# Logic behind cubic resolution of $x^4+px^2+qx+r=0$

I'm reading "A course in algebra" by E. B. Vinberg to have a basic understanding of algebra.

In its Chapter 3 there's an interesting demonstration on cubic resolution of equation $x^4+px^2+qx+r=0$.

1. Define $\sigma_i$ the elementary symmetric polynomials: $$\sigma_i=\sum_{k_1<k_2<\cdots<k_i}x_{k_1}x_{k_2}\cdots x_{k_i}$$

2. Define $h_i(x_1,x_2,x_3,x_4)$:
$$\left\{ \begin{array}{ll} h_1=x_1x_2+x_3x_4 \\ h_2=x_1x_3+x_2x_4\\ h_3=x_1x_4+x_2x_3 \end{array} \right.$$

Then one has:
$$\left\{ \begin{array}{ll} h_1+h_2+h_3&=\sigma_2 \\ h_1h_2+h_1h_3+h_2h_3&=\sigma_1\sigma_3-4\sigma_4\\ h_1h_2h_3&=\sigma_1^2\sigma_4+\sigma_3^2-4\sigma_2\sigma_4 \end{array} \right.$$

3. Let $c_1$, $c_2$, $c_3$, $c_4$ be the roots of equation $x^4+px^2+qx+r=0$.

Then $$\left\{ \begin{array}{ll} \sigma_1(c_1,c_2,c_3,c_4)=0\\ \sigma_2(c_1,c_2,c_3,c_4)=p\\ \sigma_3(c_1,c_2,c_3,c_4)=-q\\ \sigma_4(c_1,c_2,c_3,c_4)=r \end{array} \right.$$

4. Let $d_i=h_i(c_1,c_2,c_3,c_4)$.

So $d_i$ are the roots of equation $y^3+a_1y^2+a_2y+a_3=0$ where $$\left\{ \begin{array}{ll} a_1=-(d_1+d_2+d_3)\\ a_2=d_1d_2+d_1d_3+d_2d_3\\ a_3=-d_1d_2d_3 \end{array} \right.$$

From step 2 one has $a_1=-p$, $a_2=-4r$, $a_3=4pr-q^2$.

So the equations has the form $y^3-py^2-4ry+(4pr-q^2)=0$. This is assumed solvable.

5. Define $$\left\{ \begin{array}{ll} b_1=c_1+c_2-c_3-c_4\\ b_2=c_1-c_2+c_3-c_4\\ b_3=c_1-c_2-c_3+c_4 \end{array} \right.$$

So one has $$\left\{ \begin{array}{ll} b_1^2=4(d_1-p)\\ b_2^2=4(d_2-p)\\ b_3^2=4(d_3-p)\\ b_1b_2b_3=-8q \end{array} \right.$$

From here it leads to $$\left\{ \begin{array}{ll} c_1+c_2+c_3+c_4&=0\\ c_1+c_2-c_3-c_4&=b_1&=2\sqrt{d_1-p_1}\\ c_1-c_2+c_3-c_4&=b_2&=2\sqrt{d_2-p_1}\\ c_1-c_2-c_3+c_4&=b_3&=2\sqrt{d_3-p_1} \end{array} \right.$$

So it's solved: $$c_{1,2,3,4}=\frac{1}{2}(\pm\sqrt{d_1-p_1}\pm\sqrt{d_2-p_1}\pm\sqrt{d_3-p_1})$$ , with number of minuses being even, and the value of the square roots chosen so that $b_1b_2b_3=-8q$.

Now, all these steps I understand, and they are quite smart. However, if we look at step 5, is it just some math trick (being sensitive to math forms) or there's something, some higher logic behind it?

The logic before step 5, to put it simple, is to get $d_i$ via $\sigma_i$, using the symmetry, without knowing exactly $c_i$. This is quite smart. So step 5 the idea is to get $c_i$ from $d_i$.

How ppl get the formula in step 5 to achieve that? is it just by try and error, by luck?

Or, there's some logic behind it? I suspect so as $b_i$ look similar to $d_i$, there seems also some "indirect symmetry" behind the construction of $b_i$.

• Yes, there is some logic behind it — it is explained by Galois theory... – Grigory M May 30 '14 at 9:00
• @GrigoryM mind give a bit more details? e.g. from which book etc... – athos May 30 '14 at 9:34
• ...Maybe Dummit & Foote, ch. 14 (see esp. sec. 6–7) is not a bad reference (but I hope someone will give a better reference) – Grigory M May 30 '14 at 10:51
• But the trick predates Galois theory, so how did the person who came upon the trick for the first time, find it? Just sheer luck? – Raskolnikov May 31 '14 at 5:29
• @Raskolnikov: No luck needed, just simple manipulations; see my answer. =) – user21820 May 31 '14 at 6:58

This is only part of the way to a real answer, but I believe the point of the $b_i$ is that they are linear polynomials in the subfield $\mathbb{Q}(c_1,c_2,c_3,c_4)^{A_4}$ of the splitting field $\mathbb{Q}(c_1,c_2,c_3,c_4)$ of your initial polynomial that is left invariant under even permutations of the roots $c_j$. I believe the vector space $\mathbb{Q}b_1 + \mathbb{Q}b_2 + \mathbb{Q}b_3$ spanned by them should be exactly the linear polynomials (no constant term) invariant under those permutations. (Note that they each have exactly two minus signs.)

