## 305 -Extensions by radicals (2)

Last lecture we defined ${{mathbb F}^{p(x)}}$ where ${{mathbb F}}$ is a subfield of a field ${{mathbb K},}$ all the roots of the polynomial ${p(x)}$ are in ${{mathbb K},}$ and all the coefficients of ${p(x)}$ are in ${{mathbb F}.}$ Namely, if ${r_1,dots,r_n}$ are the roots of ${p,}$ then ${{mathbb F}^{p(x)}={mathbb F}(r_1,dots,r_n),}$ the field generated by ${r_1,dots,r_n}$ over ${{mathbb F}.}$

The typical examples we will consider are those where ${{mathbb F}={mathbb Q},}$ ${{mathbb K}={mathbb C},}$ and the coefficients of ${p(x)}$ are rational or in fact, integers.

For example, if ${p(x)=x^2+2x+2,}$ then ${{mathbb Q}^{p(x)}={mathbb Q}(i),}$ because the roots of ${p}$ are ${-1pm i}$ and ${{mathbb Q}(-1+i,-1-i)={mathbb Q}(i).}$ To see this last equality, notice:

1. ${{mathbb Q}(-1+i,-1-i)}$ contains ${{mathbb Q};}$
2. ${{mathbb Q}(-1+i,-1-i)}$ contains ${i,}$ since ${i=(-1+i)+1,}$ both ${-1+i}$ and ${1}$ are in ${{mathbb Q}(-1+i,-1-i),}$ and ${{mathbb Q}(-1+i,-1-i)}$ is closed under addition.

From this it follows that ${{mathbb Q}(-1+i,-1-i)}$ contains ${{mathbb Q}(i)}$ since, as shown in the proof of Corollary 4 from last lecture , ${{mathbb Q}(i)}$ is contained in any field that contains both ${{mathbb Q}}$ and ${i.}$

Conversely:

1. ${{mathbb Q}(i)}$ contains ${{mathbb Q};}$
2. ${{mathbb Q}(i)}$ contains ${-1+i}$ since it contains both ${-1}$ and ${i}$ and is closed under addition, and it also contains ${-1-i}$ since ${-iin{mathbb Q}(i),}$ because any field is closed under additive inverses.

From this it follows that ${{mathbb Q}(i)}$ is a field that contains ${{mathbb Q}}$ and the numbers ${-1pm i,}$ so it contains the field ${{mathbb Q}(-1+i,-1-i),}$ and we are done.

If ${{mathbb F}}$ is a field, the collection of polynomials with coefficients in ${{mathbb F}}$ is an important example of a mathematical structure that we have found before:

Definition 1 A commutative ring is a structure ${(R,+,times,0,1)}$ that satisfies all the field axioms except (perhaps) for the existence of multiplicative inverses of nonzero elements.

See lecture 4.1 for the definition of field.

For example, ${{mathbb Z}}$ is a commutative ring, as are all the sets ${{mathbb Z}_n}$ for ${nin{mathbb Z}^+.}$ Any field is a commutative ring, of course.

There are also noncommutative rings, and we will encounter them later.

Definition 2 Let ${{mathbb F}}$ be a field. Then ${{mathbb F}[x]}$ is the set of polynomials with coefficients in ${{mathbb F}.}$

We think of polynomials as purely formal expressions (rather than functions), so two polynomials are equal iff their coefficients are equal.

We turn ${{mathbb F}}$ into a commutative ring by adding and multiplying polynomials as usual: If ${p(x)=a_0+a_1x+dots+a_nx^n}$ and ${q(x)=b_0+b_1x+dots+b_nx^n}$ where ${a_0,dots,a_n,b_0,dots,b_nin{mathbb F}}$ (we allow for the possibility that ${a_n-0}$ or ${b_n=0}$), then

$displaystyle (p+q)(x)=sum_{j=0}^n(a_j+b_j)x^j$

and

$displaystyle (pq)(x)=sum_{j=0}^{2n}left(sum_{k=0}^ja_kb_{j-k}right)x^j.$

The zero element of ${{mathbb F}[x]}$ is the polynomial with all its coefficients equal to zero, and the one element is the polynomial whose constant coefficient is 1 and all its other coefficients are zero.

Notice that ${x}$ in the definition above is a “variable,” not an element of ${{mathbb F},}$ and that the definition is purely formal, i.e., a polynomial is not a function.

There are several ways of making the above rigorous, similar to the way we identified ${{mathbb C}}$ and ${{mathbb R}times{mathbb R}.}$ For example, we could identify ${{mathbb F}[x]}$ with a subset ${{mathbb F}^{mathbb N},}$ the set of infinite sequences ${(a_0,a_1,dots)}$ of elements of ${{mathbb F},}$ namely, ${{mathbb F}[x]}$ is the set of all those sequences such that ${a_j=0}$ for all but finitely many values of ${j.}$ Of course, we think of ${(a_0,a_1,dots)}$ as ${a_0+a_1x+a_2x^2+dots}$ In this approach there is not even a variable ${x}$ anywhere. This extra layer of rigor is not needed in what follows, so we can safely ignore it.

