305 -Extensions by radicals (3)

3. Examples

Last lecture we defined what it means that a polynomial with coefficients in a field {{\mathbb F}} is solvable by radicals over {{\mathbb F}}. Namely, there is a tower of extensions

\displaystyle {\mathbb F}(t_1,\dots,t_k):{\mathbb F}(t_1,\dots,t_{k-1}):\dots:{\mathbb F}(t_1):{\mathbb F}

where for each {j,} {1\le j\le k,} there is a positive integer {m_j} such that {t_j^{m_j}\in{\mathbb F}(t_1,\dots,t_{j-1}),} and such that {{\mathbb F}^{p(x)}\subseteq{\mathbb F}(t_1,\dots,t_k).}

An obvious class of examples comes from considering polynomials of the form {p(x)=x^n-r\in{\mathbb Q}[x].}

  1. If {r=0,} then {{\mathbb Q}^{p(x)}={\mathbb Q}.}
  2. If {r>0,} the roots of {p(x)} are all the numbers of the form {\root n\of r\zeta_n^l,} {0\le l<n,} where {\zeta_n=\cos(2\pi/n)+i\sin(2\pi/n).} (Remember the convention that if {0<r} then {\root n\of r} denotes the unique positive real whose {n}-th power is {r.}) One obvious way of adding the roots of {p(x)} to {{\mathbb Q}} consists in first adding {w=\root n\of r,} and then adding {\zeta_n.} Clearly, any field containing these two numbers will contain all the numbers of the form {w\zeta_n^l,} so this will indeed give us an extension containing {{\mathbb Q}^{p(x)}.} This shows that {p(x)} is solvable by radicals, since we can form the tower

    \displaystyle {\mathbb Q}(t_1,t_2):{\mathbb Q}(t_1):{\mathbb Q},

    where {t_1=w,} so {t_1^n\in{\mathbb Q}} and we can take {m_1=n,} and {t_2=\zeta_n,} so {t_2^n=1\in{\mathbb Q}(t_1),} so we can take {m_2=n.} Since {{\mathbb Q}^{p(x)}\subseteq{\mathbb Q}(t_1,t_2),} we are done.

  3. If {r<0,} the roots of {p(x)} are the numbers of the form {\root n\of {|r|}\zeta_{2n}^{2l+1},} {0\le l<n,} where {\zeta_{2n}=\cos(\pi/n)+i\sin(\pi/n).} This is because all the numbers {\zeta_{2n}^{2l+1},} {0\le l<n,} are different, since they are {2n}-th roots of unity, and {(\zeta_{2n}^{2l+1})^n=\cos(n(2l+1)\pi/n)+i\sin(n(2l+1)\pi/n)} {=\cos((2l+1)\pi)+i\sin((2l+1)\pi)=-1.} Again, {{\mathbb Q}^{p(x)}} is an extension by radicals, as witnessed by the tower

    \displaystyle {\mathbb Q}(t_1,t_2):{\mathbb Q}(t_1):{\mathbb Q},

    where {t_1=\root n\of{|r|},} {m_1=n,} and {t_2=\zeta_{2n},} {m_2=n.}

Proposition 1 If {p(x)\in{\mathbb Q}[x]} is a quadratic polynomial, then {p(x)} is solvable by radicals over {{\mathbb Q}.}

Proof: Let {p(x)=ax^2+bx+c,} where {a,b,c\in{\mathbb Q}} and {a\ne0.} Let {t=\sqrt{b^2-4ac}.} Remember our convention that if {b^2-4ac<0,} this is the number {i\sqrt{4ac-b^2}.}

Consider the tower

\displaystyle {\mathbb Q}(t):{\mathbb Q}.

Note that {{\mathbb Q}^{p(x)}\subseteq {\mathbb Q}(t),} since the roots of {p(x)} are the numbers {(-b\pm t)/(2a).} Also, this is an extension by radicals, since {t^2\in{\mathbb Q}.} \Box

Proposition 2 If {s(x)\in{\mathbb Q}[x]} is a cubic polynomial, then {s(x)} is solvable by radicals over {{\mathbb Q}.}

Proof: The argument is similar to the previous one. We use Cardano’s method rather than the quadratic formula, see lecture 2.1. I briefly go through the steps:

Say that {s(x)=\alpha x^3+\beta x^2+\gamma x+\delta,} where {\alpha,\dots,\delta\in{\mathbb Q}} and {\alpha\ne0.}

Let {s_1(x)=x^3+ax^2+bx+c,} where {a=\beta/\alpha,} {b=\gamma/\alpha} and {c=\delta/\alpha,} so {s_1(x)\in{\mathbb Q}[x].} Since the roots of {s(x)} and the roots of {s_1(x)} are the same, it suffices to show that {s_1(x)} is solvable by radicals.

