You are here
HomeGaloistheoretic derivation of the quartic formula
Primary tabs
Galoistheoretic derivation of the quartic formula
Let $x^{4}+ax^{3}+bx^{2}+cx+d$ be a general polynomial with four roots $r_{1},r_{2},r_{3},r_{4}$, so $(xr_{1})(xr_{2})(xr_{3})(xr_{4})=x^{4}+ax^{3}+bx^{2}+cx+d$. The goal is to exhibit the field extension $\mathbb{C}(r_{1},r_{2},r_{3},r_{4})/\mathbb{C}(a,b,c,d)$ as a radical extension, thereby expressing $r_{1},r_{2},r_{3},r_{4}$ in terms of $a,b,c,d$ by radicals.
Write $N$ for $\mathbb{C}(r_{1},r_{2},r_{3},r_{4})$ and $F$ for $\mathbb{C}(a,b,c,d)$. The Galois group $\operatorname{Gal}(N/F)$ is the symmetric group $S_{4}$, the permutation group on the four elements $\{r_{1},r_{2},r_{3},r_{4}\}$, which has a composition series
$1\lhd\mathbb{Z}/2\lhd V_{4}\lhd A_{4}\lhd S_{4},$ 
where:

$A_{4}$ is the alternating group in $S_{4}$, consisting of the even permutations.

$V_{4}=\{1,(12)(34),(13)(24),(14)(23)\}$ is the Klein fourgroup.

$\mathbb{Z}/2$ is the two–element subgroup $\{1,(12)(34)\}$ of $V_{4}$.
Under the Galois correspondence, each of these subgroups corresponds to an intermediate field of the extension $N/F$. We denote these fixed fields by (in increasing order) $K$, $L$, and $M$.
We thus have a tower of field extensions, and corresponding automorphism groups:
$\xymatrix{\text{Subgroup}&\text{Fixed field}\\ 1&N\ar@{}[d]\\ \mathbb{Z}/2&M\ar@{}[d]\\ V&L\ar@{}[d]\\ A_{4}&K\ar@{}[d]\\ S_{4}&F}$ 
By Galois theory, or Kummer theory, each field in this diagram is a radical extension of the one below it, and our job is done if we explicitly find what the radical extension is in each case.
We start with $K/F$. The index of $A_{4}$ in $S_{4}$ is two, so $K/F$ is a degree two extension. We have to find an element of $K$ that is not in $F$. The easiest such element to take is the element $\Delta$ obtained by taking the products of the differences of the roots, namely,
$\Delta:=\prod_{{1\leq i<j\leq 4}}(r_{i}r_{j})=(r_{1}r_{2})(r_{1}r_{3})(r_{1% }r_{4})(r_{2}r_{3})(r_{2}r_{4})(r_{3}r_{4}).$ 
Observe that $\Delta$ is fixed by any even permutation of the roots $r_{i}$, but that $\sigma(\Delta)=\Delta$ for any odd permutation $\sigma$. Accordingly, $\Delta^{2}$ is actually fixed by all of $S_{4}$, so:

$\Delta\in K$, but $\Delta\notin F$.

$\Delta^{2}\in F$.