It is important, however, to keep in mind that two polynomials are equal iff their coefficients are equal. This is not the same as saying that ${p(x)=q(x)}$ if and only if, when evaluated in the elements of ${{mathbb F},}$ ${p}$ and ${q}$ take the same values. For example, let ${{mathbb F}={mathbb Z}_2}$ and let ${p(x)=x+1}$ and ${q(x)=x^2+1.}$ Then, as functions, ${p=q}$ since ${p(0)=1=q(0)}$ and ${p(1)=0=q(1).}$ However, as polynomials, ${pne q}$ since the coefficients of ${p}$ are (in increasing order of degree) ${1,1,0,0,dots}$ while the coefficients of ${q}$ are ${1,0,1,0,0,dots}$ and these two sequences are different.

Since we also want to consider polynomials as functions, we introduce the evaluation map:

Definition 3 Let ${{mathbb K}:{mathbb F}.}$ The evaluation map from polynomials over ${{mathbb F}}$ to elements of ${{mathbb K}}$ is the function ${{rm eval}:{mathbb F}[x]times{mathbb K}rightarrow{mathbb K}}$ given by ${{rm eval}(p(x),r)=p(r).}$ We will usually just use the notation ${p(r)}$ without mentioning the function ${{rm eval}}$ unless the distinction is important at the moment.

Lemma 4 Let ${p,q}$ be polynomials in ${{mathbb C}[x].}$ Let ${n}$ be the largest of their degrees. If ${p(r)=q(r)}$ for more than ${n}$ many complex numbers ${r,}$ then ${p=q}$ as polynomials. ${Box}$

I’ll leave the proof of Lemma 4 as an exercise. As a follow up exercise, it is useful to think whether the same result is true if a finite field ${{mathbb F}}$ is used instead of ${{mathbb C}.}$

Since we are interested in the question of whether there is a formula for the solutions of a polynomial equation, i.e., whether there is a formula that identifies from a polynomial ${p(x)}$ the numbers ${r}$ that are solutions of ${{rm eval}(p,r)=0,}$ the study of field extensions is important, since the roots of a polynomial with coefficients in a field ${{mathbb F}}$ will typically not be elements of ${{mathbb F}.}$ For example:

1. Let ${p(x)in{mathbb Q}[x]}$ be the polynomial ${p(x)=x^2-2.}$ Its roots are ${pmsqrt2}$ which are irrational numbers.
2. Let ${p(x)in{mathbb R}[x]}$ be the polynomial ${p(x)=x^2+1.}$ Its roots are ${pm i.}$
3. Let ${p(x)=x^2+x+1in{mathbb Z}_2[x].}$ This polynomial has no roots in ${{mathbb Z}_2,}$ since ${p(0)=p(1)=1.}$ Recall that in lecture 4.3 we constructed a field ${{mathbb F}_4}$ of 4 elements. Using the same notation we used there, ${{mathbb F}_4={0,1,a,b},}$ where ${a+1=a^2=b}$ (this easily determines all the other products and sums in ${{mathbb F}_4.}$) Notice that ${p(a)=a^2+a+1=(a+1)+(a+1)=0}$ and ${p(b)=b^2+b+1=(a+1)^2+(a+1)+1}$ ${=(a^2+1)+(a+1)+1=p(a)+1+1=0,}$ so the roots of ${p}$ are ${a}$ and ${b}$ (which do not belong to ${{mathbb Z}_2.}$)

The extensions that are most relevant to us in the context of this question are those that are obtained by radicals, since the formulas we are looking for (or trying to show that do not exist) are those obtained from the coefficients of ${p(x)in{mathbb F}[x]}$ and the elements of ${{mathbb F}}$ by means of ${+,-,times,div}$ and the extraction of ${n}$-th roots of elements of ${{mathbb F}.}$

We need to be careful here, since there are ${n}$ possible choices for the ${n}$-th roots of a number ${r.}$ For example, if ${r}$ is a complex number (even if ${r}$ is real), the ${n}$-th roots of ${r}$ are the roots of the polynomial ${p(x)=x^n-r,}$ i.e., the solutions of ${{rm eval}(p,r)=0.}$ If ${r=|r|(costheta+isintheta),}$ then these solutions are given by

$displaystyle w=root nof{|r|}(cos(theta/n)+isin(theta/n)),wzeta_n,wzeta_n^2,dots,wzeta_n^{n-1},$

where ${displaystyle zeta_n=cosleft(frac{2pi}nright)+isinleft(frac{2pi}nright).}$

