Advanced Algebra I
field extensions and algebraic closure
In this section, we are going to prove the existence and uniqueness of algebraic closure. As a consequence, we are able to show the existence and uniqueness of splitting fields.
Proposition 0.1. Let F be a field. The following are equivalent:
(1) Every polynomial of F [x] of degree ≥ 1 has a root in F . (2) Every polynomial of F [x] of degree ≥ 1 has all the roots in F . (3) Every irreducible polynomial in F [x] has degree ≤ 1
(4) If E is an algebraic extension over F , then E = F .
(5) There is a subfield K ⊂ F such that F is algebraic over K and every polynomial in K[x] splits in F [x].
Definition 0.2. A field F satisfying above conditions is said to be algebraically closed.
Sketch of the proof of the Proposition. (1) ⇒ (2) by induction on de- gree. And hence (1) ⇔ (2) are equivalent. It’s easy to see that (2) ⇔ (3). We now look at (3) and (4). If E is an algebraic ex- tension. Pick u ∈ E algebraic over F with minimal polynomial p(x).
By (3), p(x) has degree 1, hence [E : F ] = deg(p(x)) = 1. In particular, E = F . Conversely, if there is an irreducible polynomial p(x) of degree
> 1, then K[x]/(p(x)) gives an algebraic extension of degree deg(p(x)).
This leads to a contradiction, hence (4) implies (3).
Lastly, it’s clear that (3) implies (5) by picking K = F . We now prove that (5) ⇒ (4). Let E be an algebraic extension over F . For any u ∈ E, u is algebraic over K as well. Let pF(x), pK(x) be the minimal polynomial of u over F, K respectively. By viewing pK(x) as a polynomial in F , then one has pF(x)|pK(x) ∈ F [x]. However, pK(x) splits in F [x]. It follows that pF(x) has degree 1. And hence u ∈ F .
Thus E = F . ¤
We can also define the notion of algebraic closure.
Proposition 0.3. Let F/K be an extension. The following are equiv- alent.
(1) F/K is algebraic, and F is algebraically closed.
(2) F/K is algebraic, and every polynomial in K[x] splits in F [x].
(3) F is a splitting field of all polynomials of K.
Definition 0.4. F is said to be an algebraical closure of K if F/K satisfies the above conditions.
Proof. The proof is an easy consequence of the Prop. 0.1, we leave it
to the readers. ¤
Theorem 0.5. Algebraic closure exists.
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The following is due to M. Artin as it appeared in [Lang, Algebra].
Proof. Let K be a field.
Step 1. There is an extension E1over K such that every polynomial of degree ≥ 1 has a root in E1.
To this end, let S be the set of all polynomials of degree ≥ 1. We consider K[S] to be the polynomial ring with indeterminates xf, for f ∈ S. Consider now an ideal I =< f (xf) >f ∈S. We claim that I 6= K[S], hence I ⊂ m for some maximal ideal m. The field K[S]/m gives an extension E1 over K. Now, for every f (x) ∈ K[x], one sees that f (xf) = f (xf) = 0 ∈ E. Hence f (x) has a root xf in E1.
It remains to show that I 6= K[S]. Suppose on the contrary that I = K[S], in particular, 1 ∈ I. We may write
1 = Xr
i=1
g(X)fi(xfi).
One can construct an algebraic extension F/K such that each fi has a root ui in F . Substitute xfi by ui in F , one has
1 = Xr
i=1
g(X)fi(ui) = 0 ∈ F,
which is the required contradiction.
Step 2. Inductively, one has K = E0 ⊂ E1 ⊂ E2.... Let E = ∪Ei, then E is a field extension over K. And E is algebraically closed.
To see this, for any polynomial f (x) = P
aixi ∈ E[x], ai ∈ Eji for some ji. One can pick J maximal among ji so that ai ∈ EJ for all i. Hence f (x) ∈ EJ. By construction, f (x) has a root in EJ+1, and inductively, f (x) has all its root in EJ+d, where d = deg(f (x)).
Therefore, f (x) has all its root in E.