Let {s_2(x)=x^3+px+q,} where {\displaystyle p=b-\frac{a^2}3} and {\displaystyle q=\frac{2a^3}{27}-\frac{ab}3+c.} Notice {s_2(x)\in{\mathbb Q}[x].} Also, for any complex number {r,} {r} is a root of {s_2(x)} iff {\displaystyle r-\frac a3} is a root of {s_1(x).} Since {a\in{\mathbb Q},} any field that contains {r} will contain {\displaystyle r-\frac a3,} and vice versa. It follows that it suffices to show that {s_2(x)} is solvable by radicals.

Let {\displaystyle s_3(x)=x^2+qx-\frac{p^3}{27},} so {s_3(x)\in{\mathbb Q}[x].} As shown in Proposition 1, {s_3(x)} is solvable by radicals and there is a tower {{\mathbb Q}(t_1):{\mathbb Q}} witnessing this, with {m_1=2.} This is to say {t_1^2\in{\mathbb Q}} and {{\mathbb Q}^{s_3(x)}\subseteq{\mathbb Q}(t_1).} (According to the proof of Proposition 1, {\displaystyle t_1=\sqrt{q^2+\frac{4p^3}{27}},} but this is not important right now.)

Let {r} be a root of {s_3(x),} so {r\in{\mathbb Q}(t_1).} Let {w} be a cubic root of {r.} The other two cubic roots of {r} are {w\zeta_3} and {w\zeta_3^2,} where {\zeta_3=\cos(2\pi/3)+i\sin(2\pi/3).} Hence, we can extend our tower to

\displaystyle {\mathbb Q}(t_1,t_2,t_3):{\mathbb Q}(t_1,t_2):{\mathbb Q}(t_1):{\mathbb Q},

where {t_2=w,} {m_2=3,} and {t_3=\zeta_3,} {m_3=3.}

Finally, notice that the roots of {s_2(x)} are the three numbers {\displaystyle w-\frac p{3w},} {\displaystyle w\zeta_3-\frac p{3w\zeta_3},} and {\displaystyle w\zeta_3^2-\frac p{3w\zeta_3^2}.} These three numbers belong to {{\mathbb Q}(t_1,t_2,t_3),} and it follows that {{\mathbb Q}^{s_2(x)}\subseteq {\mathbb Q}(t_1,t_2,t_3),} so {s_2(x)} is solvable by radicals and therefore so is {s(x).} \Box

Proposition 3 If {s(x)\in{\mathbb Q}[x]} is a quartic polynomial, then {s(x)} is solvable by radicals over {{\mathbb Q}}.

Proof: This is similar to the previous case, but now we use Ferrari’s method, see lecture 2.2.

Briefly, just as in the previous case, we may assume {s(x)=x^4+px^2+qx+r\in{\mathbb Q}[x].}

Let {s_1(x)=x^3-px^2-4rx+(4pr-q^2)\in{\mathbb Q}[x].} By Proposition 2, {s_1(x)} is solvable by radicals over {{\mathbb Q}} and, in fact, there is a tower

\displaystyle {\mathbb Q}(t_1,t_2,t_3):{\mathbb Q}(t_1,t_2):{\mathbb Q}(t_1):{\mathbb Q},

with {m_1=2,} {m_2=3,} {m_3=3} witnessing this. That is to say, {{\mathbb Q}^{s_1(x)}\subseteq {\mathbb Q}(t_1,t_2,t_3).}

Fix a root {u} of {s_1.} Extend the tower above as

\displaystyle {\mathbb Q}(t_1,\dots,t_5):{\mathbb Q}(t_1,\dots,t_4):{\mathbb Q}(t_1,t_2,t_3),

where {t_4^2=u-p} and {\displaystyle t_5^2=\frac{u^2}4-r.} This is still an extension by radicals, since {u\in{\mathbb Q}(t_1,t_2,t_3).} Also, {\displaystyle t_4t_5=\pm\frac q2.} Using that {-t_5\in{\mathbb Q}(t_1,\dots,t_5),} we may assume that {\displaystyle t_4t_5=-\frac q2,} since otherwise we may replace {t_5} with {t_5'} and now the equation holds.

