User:Tripler06/sandbox

Splitting Fields
Let $$F$$ be a field and $$p(x)$$ be a nonconstant polynomial in $$F[x]$$. We already know that we can find a field extension of $$F$$ that contains a root of $$p(x)$$. However, we would like to know whether an extension $$E$$ of $$F$$ containing all of the roots of $$p(x)$$ exists. In other words, can we find a field extension of $$F$$ such that $$p(x)$$ factors into a product of linear polynomials? What is the "smallest" extension containing all the roots?

Existence of Splitting Fields
We will see that rather than looking at arbitrary field extension, splitting fields will be the things to consider. First we need to know that they always exist.Proof. This is a largely uninteresting case of proof by induction. We will induct on the degree of $$f(x)$$. If $$f(x)$$ is linear, then clearly its roots (in fact just the one root) is contained in $$F$$ so $$F$$ itself is a splitting field. Suppose $$\deg(f(x)) > 1$$. If $$f(x)$$ splits into the product of linear terms, then again all the roots are contained in $$F$$, so we already have a splitting field. So suppose $$f(x)$$ has an irreducible factor of degree at least 2. Then there exists a field extension $$E_1$$ containing a root $$\alpha$$ of $$f(x)$$. Then in $$E_1$$, we can factorise the polynomial into $$f(x) = (x - \alpha)f_1(x)$$ where $$f_1(x)$$ is a polynomial of degree $$\deg(f(x)) - 1$$. Then by induction there exists $$E_2$$ a field extension of $$E_1$$ that is a splitting field of $$f_1(x)$$. Therefore $$E_2$$ is a field extension of $$F$$ that contains all the roots of $$f(x)$$. Taking the intersection of all subfields of $$E_2$$ containing $$F$$ and the roots of $$f(x)$$ gives us $$E$$, a splitting field of $$f(x)$$. $$\blacksquare$$

Uniqueness of Splitting Fields
Above we were careful to say a splitting field of $$f(x)$$. In fact, this was an unnecessary precaution since the splitting field of a polynomial is unique up to isomorphism. This follows from a generalisation of Theorem 4.1.4, where we claim the theorem holds even if we adjoin all the roots of the polynomial, instead of just one.Proof. This is once again a proof by induction on the degree of $$f(x)$$. If $$f(x)$$ is of degree 1 or indeed splits into factors of degree 1 then the splitting field of $$f(x)$$ is $$F$$ so we can take $$\sigma = \varphi$$. Thus suppose $$f(x)$$ has an irreducible factor $$p(x)$$ of degree at least 2 so $$\widetilde{p}(x) = \varphi_*(p(x))$$is an irreducible factor of $$\widetilde{f}(x)$$. Then by the previous theorem we know $$\varphi$$ extends to an isomorphism $$\psi: F(\alpha) \to \widetilde{F}(\beta)$$ where $$\alpha$$ is a root of $$p(x)$$ and $$\beta$$ is a root of $$\widetilde{p}(x)$$. Therefore over $$F(\alpha)$$ and $$\widetilde{F}(\beta)$$ respectively we can write $$f(x) = (x - \alpha) f_1(x)$$ and $$\widetilde{f}(x) = (x - \beta)\widetilde{f}_1(x)$$. Notice that $$E$$ is a splitting field of $$f_1(x)$$ over $$F(\alpha)$$. Indeed if a splitting field was strictly contained within $$E$$, then it would contain all the roots of $$f_1(x)$$ and $$\alpha$$ and hence would contain all the roots of $$f(x)$$. But this would contradict $$E$$ being a splitting field of $$f(x)$$. Of course the same holds true for $$\widetilde{f}_1(x)$$ over $$\widetilde{F}(\beta)$$. Since $$f_1(x)$$ and $$\widetilde{f}_1(x)$$ have degree strictly less than $$\deg(f(x))$$, by induction we can assume that the statement of theorem holds for them. In particular, $$\psi$$ extends to an isomorphism $$\sigma: E \to \widetilde{E}$$. But since $$\psi$$ was an extension of $$\varphi$$, $$\sigma$$ must also be an extension of $$\varphi$$ concluding the proof. $$\blacksquare$$Proof. Apply Theorem 4.1.7 to the case with $$F = \widetilde{F}$$ and $$\varphi$$ as the identity map. $$\blacksquare$$

