Advanced Algebra I
field extensions and algebraic closure
Let K be a subfield of F , then we say that F is an extension over K and denote it by F/K. Recall that F can be viewed as a vector space over K. We say that the extension F/K is finite of infinite according the dimension of F as a vector space over K.
Let F/K be an extension, an element u ∈ F is said to be algebraic over K if there is a non-zero polynomial f (x) ∈ K[x] such that f (u) = 0. In other words, the ring homomorphism
ϕ : K[x] → F, f (x) 7→ f (u)
has a non-zero kernel. Let I be the kernel. Since K[x] is a PID, I = (p(x)) for some p(x). Let K[u] be the image of ϕ, then
K[x]/(p(x)) ∼= K[u] ⊂ F.
It’s easy that (p(x)) is a prime ideal , that is, p(x) is irreducible. We may assume that p(x) has leading coefficient 1. Such p(x) is called the minimal polynomial of u over K.
We say that F/K is algebraic if every element of F is algebraic over K.
Let’s recall some more properties. If F/K, then we denote [F : K]
to be the dimension dimKF .
Proposition 0.1. If E/F and F/K, then [E : F ][F : K] = [E : K].
Sketch of the proof. Let {ui}i∈I be a basis of E/F and {vj}j∈J be a basis of F/K. Then one can prove that {uivj}(i,j)∈I×J is a basis of E/K. Hence
[E : K] = |I × J| = |I| · |J| = [E : F ] · [F : K].
¤ Proposition 0.2. Suppose that we have a tower of fields K ⊂ F ⊂ E.
Then E is finite over K if and only if E is finite over F and F is finite over K.
Proof. Easy corollary of the previous proposition. ¤ Proposition 0.3. If F/K is finite, then F/K is algebraic.
Proof. suppose that [F : K] = n. For any u 6= 0 ∈ F , then {1, u, ..., un} is linearly dependent over K. Thus there are a0, ..., an ∈ K non all zero such that Pn
i=0aiui = 0. It follows that u satisfies the polynomial f (x) =Pn
i=0aixi ∈ K[x]. ¤
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Let F/K be an extension, and u ∈ F . We denote by K(u) the smallest subfield of F containing K and u. It’s easy to see that
K(u) = {f (u)
g(u)|f (x), g(x) ∈ K[x], g(u) 6= 0}.
Similarly, for S ⊂ F , we denote by K(S) the smallest subfield contain- ing both K and S. If F = K(S) for a finite set S, then F is said to be finitely generated over K.
Proposition 0.4. Let F/K be an extension. Then u ∈ F is algebraic over K if and only if K(u) = K[u]. And in the algebraic case, [K[u] : K] = deg(p(x)), where p(x) is the minimal polynomial.
Sketch of the proof. If u ∈ F is algebraic over K, let p(x) be the min- imal polynomial. One sees that g(u) 6= 0 if and only (g(x), p(x)) = 1.
There are s(x), t(x) such that
1 = s(x)g(x) + t(x)p(x),
hence 1 = s(u)g(u). One has f (u)g(u) = f (u)s(u) and hence K(u) ⊂ K[u].
Conversely, 1u ∈ K(u) = K[u]. Thus u1 = f (u) for some f (x) ∈ K[x].
One sees that u satisfies xf (x) − 1. ¤
Proposition 0.5. F/K is finite if and only if F/K is finitely generated and algebraic.
Sketch of the proof. If F/K is finite, let {u1, ..., un} be a basis of F/K, then F = K(u1, ..., un) hence is finitely generated.
Conversely, suppose that F = K(u1, ..., un) is algebraic over K. In particular, each ui is algebraic over K. In particular, u1 is algebraic over K, u2 is algebraic over K(u1), and so on. Then one has that [K(u1, ..., un) : K] = [K(u1, ..., un) : K(u1, ..., un−1)]·[K(u1, ..., un−1) : K]
is finite by induction. ¤
Proposition 0.6. Suppose that we have a tower of fields K ⊂ F ⊂ E.
Then E is algebraic over K if and only if E is algebraic over F and F is algebraic over K.
Sketch of the proof. We will only prove that E is algebraic over F and F is algebraic over K implies that E is algebraic over K. The remaining statement are easy.
Pick any u ∈ E. Since u is algebraic over F , let f (x) = P
aixi be the minimal polynomial of u over F .
We then consider the field F0 := K(a0, ..., an). It’s clear that u satisfies a polynomial f (x) ∈ F0. It follows that u ∈ F0(u) which is finite over K. Therefore, u is algebraic over K. ¤ Let L/K and M/K are extensions over K and both L, M are con- tained in a field F . We denote by LM the smallest subfield containing both L and M. LM is called the compositum of L and M.
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A useful remark is that if L = K(S) for some S ⊂ L, then LM = M(S).
For a certain property of field extension, denoted C, we are interested whether C is preserved after extension, lifting or compositum. More precisely, we would like to know a property C satisfying the following conditions:
(1) (extension) Both E/F and F/K are C if and only if E/K is C.
(2) (lifting/ base change) If E/K is C, then EF/F is C.
(3) (compositum) If both E/K, F/K are C, then EF/K is C.
Proposition 0.7. The property of being finite or algebraic satisfying the above three.
Sketch of the proof. It’s easy to that being finite and finitely generated satisfies the above three statement. Hence so does being algebraic. ¤