Then {s(x)} factors as a product of two quadratics in {{\mathbb Q}(t_1,\dots,t_5).} (But these quadratics do not necessarily have coefficients in {{\mathbb Q}(t_1,\dots,t_4),} much less in {{\mathbb Q}.}) Using (the proof of) Proposition 1, we can further extend the tower as

\displaystyle {\mathbb Q}(t_1,\dots,t_7):{\mathbb Q}(t_1,\dots,t_6):{\mathbb Q}(t_1,\dots,t_5),

where {t_6^2,t_7^2\in{\mathbb Q}(t_1,\dots,t_5)} and both quadratics factor in {{\mathbb Q}(t_1,\dots,t_7).} This shows that {{\mathbb Q}^{s(x)}} is solvable by radicals. \Box

Notice that we could have made the tower above slightly shorter, since we did not need all cubic roots of {s_1(x),} any of them would have worked.

We have shown that any polynomial with coefficients in {{\mathbb Q}} of degree at most 4 is solvable by radicals. In fact, the formulas we have for the roots of these polynomials allowed us to find the required tower of extensions in a “uniform” way.

The method above is fairly general. For example, we have the same result if instead of a polynomial over {{\mathbb Q}} we had started with a polynomial in {{\mathbb F}[x]} for any other subfield {{\mathbb F}} of {{\mathbb C}.}

On the other hand, if a polynomial {p(x)\in{\mathbb F}[x]} is not solvable by radicals, then it is certainly the case that there is no “formula” that from the coefficients of the polynomial returns the values of its roots, as long as by formula we understand any expression formed from these coefficients (and other numbers in the field {{\mathbb F}}) by means of the elementary operations and the extraction of {n}-th roots.

A final remark is in order. The method above also works for other fields; for example, for almost all finite fields. There are two problems, though. The first is that to even define {{\mathbb F}^{p(x)}} we need to begin with an extension field of {{\mathbb F}} in which {p(x)} factors. (This field would play the role of {{\mathbb C}}.) Such a field always exists, in fact, a version of the fundamental theorem of algebra holds for any field: For any {{\mathbb F}} there is an extension {{\mathbb K}} of {{\mathbb F}} such that any polynomial with coefficients in {{\mathbb K}} (in particular, any polynomial with coefficients in {{\mathbb F}}) factors in {{\mathbb K}.} However, in this course we do not have the tools to show that this is the case.

The second problem is more subtle. Consider, for example, the quadratic {p(x)=x^2+x+1\in{\mathbb Z}_2[x].} We cannot do as usual to solve the equation {p(r)=0} (i.e., {{\rm eval}(p,r)=0})—not even formally, without worrying about which field the solutions live in.

I mean, we cannot use the quadratic formula to solve this equation. This is because {2=0} in {{\mathbb Z}_2,} so the expression {\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}} makes no sense, since it requires that we divide by 2.

Nevertheless, it turns out that {p(x)} is solvable by radicals over {{\mathbb Z}_2}, because all the nonzero elements of {{\mathbb F}_4,} the field with 4 elements, satisfy the equation {r^3-1=0} and, as mentioned in example 3 following Lemma 4 of the notes for last lecture, we know that the roots of {p(x)} are simply the two new elements of {{\mathbb F}_4.}

Similarly, if we begin with a field {{\mathbb F}} of characteristic 2 or 3, we cannot use Cardano’s method to show that a polynomial {p(x)\in{\mathbb F}[x]} of degree 3 is solvable by radicals over {{\mathbb F},} since this method requires that at some point we divide some element of the field (or of an extension) by 2 and another by 3. The same problem occurs with Ferrari’s method for equations of degree 4.

Unfortunately, it would take us far afield to see whether Propositions 1, 2, and 3 hold for all fields (finite or infinite, and of any characteristic), so I will live this question unanswered.

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