Field Extension & Splitting Field Examples | Adjoin Roots: ℚ(√2), ℚ(√3), ℚ(√2,√3)=ℚ(√2)(√3)

[@uzferry5524]

thank you, I really appreciate the slower pace and all the examples.

[@billkinneymath]

You're welcome! Thanks for watching!

[@girlsinacoma]

Gallian book seems to gloss over these examples in extension field chapter. Thanks for diving in.

[@billkinneymath]

You're welcome! Yes, I think his philosophy is that the readers should check it on their own.

[@emmettdja]

It is also cool to note that Q adjoin sqrt(3) adjoin sqrt(2) is a splitting field for x^2-6 since sqrt(6) is in that field. sqrt(2) * sqrt(3) = sqrt(6). it is also minimal, 4 dimensional vector space.

[@billkinneymath]

Very cool!

[@wenzhang365]

This really helps. BTW, there seems to be an 8 min segment missing at 29:00. Thank you

[@billkinneymath]

So glad it helped! Yes, I edited that part out because I was talking with the students about some other things during that time.

[@wenzhang365]

@@billkinneymath Thank you. I didn't miss anything.

[@GiovannaIwishyou]

This is such a good content, too bad I couldn't find something like this 9 years ago (2014 when I had abstract algebra last time as the undergraduate student )😅

[@billkinneymath]

Glad you are able to enjoy it now anyway!

[@arcturusgd]

I've come across this before Because of a specific video about extending sets. Back then it was impossible, but now it is. This was a different thing but it s like joining or adding a field of new elements K to old elements M For set A, A(M) = {K1, K2, K3...Kn, MK1, MK2, MK3, MKn...} But also can be Kn + Km*M

[@doodleyeon]

Thank you so much for this great content! Question: Why is Q(2^(1/3))={a + 2^(1/3)*b + 2^(2/3)*c : a, b, c in Q}? I understand this is true when I factor x^3 - 2 but why is it Q(2^(1/3)), and not Q(2^(1/3), 2^(2/3))?

[@billkinneymath]

You're welcome! Those fields are actually equal to each other. The reason is that, Q(2^(1/3)), being a field and containing 2^(1/3), also contains its square 2^(2/3) (by closure under multiplication). So, 2^(2/3) has "already" been adjoined by adjoining 2^(1/3) to Q.

[@doodleyeon]

@@billkinneymath Oh I cant believe I missed that! Then can the said simple extension field be written like this: Q(2^(1/3))={a + 2^(1/3)*b : a, b in Q} instead of in my previous comment?

[@billkinneymath]

@@doodleyeon No. The reason is that the degree of the minimal polynomial of 2^(1/3) is 3. So as a vector space over Q, elements have to be written as linear combinations of (2^(1/3))^0 = 1, 2^(1/3), and (2^(1/3))^2 = 2^(2/3).