As this shows, it is more accurate to say that ${(wzeta_n^j)^n=r}$ that to claim that ${root nof r=wzeta_n^j,}$ although we do refer to ${wzeta_n^j}$ as an ${n}$-th root of ${r.}$ We will continue with the convention that ${root nof s}$ is only used when ${s}$ is a nonnegative real number, in which case ${root nof s}$ denotes the unique nonnegative real whose ${n}$-th power is ${s.}$

(This is necessary in order to be precise, but slightly annoying, since we cannot even write the quadratic formula the way we are used to. If we ever abuse language and write ${sqrt r}$ for some negative ${rin{mathbb R},}$ we mean ${isqrt{-r}.}$)

Definition 5 Suppose ${{mathbb F}}$ is a field, ${p(x)in {mathbb F}[x],}$ and ${{mathbb K}:{mathbb F}^{p(x)},}$ i.e., all the roots of ${p(x)}$ are in ${{mathbb K}}$ and ${{mathbb K}}$ is an extension field of ${{mathbb F}.}$

We say that the equation ${p(x)=0}$ (more precisely, ${{rm eval}(p,r)=0}$) is solvable by radicals over ${{mathbb F}}$ iff there are numbers ${t_1,dots,t_k}$ in ${{mathbb K}}$ and (not necessarily distinct) positive integers ${m_1,dots,m_k}$ such that

1. ${{mathbb F}^{p(x)}subseteq {mathbb F}(t_1,dots,t_k),}$
2. ${t_1^{m_1}in{mathbb F},}$ and
3. For all ${j}$ with ${1 ${t_j^{m_j}in {mathbb F}(t_1,dots,t_{j-1}).}$

We also say that ${{mathbb F}^{p(x)}}$ is an extension by radicals, although this is perhaps more accurate of ${{mathbb F}(t_1,dots,t_k).}$

The intention of this definition is that (as one can easily verify) all the elements of a field extension ${{mathbb H}(s)}$ are of the form ${q(s)/t(s)}$ for some polynomials ${q(x),t(x)in{mathbb H}[x],}$ and if ${s}$ is an ${n}$-th root of some element of ${{mathbb H},}$ then certainly any such expression ${q(s)/t(s)}$ qualifies as a “formula by radicals.” More generally, the same applies to any element of ${{mathbb F}(t_1,dots,t_k)}$ as in Definition 5.

For example, consider ${p(x)=x^4+x^2+2in{mathbb Q}[x].}$ Let ${rin{mathbb C}}$ be a root of ${p.}$ Then ${(r^2)^2+(r^2)+2=0,}$ so ${r^2=-1pm i,}$ and ${r}$ is a square root of either ${-1+i}$ or ${-1-i.}$ It follows that ${{mathbb Q}^{p(x)}}$ is an extension by radicals, since one can form the following tower of extensions:

1. ${{mathbb Q}(i):{mathbb Q},}$
2. ${{mathbb Q}(i,r_1)=({mathbb Q}(i))(r_1):{mathbb Q}(i),}$ where ${r_1}$ is a square root of ${-1+i,}$
3. ${{mathbb Q}(i,r_1,r_2)=({mathbb Q}(i,r_1))(r_2):{mathbb Q}(i,r_1),}$ where ${r_2}$ is another square root of ${-1+i,}$
4. ${{mathbb Q}(i,r_1,r_2,r_3):{mathbb Q}(i,r_1,r_2),}$ where ${r_3}$ is a square root of ${-1-i,}$ and
5. ${{mathbb Q}(i,r_1,r_2,r_3,r_4):{mathbb Q}(i,r_1,r_2,r_3),}$ where ${r_4}$ is another square root of ${-1-i.}$

Notice that since ${{mathbb Q}^{p(x)}={mathbb Q}(r_1,r_2,r_3,r_4),}$ then certainly ${{mathbb Q}^{p(x)}subseteq {mathbb Q}(i,r_1,r_2,r_3,r_4).}$ Moreover, this example satisfies Definition 5 with ${t_1=i,}$ ${t_2=r_1,}$ ${t_3=r_2,}$ ${t_4=r_3,}$ and ${t_5=r_4,}$ and ${m_1=dots=m_5=2.}$

Notice also that we needed to add ${i}$ at the beginning, since otherwise we could not add square roots of ${-1pm i,}$ since these elements ${-1pm i}$ are not even present in our field ${{mathbb Q}.}$

This example also illustrates that we can witness that a field is an extension by radicals in more ways than one, since ${(-1+i)^2=-2i,}$ and ${(-1-i)^2=2i,}$ so ${(-1pm i)^4=-4,}$ so we could just as well begin by first adding the fourth roots of -4, which would put ${-1pm i}$ into our field, and then adding the square roots of these elements.

Another point to notice is that we did not take particular care in finding a tower of extensions of minimal length or anything like that, and we do not need at the end to check whether or not ${{mathbb F}(t_1,dots,t_k)}$ and ${{mathbb F}^{p(x)}}$ coincide, since all we need is containment rather than equality.

Typeset using LaTeX2WP. Here is a printable version of this post.