assure you, that it isn't always more efficient tool of communication than Russian.

@alexanderchekmenev1217 3 жыл бұрын

I confirm it as a Russian ) In Russian if you want to distinguish "a" and "the" scenarios you can use pronouns or adjectives as in your examples. So other parts of speech kind of act as articles. And that's the reason why articles are difficult for Russians: we are used to using other parts of speech as articles (or use nothing it the context is clear).

@Suav58 3 жыл бұрын

@@alexanderchekmenev1217 Thank you. Native English speakers tend to drop an article, when the context is clear too, while those, for whom English is an acquired communication skill seem to be more rigid in using them.

@user-jc2lz6jb2e 3 жыл бұрын

11:31 the roots should be the odd powers of alpha.

@Rocktru2000 3 жыл бұрын

Thank you, I was kind of lost in that part

@hyperduality2838 3 жыл бұрын

Signals are dual to noise -- signal to noise ratio in electronics. Alternating currents AC are dual to direct currents DC. Antinomy (duality) is two truths that contradict (ambiguity) each other -- Immanuel Kant. "Always two there are" -- Yoda.

@user-jc2lz6jb2e 3 жыл бұрын

cos(2pi/7) is always the example used of a degree 3 Galois extension of Q, no matter the book. Almost feels like it's the only one.

@vishalraman-piano5499 3 жыл бұрын

Clearly not, since you can take the other roots of the degree 3 polynomial. We discussed in the previous lecture that cos(4pi/7) and cos(6pi/7) work for the same reason.

@user-jc2lz6jb2e 3 жыл бұрын

@@vishalraman-piano5499 well yeah, but either way, you're still getting the same field, hence it's the one degree 3 example of a Galois extension always given.

@boboryan1012 2 жыл бұрын

@@user-jc2lz6jb2e because there is one group of order 3 up to isomorphism.

@sidkulkarni5599 3 жыл бұрын

22:30 thanks for the laugh, an excellent bonus to a great lecture!

@mina_mozna 2 жыл бұрын

This is a great lecture, thank you Dr.

@rserserserse Жыл бұрын

Great lecture

@baronvonbeandip Жыл бұрын

I'm so glad for the internet and, in part, for COVID specifically because it's forced people to put this material into videos on the internet, like they should've always done. Having to travel 4 hours to the nearest college campus that teaches algebra is a huge problem when in-person lectures are mandatory. I'd rather have lectures all day, instead of the meager hour you'd normally get. Wish institutions would get the hint.

@NightWanderer31415 3 жыл бұрын

lol at electrical engineers getting their signs wrong. Great lecture.

@shunansun899 3 жыл бұрын

Your explications are very clear. Thank you.

@swaruppaul4417 3 жыл бұрын

Really this is an overwhelming experience to get to learn from the master. Thank you so much Professor and can you please suggest, which book should I use as a text for this course?

@cogito2958 3 жыл бұрын

He recommends Galois theory by E Artin in the introduction video.

@swaruppaul4417 3 жыл бұрын

Oh, okay...! Thanks..😊

@Darrida Жыл бұрын

You very attentuve, fir me as native Russian it is hard to distinguish "the splitting field" and "a splitting field" . Because we have words ending instead of a and the

@hausdorffm 2 жыл бұрын

2:09 The notion of roots of polynomial has ambiguity. For example, consider a polynomial p(x) = x^2 + 1. In (Z/2Z)[x], p(x) = (x+1)^2, so all roots seem to be -1. On the other hand, In C[x], p(x) = (x+i)(x-i), i.e., the roots of p(x) seem to be -i and i where I is the square root of minus 1. Therefore, notion of roots of polynomial depends on the field of coefficients. But the condition that p(x) factors into linear factors is more clear what this means. So, at 2:09, the word “roots” are excluded from the definition of splitting fields. 10:15 For any cubic polynomial p(x) over a field Q, if an extension field Q[x]/(p(x)) contains two roots of p(x), then automatically the remained root also is in Q[x]/(p(x)). This is because the sum of three roots is a coefficient of p(x) which is in Q. Now, I do not understand where the polynomial comes from, but now, p(x) = 8x^3 + 4x^2 -4x -1 has three roots cos(2pi/7), cos(4pi/7), cos(6pi/7) satisfying cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -4/8. And thus cos(6pi/7) = 4/8 - cos(2pi/7) - cos(4pi/7), the left hand side is in the field Q[x]/(p(x)) as explained at 10:15 and hence all roots of p(x) are contained in Q[x]/(p(x)). To show that Q[x]/(p(x)) is the minimal among all fields containing roots of p(x), suppose that E is another such a filed and consider the map Q[x] -> E which sends x to cos(6pi/7). Because p(x) is an irreducible (minimal) polynomial of cos(6pi/7) over Q, the image is isomorphic to Q[x]/(p(x)) and this gives an inclusion. Hence, Q[x]/(p(x)) is minimal among them. 2:49 In the definition of splitting field, I am not sure why “LINEAR FACTORS” is used. I guess “L generated by roots of p(x) over K” means that L is a minimal field among all fields containing both K and roots of p(x). 12:09 If "a" is a root of x^4+1 = 0, then Q(a)=Q(a^3) =Q(a^5) = Q(a^7) = Q[x]/(x^4+1). 15:18 typo? K_2 = K_0[x]/() ... but it will be K_1[x]/() 15:25~23:00 About uniqueness of splitting fields of p(x) over a filed K. I do not understand yet. 21:18 Isomorphisms of splitting fields is a little stronger than fields. In fact, if there is two extensions of fields K -> K[i];x -> x and K ->K[j];x -> jx, then maps K[i] -> K[j] defined by i-> j and i-> -j are not isomorphism as extension but isomorphism as field. I cannot understand why uniqueness does not follow straightforward. Is “a” splitting field defined by “the” minimal field which contains both K and all roots of a fixed p(x) of K[x]?

@alexmueller7864 3 жыл бұрын

I'm showing 21:33 to my dad. I studied math in college and he's an electrical engineer haha.

@hyperduality2838 3 жыл бұрын

@Treviscoe Жыл бұрын

Wow, a Richard Borcherds lecture!

@vs-cw1wc 3 жыл бұрын

22:50 Why is this a minor problem? It seems to me that the ambiguity hints at the possibility of having automorphisms of the splitting field, which is key to Galois theory.

@hyperduality2838 3 жыл бұрын