Abstract Algebra, Lecture 34A: Field Extension and Splitting Field Examples (including x^3+1 over Q)

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Abstract Algebra, Lecture 34A. Field Theory and Galois Theory, Part 4.

(0:00) Lecture plan and plan to get to the basics of Galois theory for the last couple lectures.
(0:54) Baseball story from my past.
(3:28) Example to verify the equality of field extensions in a certain situation and compute the dimension of the field extension as a vector space over the rationals Q.
(17:50) Theorem about relating elements in a simple field extension to linear combinations of the zero that is adjoined and relate to the example (and draw a tower of fields).
(21:19) Splitting field for x^3 + 1 over Q.
(32:00) Start the example of the splitting field of x^4 + 1 over Q and of x^4 + 1 over R.

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When explicitly writing out what the elements of Q(sqrt(2), sqrt(3)), it should be {a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6)|a, b, c, d in Q}. You wrote the first term as (a+c) rather than just a at 17:47

danielbradley

1/2 is already in the field of the rationals so why can't you just adjoin sqrt(3)i/2 instead of 1/2+sqrt(3)i/2 ? (or even just sqrt(3)i)

barrettbarry

Thompson Linda Miller Michelle Garcia Shirley

XrcyhikUbhdfbjdf

Hi, thanks for your excellent lectures and for putting them out for the benefit of the public!
I have a question:
when proving sqrt(2) and sqrt(3) are in Q(sqrt(2)+sqrt(3)) you start with 1/(sqrt(2)+sqrt(3)) and then multiplied both the numerator and the denominator by (sqrt(2)-sqrt(3)) - but you haven't yet proved (sqrt(2)-sqrt(3)) is in Q(sqrt(2)+sqrt(3)), so I don't understand how this proves your claim - because as I understand it, you are seeking to show that by using operations available in Q(sqrt(2)+sqrt(3)) *alone* you can get sqrt(2) and sqrt(3). What am I missing?

sarelaiber

And another question:
At 28:45 you say that the simplest form of the field in Q(0.5+(sqrt(3)/2)i), but since 0.5 is rational, can't we say that a simpler form of the field is Q((sqrt(3)/2)i), and even simpler: Q(sqrt(3)i)?
(since we can get, by closure, from sqrt(3)i to 0.5+(sqrt(3)/2)i)

sarelaiber

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