## 305 -5. Extensions by radicals

(This post was typeset using Luca Trevisan‘s LaTeX2WP program.)

Last lecture we characterized subfields and used the characterization to provide many new examples of fields. Now we start to explore systematically which subfields of the complex numbers are suitable to study the question of which polynomial equations can be solved.

1. Extensions

Definition 1 If ${{\mathbb K}}$ is a field and ${{\mathbb F}}$ is a subfield, we write ${{\mathbb K}:{\mathbb F}}$ and also say that ${{\mathbb K}}$ is an extension of ${{\mathbb F}.}$

For example, ${{\mathbb C}:{\mathbb R}:{\mathbb Q}.}$

Definition 2 If ${{\mathbb K}}$ is a field, ${{\mathbb K}:{\mathbb F},}$ and ${S\subset{\mathbb K},}$ we denote by ${{\mathbb F}(S)}$ the smallest subfield of ${{\mathbb K}}$ that contains both ${{\mathbb F}}$ and ${S.}$ If ${S=\{s_1,\dots,s_n\},}$ we write ${{\mathbb F}(s_1,\dots,s_n).}$ We say that ${{\mathbb F}(S)}$ is the subfield of ${{\mathbb K}}$ generated by ${S}$ over ${{\mathbb F}.}$

For example, last lecture we showed that ${{\mathbb Q}(\sqrt 2)=\{a+b\sqrt2 : a,b\in{\mathbb Q}\}.}$

“Smallest” in the definition above refers to containment. Put another way, we are claiming that there is a field ${{\mathbb F}(S)}$ that contains ${{\mathbb F}}$ and ${S}$ and such that given any subfield ${{\mathbb H}}$ of ${{\mathbb K}}$ that also contains ${{\mathbb F}}$ and ${S,}$ then ${{\mathbb H}\supseteq{\mathbb F}(S).}$ This requires an argument.

Theorem 3 Let ${{\mathbb K}}$ be a field. Let ${{\mathcal T}}$ be a nonempty collection of subfields of ${{\mathbb K}.}$ Then ${\bigcap{\mathcal T}=\bigcap\{{\mathbb F}:{\mathbb F}\in{\mathcal T}\}}$ is a field.

Proof: We use the characterization from last lecture to show that ${\bigcap{\mathcal T}}$ is a subfield of ${{\mathbb K}.}$ It is straightforward to check that this set is closed under addition and multiplication (since all the fields in ${{\mathcal T}}$ are), and that it contains additive and multiplicative inverses of all its (nonzero) elements. It also has at least two elements, since ${0,1\in{\mathbb F}}$ for any field ${{\mathbb F}\subseteq{\mathbb K}.}$ $\Box$

Corollary 4 If ${{\mathbb K}}$ is a field, ${{\mathbb K}:{\mathbb F},}$ and ${S\subset{\mathbb K},}$ then ${{\mathbb F}(S)}$ exists.

Proof: Let ${{\mathcal T}}$ be the collection of subfields of ${{\mathbb K}}$ that contain both ${{\mathbb F}}$ and ${S.}$ Note that ${{\mathcal T}\ne\emptyset}$ since ${{\mathbb K}\in{\mathcal T}.}$ Then ${\bigcap{\mathcal T}}$ is a field, it contains both ${{\mathbb F}}$ and ${S,}$ and is contained in any subfield of ${{\mathbb K}}$ that contains both ${{\mathbb F}}$ and ${S,}$ by definition of ${{\mathbb T}.}$ Therefore, ${\bigcap{\mathcal T}={\mathbb F}(S).}$ $\Box$

This is a characterization “from above.” As in the case of ${{\mathbb Q}(\sqrt2),}$ in many concrete examples it is possible to give a simple description (a characterization “from below”) of ${{\mathbb F}(S),}$ but this is in general a difficult problem.

Proposition 5 Let ${{\mathbb F}}$ be a subfield of ${{\mathbb C}.}$ Then ${{\mathbb F}:{\mathbb Q}.}$

Proof: We already know that ${0,1\in{\mathbb F}.}$ By induction, ${n\in{\mathbb F}}$ for all ${n\in{\mathbb N},}$ since ${{\mathbb F}}$ is closed under addition. Since ${{\mathbb F}}$ is closed under additive inverses, ${{\mathbb Z}\subseteq{\mathbb F}.}$ Since ${{\mathbb F}}$ is closed under multiplicative inverses (of nonzero elements) and under multiplication, then ${{\mathbb Q}\subseteq{\mathbb F}.}$ $\Box$

Recall Gauß’ fundamental theorem of algebra, that we will take for granted in what follows. (That we take it for granted does not mean that this is an easy result. All its known proofs use deep facts about ${{\mathbb C},}$ either about its algebraic structure or its analytic structure.)

Theorem 6 (Gauß) Let ${p(x)}$ be a nonconstant polynomial with complex coefficients. Then ${p}$ has at least one complex root, i.e., there is some ${r\in{\mathbb C}}$ such that ${p(r)=0.}$ By induction, it follows that if ${p}$ has degree ${n,}$ then it has exactly ${n}$ roots, taking into account their multiplicity. ${\Box}$

Definition 7 Let ${{\mathbb F}}$ be a subfield of ${{\mathbb C}}$ and let ${p(x)}$ be a nonconstant polynomial with coefficients in ${{\mathbb F}}$. Then ${{\mathbb F}^{p(x)}}$ denotes the smallest subfield of ${{\mathbb C}}$ that contains ${{\mathbb F}}$ and the roots of ${p.}$

We will usually use this notation with ${{\mathbb Q}}$ in place of ${{\mathbb F}.}$ In this case, thanks to Proposition 5, we have that ${{\mathbb Q}^{p(x)}}$ is simply the smallest subfield of ${{\mathbb C}}$ that contains the roots of ${p.}$

For example, if ${p(x)=x^2-2,}$ then ${{\mathbb Q}^{p(x)}={\mathbb Q}(\sqrt2).}$

Clearly, ${{\mathbb F}^{p(x)}:{\mathbb F}.}$ This extension has several interesting properties. For example, it is a vector space over ${{\mathbb F}.}$ Next lecture we will study the important case when ${{\mathbb F}^{p(x)}}$ is an extension by radicals.

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