$K=F[\Delta]$ = $F[\sqrt{\Delta^{2}}]$, thus exhibiting $K/F$ as a radical extension.
The element $\Delta^{2}\in F$ is called the discriminant of the polynomial. An explicit formula for $\Delta^{2}$ can be found using the reduction algorithm for symmetric polynomials, and, although it is not needed for our purposes, we list it here for reference:
$\displaystyle\Delta^{2}$  $\displaystyle=$  $\displaystyle 256d^{3}d^{2}(27a^{4}144a^{2}b+128b^{2}+192ac)$  
$\displaystyle c^{2}(27c^{2}18abc+4a^{3}c+4b^{3}a^{2}b^{2})$  
$\displaystyle 2d(abc(9a^{2}40b)2b^{3}(a^{2}4b)3c^{2}(a^{2}24b)).$ 
Next up is the extension $L/K$, which has degree 3 since $[A_{4}:V_{4}]=3$. We have to find an element of $N$ which is fixed by $V_{4}$ but not by $A_{4}$. Luckily, the form of $V_{4}$ almost cries out that the following elements be used:
$\displaystyle t_{1}$  $\displaystyle:=$  $\displaystyle(r_{1}+r_{2})(r_{3}+r_{4})$  
$\displaystyle t_{2}$  $\displaystyle:=$  $\displaystyle(r_{1}+r_{3})(r_{2}+r_{4})$  
$\displaystyle t_{3}$  $\displaystyle:=$  $\displaystyle(r_{1}+r_{4})(r_{2}+r_{3})$ 
These three elements of $N$ are fixed by everything in $V_{4}$, but not by everything in $A_{4}$. They are therefore elements of $L$ that are not in $K$. Moreover, every permutation in $S_{4}$ permutes the set $\{t_{1},t_{2},t_{3}\}$, so the cubic polynomial
$\Phi(x):=(xt_{1})(xt_{2})(xt_{3})$ 
actually has coefficients in $F$! In fancier language, the cubic polynomial $\Phi(x)$ defines a cubic extension $E$ of $F$ which is linearly disjoint from $K$, with the composite extension $EK$ equal to $L$. The polynomial $\Phi(x)$ is called the resolvent cubic of the quartic polynomial $x^{4}+ax^{3}+bx^{2}+cx+d$. The coefficients of $\Phi(x)$ can be found fairly easily using (again) the reduction algorithm for symmetric polynomials, which yields
$\Phi(x)=x^{3}2bx^{2}+(b^{2}+ac4d)x+(c^{2}+a^{2}dabc).$  (1) 
Using the cubic formula, one can find radical expressions for the three roots of this polynomial, which are $t_{1}$, $t_{2}$, and $t_{3}$, and henceforth we assume radical expressions for these three quantities are known. We also have $L=K[t_{1}]$, which in light of what we just said, exhibits $L/K$ as an explicit radical extension.
The remaining extensions are easier and the reader who has followed to this point should have no trouble with the rest. For the degree two extension $M/L$, we require an element of $M$ that is not in $L$; one convenient such element is $r_{1}+r_{2}$, which is a root of the quadratic polynomial
$(x(r_{1}+r_{2}))(x(r_{3}+r_{4}))=x^{2}+ax+t_{1}\in L[x]$  (2) 
and therefore equals $(a+\sqrt{a^{2}4t_{1}})/2$. Hence $M=L[r_{1}+r_{2}]=L[(a+\sqrt{a^{2}4t_{1}})/2]$ is a radical extension of $L$.
Finally, for the extension $N/M$, an element of $N$ that is not in $M$ is of course $r_{1}$, which is a root of the quadratic polynomial
$(xr_{1})(xr_{2})=x^{2}(r_{1}+r_{2})x+r_{1}r_{2}.$  (3) 
Now, $r_{1}+r_{2}$ is known from the previous paragraph, so it remains to find an expression for $r_{1}r_{2}$. Note that $r_{1}r_{2}$ is fixed by $(12)(34)$, so it is in $M$ but not in $L$. To find it, use the equation $(t_{2}+t_{3}t_{1})/2=r_{1}r_{2}+r_{3}r_{4}$, which gives
$(xr_{1}r_{2})(xr_{3}r_{4})=x^{2}\frac{(t_{2}+t_{3}t_{1})}{2}x+d$ 
and, upon solving for $r_{1}r_{2}$ with the quadratic formula, yields
$\displaystyle r_{1}r_{2}$  $\displaystyle=$  $\displaystyle\frac{(t_{2}+t_{3}t_{1})+\sqrt{(t_{2}+t_{3}t_{1})^{2}16d}}{4}$  (4)  
$\displaystyle r_{3}r_{4}$  $\displaystyle=$  $\displaystyle\frac{(t_{2}+t_{3}t_{1})\sqrt{(t_{2}+t_{3}t_{1})^{2}16d}}{4}$  (5) 
We can then use this expression, combined with Equation (3), to solve for $r_{1}$ using the quadratic formula. Perhaps, at this point, our poor reader needs a summary of the procedure, so we give one here:
1. Find $t_{1}$, $t_{2}$, and $t_{3}$ by solving the resolvent cubic (Equation (1)) using the cubic formula,
2. From Equation (2), obtain
$\displaystyle r_{1}+r_{2}$ $\displaystyle=$ $\displaystyle\frac{(a+\sqrt{a^{2}4t_{1}})}{2}$ $\displaystyle r_{3}+r_{4}$ $\displaystyle=$ $\displaystyle\frac{(a\sqrt{a^{2}4t_{1}})}{2}$ 3. Using Equation (3), write
$\displaystyle r_{1}$ $\displaystyle=$ $\displaystyle\frac{(r_{1}+r_{2})+\sqrt{(r_{1}+r_{2})^{2}4(r_{1}r_{2})}}{2}$ $\displaystyle r_{2}$ $\displaystyle=$ $\displaystyle\frac{(r_{1}+r_{2})\sqrt{(r_{1}+r_{2})^{2}4(r_{1}r_{2})}}{2}$ $\displaystyle r_{3}$ $\displaystyle=$ $\displaystyle\frac{(r_{3}+r_{4})+\sqrt{(r_{3}+r_{4})^{2}4(r_{3}r_{4})}}{2}$ $\displaystyle r_{4}$ $\displaystyle=$ $\displaystyle\frac{(r_{3}+r_{4})\sqrt{(r_{3}+r_{4})^{2}4(r_{3}r_{4})}}{2}$ where the expressions $r_{1}+r_{2}$ and $r_{3}+r_{4}$ are derived in the previous step, and the expressions $r_{1}r_{2}$ and $r_{3}r_{4}$ come from Equation (4) and (5).
4. Now the roots $r_{1},r_{2},r_{3},r_{4}$ of the quartic polynomial $x^{4}+ax^{3}+bx^{2}+cx+d$ have been found, and we are done!
Mathematics Subject Classification
12D10 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new problem: Geometry by parag
Aug 24
new question: Scheduling Algorithm by ncovella
new question: Scheduling Algorithm by ncovella
Comments
Inconsistent results
I've been trying to write a program that does this automatically for my calculator, but have encountered a problem. Which root I choose from the resolvent cubic to be root 1 (t1) affects the result, and it's a different root that's needed for some equations than in others.
I solve the resolvent using the equations from "galois theoretic derivation of the cubic equation", so my t1 is r1 from that page, t2 = r2 etc.
Is there any way to know which root yields the correct result?
Re: Inconsistent results
I wonder if there is a more simple but concise expression for the roots. Even for cubic equations, the roots are more simpler than this.
Re: Inconsistent results
I'm getting the same problemany help would be most appreciated.
I have noticed this: the root of the resolvent cubic that I choose makes a difference in which roots are defined as r1,r2,r3,r4. I solve for values of (r1+r2), (r3+r4), (r1*r2), and (r3*r4) that all look reasonable. It's just that sometimes I have a mislabeling. r1 + r2 = (r1+r2), as expected, but r1 * r2 = (r3*r4), and not (r1*r2). When you attempt to solve the resulting quadratics, chaos ensuesyou're not solving ones that are meaningful in any way.
These Galoistheoretic derivations are really neat, but there's still some ambiguity in them that's keeping me from programming them algorithmically.
Thank you. :)