Note also that the $d_i$ are quadratic polynomials in the roots $c_j$, and also invariant under even permutations of the roots. One might guess from this that an expression for the $d_i$ in terms of the $b_i$ and other known quantities (and vice versa) might be found. So you might be led to the $b_i$ by asking what set of linear polynomials in the roots "has the same symmetries" as the $d_i$.

A more detailed account (involving more knowledge of field theory than I can actually recall at the moment) can be found, for instance, here: http://www.jmilne.org/math/CourseNotes/FT.pdf

The part you want is around page 50, but you will need to have read the earlier material to make much sense of it.

• i believe this is the one explaining it. let me take some time to digest. thank you! – athos Jun 3 '14 at 2:39
• @athos: I took a look at the PDF but page 50 definitely doesn't explain what you're after... – user21820 Jun 3 '14 at 6:34
• I concur that the same expressions don't occur, so it's in that sense not the same solution of the quartic. Yours is a much more thorough and probably more useful solution. – jdc Jun 4 '14 at 16:02

There wasn't any luck. The hardest part was actually in coming up with $d_1,d_2,d_3$. After that it is just a matter of manipulation to make use of the fact that $c_1 + c_2 + c_3 + c_4 = 0$ because the quartic is depressed.

$d_1 = c_1 c_2 + c_3 c_4 = p - ( c_1 c_3 + c_1 c_4 + c_2 c_3 + c_2 c_4 ) = p - (c_1+c_2)(c_3+c_4) = p + (c_1+c_2)^2$

$d_2 = c_1 c_3 + c_2 c_4 = p - ( c_1 c_2 + c_1 c_4 + c_3 c_2 + c_3 c_4 ) = p - (c_1+c_3)(c_2+c_4) = p + (c_1+c_3)^2$

$d_3 = c_2 c_3 + c_1 c_4 = p - ( c_1 c_2 + c_1 c_3 + c_4 c_2 + c_4 c_3 ) = p - (c_1+c_4)(c_2+c_3) = p + (c_1+c_4)^2$

From here it is clear that we can get $c_1,c_2,c_3,c_4$ because:

$c_1+c_2 = \sqrt{d_1-p}$ where the sign of the $\sqrt{}$ is unknown

$c_1+c_3 = \sqrt{d_2-p}$ where the sign of the $\sqrt{}$ is unknown

$c_1+c_4 = \sqrt{d_3-p}$ where the sign of the $\sqrt{}$ is such that:

$\sqrt{d_1-p} \sqrt{d_2-p} \sqrt{d_3-p} = c_1^2 (c_1+c_2+c_3+c_4) + (c_1c_2c_3+c_1c_2c_4+c_1c_3c_4+c_2c_3c_4)$

$= -q$ because $c_1 + c_2 + c_3 + c_4 = 0$

$c_1 + c_2 + c_3 + c_4 = 0$ (again)

The sum of the first three minus the fourth gives:

$2 c_1 = \sqrt{d_1-p} + \sqrt{d_2-p} + \sqrt{d_3-p}$

You can check that your textbook arranged the above steps in quite a backward way, which doesn't help in understanding where the various expressions came from.

Notes

Similar manipulations also give rise to the fact that the resolvent cubic $y \mapsto y^3 − p y^2 - 4r y + (4pr−q^2)$ has the same discriminant as the original quartic.

$d_1 - d_2 = ( c_1 c_2 + c_3 c_4 ) - ( c_1 c_3 + c_2 c_4 ) = (c_1-c_4)(c_2-c_3)$

$d_1 - d_3 = ( c_1 c_2 + c_3 c_4 ) - ( c_1 c_4 + c_2 c_3 ) = (c_1-c_3)(c_2-c_4)$

$d_2 - d_3 = ( c_1 c_3 + c_2 c_4 ) - ( c_1 c_4 + c_2 c_3 ) = (c_1-c_2)(c_3-c_4)$

Then it is natural to consider the product of both sides which gives the result.

Alternatives

There is another resolvent cubic where the above steps are actually much easier to see:

Let $d_1 = - (c_1+c_2)(c_3+c_4) = (c_1+c_2)^2$

Let $d_2 = - (c_1+c_3)(c_2+c_4) = (c_1+c_3)^2$

Let $d_3 = - (c_1+c_4)(c_2+c_3) = (c_1+c_4)^2$

Then by symmetry the symmetric polynomials on $(d_1,d_2,d_3)$ will be symmetric in $(c_1,c_2,c_3,c_4)$ as well, and from those it is easy to compute the resolvent cubic that has $(d_1,d_2,d_3)$ as roots. After that the solution is identical, and in some sense simpler than the one your textbook uses. Later on it will also turn out that the criterion to distinguish whether the Galois group of an irreducible separable quartic is $C_4$ or $D_4$ will be much simpler using this resolvent cubic than your textbook's.

• thanks for the comments. but my question is not about how to calculate $\sigma_i \rightarrow a_i \rightarrow d_i \rightarrow b_i \rightarrow c_i$, but how would the guy first came out the idea, constructed $b_i$ from nowhere? – athos Jun 3 '14 at 2:31
• @athos: I indeed answered your question... You just have to follow this method backwards to derive the expressions for the "$b$"s that you have. $b_1 = 2(c_1+c_2) = c_1+c_2-c_3-c_4$. I am quite sure that the first person who solved the quartic by this kind of method (using symmetric polynomials) did not come up with those expressions directly but indirectly through the same way I showed you. I can assure you that modern texts use techniques that were not available to the early solvers of the quartic. – user21820 Jun 3 '14 at 6:34