MATH 121 Problem Set 3 Solutions
Jason Lo
Sunday, 18 February 2007
All references to Dummit and Foote below are for the 2nd edition of the
book.
1.
See (sketches of) solutions to midterm.
2. If charF = 0, then any extension of F is separable, and so F sep = E, so
[E : F sep ] = 1 = p0 for any prime p. So we might as well assume that charF = p
is prime.
Now E/F sep is a finite and inseparbale extension. For any α ∈ E \ F sep , let
p(x) be the minimal polynomial of α over F . Then we know (Proposition 38,
section 13.5 of Dummit and Foote)
k
p(x) = psep (xp )
for some unique k ∈ Z+ and unique irreducible separable polynomial psep (x)
k
k
k
over F . Hence αp ∈ F sep , the minimal polynomial of α over F sep is xp −αp =
0, and α lies in the top end of this tower of field extensions:
k
k−1
αp ∈ F sep ⊆ F sep (αp
For each extension
k−2
) ⊆ F sep (αp
i
) ⊆ · · · ⊆ F sep (αp ) ⊆ F sep (α).
i−1
F sep (αp ) ⊆ F sep (αp
)
(1)
(2)
above, the degree of extension must be p (or else [F sep (α) : F sep ] < pk , contradicting that the degree of the minimal polynomial of α over F sep has degree
pk .
Now suppose E = F sep (α1 , · · · , αs ) where each αi ∈ E \ F sep . We know
sep
[F (α1 ) : F sep ] is a power of p. We would be done if we can show, that for
any finite extension L of F sep , and any αi , [L(αi ) : L] is a power of p. This can
be shown as follows:
m
/ L but
Write α = αi . Then let m be the least positive integer such that αp ∈
m+1
m
m+1
αp
∈ L. Then the minimal polynomial q(x) of αp over L divides xp −αp
.
So q(x) is inseparable, and so must have degree at least p. Hence q(x) must be
m+1
equal to xp − αp
. But then it can be checked easily that for each 0 < i < m,
i
i+1
[L(αp ) : L(αp )] = p. Hence [L(αi ) : L] is also a power of p.
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3. Problem 8 of section 14.1 refers to another exercise in the book, problem
18 of section 13.2, which basically has the solution laid out.
4. If every irreducible polynomial F [x] with a root in E splits over E, then for
every α ∈ E, we can just let fα (x) be its minimal polynomial over F , and then
E would be the splitting field of the family {fα }α∈E .
Now suppose E is the splitting field of a family of polynomials. Let f (x) ∈
F [x] be an irreducible polynomial with a root α ∈ E. Suppose E is contained
in the algebraic closure F̄ of F . Let β be any root of f (x) in F̄ . Then we get
an isomorphism ϕ : F (α) → F (β) ⊆ F̄ over F .
We can extend ϕ to an embedding ϕ : E → F̄ (a standard Zorn’s lemma
argument). Suppose now that E is the splitting field of the family of polynomials
{gi (x)}i over F . Then ϕ : E → F̄ merely permutes the roots of each gi , and so
im ϕ = E. Hence E ⊇ F (β) 3 β. Hence E contains every root of f (x), i.e. f (x)
splits over E, proving the first part of this problem.
Parts of the arguments above can already prove the equivalence of all three
statements.
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