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Example 1: Let $$p(x) = x^4+2x^2-8$$ be in $$\mathbb{Q}[x]$$. Then p(x) has irreducible factors $$x^2-2$$ and $$x^2+4$$. Therefore, the field $$\mathbb{Q}(\sqrt{2},i)$$ is a splitting field for p(x).

Example 2: Let $$p(x) = x^3-3$$ be in $$\mathbb{Q}[x]$$. Then p(x) has a root in the field $$\mathbb{Q}(\sqrt[3]{3})$$. However, this field is not a splitting field for p(x) since the complex cube roots of 3, $$\frac{-\sqrt[3]{3}\pm(\sqrt[6]{3})^5i}{2}$$ are not in $$\mathbb{Q}(\sqrt[3]{3})$$.

Algebraic Closures
Given a field F, the question arises as to whether or not we can find a field E such that every polynomial p(x) has a root in E. This leads us to the following theorem.

Theorem 21.11 Let E be an extension field of F. The set of elements in E that are algebraic over F form a field.

Proof. Let $$\alpha,\beta\in E$$ be algebraic over F. Then $$F(\alpha,\beta)$$ is a finite extension of F. Since every element of $$F(\alpha,\beta)$$ is algebraic over $$F,\alpha\pm\beta,\alpha\beta$$, and $$\alpha/\beta\text{ }(\beta\ne 0)$$ are all algebraic over F. Consequently, the set of elements in E that are algebraic over F forms a field.

Corollary 21.12 The set of all algebraic numbers forms a field; that is, the set of all complex numbers that are algebraic over $$\mathbb{Q}$$ makes up a field.

Let E be a field extension of a field F. We define the algebraic closure of a field F in E to be the field consisting of all elements in E that are algebraic over F. A field F is algebraically closed if every nonconstant polynomial in F[x] has a root in F.

Theorem 21.13 A field F is algebraically closed if and only if every nonconstant polynomial in F[x] factors into linear factors over F[x].

Proof. Let F be an algebraically closed field. If $$p(x)\in F[x]$$ is a nonconstant polynomial, then p(x) has a zero in F, say α. Therefore, $$x-\alpha$$ must be a factor of p(x) and so $$p(x)=(x-\alpha)q_1(x)$$, where $$deg(q_1(x)) = deg(p(x))-1$$. Continue this process with $$q_1(x)$$ to find a factorization

$$p(x)=(x-\alpha)(x-\beta)q_2(x)$$, where $$deg(q_2(x))= deg(p(x))-2$$. The process must eventually stop since the degree of p(x) is finite.

Conversely, suppose that every nonconstant polynomial p(x) in F[x] factors into linear factors. Let $$ax-b$$ be such a factor. Then $$p(b/a)=0$$. Consequently, F is algebraically closed.

Corollary 21.14 An algebraically closed field F has no proper algebraic extension E.

Proof. Let E be an algebraic extension of F; then $$F\subset E$$. For $$\alpha\in E$$, the minimal polynomial of α is $$x-\alpha$$. Therefore, $$\alpha\in F$$ and $$F=E$$.

Theorem 21.15 Every field F has a unique algebraic closure.

It is a nontrivial fact that every field has a unique algebraic closure. The proof is not extremely difficult, but requires some rather sophisticated set theory. We refer the reader to [3], [4], or [8] for a proof of this result.

We now state the Fundamental Theorem of Algebra, first proven by Gauss at the age of 22 in his doctoral thesis. This theorem states that every polynomial with coefficients in the complex numbers has a root in the complex numbers. The proof of this theorem will be given in Abstract Algebra/Galois Theory.

Theorem 21.16 (Fundamental Theorem of Algebra) The field of complex numbers is algebraically closed.