Step 3. Let Ea := {u ∈ E|u is algebraic over K}. Then Ea is an algebraic closure of K.
It’s an easy exercise to check that Ea is a field extension over K.
We leave it to the readers. It’s also clear that Ea is algebraic over K.
Hence, it suffices to check that Ea is algebraically closed.
To see this, one notices that every polynomial of K[x] splits in E and it follows that every root of K[x] is in Ea. Therefore, one has that every polynomial of K[x] splits in Ea and we are done. ¤ Remark 0.6. Let X be a set of indeterminantes and K[X] be the polynomial ring. Let m be a maximal ideal in K[X]. Then K[X]/m is a field. There is a natural embedding σ : K → K[X]/m by σ(k) = ¯k.
Ones might say that K[X]/m is an ”extension over K”, which is not completely precise cause as a set K 6⊂ K[X]/m. One can make sense of this by consider a field E ∼= K[X]/m and K ⊂ E.
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The field E can be constructed as following: Let Kc := K[X]/m − σ(K) and E = K ∪ Kc. Define on E the addition and multiplication naturally then we are there.
We next work on the uniqueness of algebraic closure. The main ingredient is the following extension theorem.
Theorem 0.7 (Extension theorem). Let σ : K → L be an embedding to an algebraically closed field L. Let E/K be an algebraic extension.
Then one can extend the embedding σ to an embedding ¯σ : E → L.
That is, there is an embedding ¯σ : E → L such that ¯σ|K = σ.
We remark that L is not necessarily an algebraic closure of K. For example, L could be something like K(x), an algebraic closure of K(x).
In order to prove the uniqueness, we need the following useful Lemma.
Sketch of the proof. The staring point is an extension to a simple ex- tension. More precisely, let u ∈ E be algebraic over K with minimal polynomial p(x). Then pσ(x) is an irreducible polynomial in σ(K)[x].
In L, Pick any root v of σ(K)[x]. This is possible since L is alge- braically closed. One claims that there is an isomorphism ( hence an embedding to L)
¯
σ : K(u) → σ(K)(v) ⊂ L extending σ. We leave the detail to the readers.
In order to work on the general case, we apply Zorn’s Lemma to the non-empty P.O. set of fields
S := {(F, τ )|K ⊂ F ⊂ E, τ : F → L, τ |K = σ}.
The ordering is given naturally as: (F1, τ1) ≤ (F2, τ2) if F1 ⊂ F2 and τ1 = τ2|F1.
By Zorn’s Lemma, there is a maximal element, say Em. It’s easy to see that Em = E. Otherwise, pick any u ∈ E, which is algebraic over K and hence over Em. There is an extension to Em(u) as we have seen in the first paragraph. This is a contradiction to the maximality of Em.
Hence Em = E. ¤
Lemma 0.8. Let E/K be an algebraic extension and σ : E → E be an embedding such that σ|K = 1K. Then σ is an isomorphism.
Proof. If E/K is finite, then injective implies isomorphic in the case of finite dimensional vector space.
In general, let’s pick any u ∈ E. It suffices to show that u is in the image of σ. To see this, let p(x) be the minimal polynomial of u over K and u = u1, u2, ..., ur be the roots of p(x) in E. Let E0 := K(u1, ..., ur).
It’s clear that for each i, σ(ui) = uj for some j. Hence σ|E0 gives an homomorphism from E0 to E0.
Now σ|E0 : E0 → E0 is an injective homomorphism of finite dimen- sional vector space E0/K. Therefore, σ|E0 is an isomorphism. In par- ticular, u is in the image of σ|E0 and therefore in the image of σ. ¤
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Corollary 0.9. Algebraic closure of a field is unique up to isomor- phism.
Proof. Suppose that E, F are algebraic closure of K. By the extension theorem, there are embedding σ : E → F and τ : F → E such that σ|K = τ |K = 1K.
Hence one has an embedding σ ◦τ : F → F , which is an isomorphism by the Lemma. Similarly, τ ◦ σ is an isomorphism. Hence E and F are
isomorphic. ¤
