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/sci/ - Science & Math


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5395620 No.5395620 [Reply] [Original]

Let <span class="math">F = K(u_1, u_2, ..., u_n)[/spoiler] be a field extension of K.

Why isn't [F : K] = n? Doesn't {u_i} form a basis for F as a vector space over K? In general, is there a way of determining [F : K] if F/K is given as above?

>> No.5395637

>>5395620
>implying /sci/ knows any math beyond calculus
lel

>> No.5395641

>>5395637
>implying i don't know all of the math

>> No.5395649

There is no way of determining [F : K] unless dx/dy > n and n is an integer. Given n is unknown, there is no way of solving this statement.

>> No.5395647 [DELETED] 

>>5395637
It's freshman algebra. Why are you even here?

>> No.5395659

>>5395649
What the...

>Carl
Oh.

>> No.5395669 [DELETED] 

>>5395649
How are you on here 24/7 to shitpost? Do you ever sleep?

>> No.5395677

So, anyone going to give an actual response to the OP?

>> No.5395688

Why would [F:K] be n?
Use only the simplest case of K as the rationals. Then if <span class="math">u_1[/spoiler] is, say, a 999th root of unity, then [F:K] will be large even without adding in <span class="math">u_2 \dots [/spoiler] because <span class="math">1, u_1, u_1^2, u_1^3, \cdots, u_1^998[/spoiler] will be the basis for a vector space over K with pretty high dimension (and some redundancy because 999 isn't prime, but just replace 999 with a high prime if you want to)

>> No.5395695

>>5395669
nobody likes evil carl sagan

>> No.5395744

>>5395620
If the {u_i} are algebraically independent, then your intuition matches the notion of transcendence degree (https://en.wikipedia.org/wiki/Transcendence_degree)), and the {u_i} form a transcendence basis.

However, transcendence degree is based on algebraic independence, while dimension of a vector space is based on linear independence. Those are generally quite different from each other.

All finite-dimensional vector spaces over K have transcendence degree 0. Proof: The powers of any transcendental element form a countably infinite, linearly independent set.