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In our Galois Theory classes we have been introduced to the idea that quartics can be solved by radicals. The course walks us through a construction that is designed to intuitively explain to us how this can be done (though does not arrive at explicit calculations of the solution). I'm wondering if someone can help fill in some gaps for me.

The construction is presented as follows:

Suppose $f(x)=x^4+px^2+qx+r \in \mathbb{Q}[x]$ is irreducible and let $\alpha_1, ... , \alpha_4$ be roots of $f$ in a splitting field $L$ over $\mathbb{Q}$.

Now define

$$\beta_1=(\alpha_1 + \alpha_2)(\alpha_3 + \alpha_4)$$

$$\beta_2=(\alpha_1 + \alpha_3)(\alpha_2 + \alpha_4)$$

$$\beta_3=(\alpha_1 + \alpha_4)(\alpha_2 + \alpha_3)$$

We then claim that $\beta_1, ... , \beta_3$ are roots of the resolvent of $f$, which is defined as $$g(x)=x^3-2px^2+(p^2-4r)x+q^2 \in \mathbb{Q}[x]$$

Question 1. Is there any simple way to show these are roots of $g$ other than by a lot of algebra (expanding out and substituting in)?

After this, it is claimed that this construction can be used to solve a quartic by radicals, and little more is said.

Now I know that cubics can be solved by radicals, so I think that if we can uncover information about the $\alpha$'s by working out the $\beta$'s (which can be done explicitly by radicals in the case of cubics), then we will be done. So this leads to my question.

Question 2. How can this set-up be used to show the solvability of the quartic by radicals? i.e. How can finding roots of the resolvent polynomial give rise to roots of the quartic?

I am not necessarily concerned with explicit formulas for the roots of the quartic, since it was made clear these are not needed for our course. I would prefer an outline of how we use the information we were presented to understand intuitively how it can be done.

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In order to get an algorithm for solving quartic polynoms, we have to show how the group $S_4$ is solvable, and then use that information to get the algorithm.

If $G$ is the subgroup of $S_4$ fixing $\beta_1$, $G = \langle (12),(34),(13)(24) \rangle$ We have the normal resolution

$G_0=\{id \} \subset G_1 = \langle (12) \rangle \subset G_2 = \langle (12)(34) \rangle \subset G_3 = \langle (12),(34),(13)(24) \rangle \subset G_4 = S_4$

where $G_1/G_0$ , $G_2/G_1$ and $G_3/G_2$ are cyclic of order $2$, and $G_4/G_3$ is cyclic of order $3$ (check this by hand).

By the fundamental theorem of Galois theory we have fields $K_i = \Bbb C(X_1,X_2,X_3,X_4)^{G_i}$ where $K_{i+1} \subset K_i$ is a normal extension with Galois group $G_{i+1}/G_i$, and where there is no intermediate fields between two consecutive $K_i$.

This is a very strong theorem : for example if you find some element $x_i \in K_i \setminus K_{i+1}$, then you know (because there is no intermediate field) that $K_i = K_{i+1}(x_i)$ : so for every polynomial expression who is fixed by $G_i$, you can express it as a polynomial expression in terms of $x_i$, and polynomials invariant by $G_{i+1}$. The method for solving the quartic makes extensive use of this :

First we have to pick some element $x_i \in K_i$ such that $K_i = K_{i+1}(x_i)$.

Since I picked $G_3$ to be the group of elements fixing $(X_1+X_2)(X_3+X_4)$, $K_3 = K_4((X_1+X_2)(X_3+X_4))$. For the others it is not difficult to find any element of $K_i \setminus K_{i+1}$, just pick any one.

If we call $p,q,r,s$ the elementary symmetric polynomials (i.e. $(X-X_1)(X-X_2)(X-X_3)(X-X_4) = X^4-pX^3+qX^2-rX+s$), then $K_i = \Bbb C(p,q,r,s,x_3,x_2, \ldots, x_i)$ so we will need an algorithm to express any polynomial expression who is invariant by $G_i$ and turn it into a polynomial in $p,q,r,s,x_j$ for $j \ge i$. (to do this you can always use brute force and linear algebra. You know such an expression has to exist so you just have to look for a big enough indeterminate polynomial so that the resulting linear system is solvable)

Next, we can compute the minimal polynomial of each $x_i$ over $K_{i+1}$, which is simply $\prod_{\sigma \in G_{i+1}/G_i} (T - \sigma(x_i))$. When we develop this, the coefficients are invariant by $G_{i+1}$ so we use the algorithm above to express its coefficients in term of $p,q,r,s,x_j$ for $j > i$.

So at this point, you have "explicit" (but very complex) polynomials saying how $x_3$ is a root of a cubic polynomial in terms of $p,q,r,s$ ; $x_2$ is a root of a quadratic polynomial in terms of $p,q,r,s,x_3$ ; $x_1$ is a root of a quadratic polynomial in terms of $p,q,r,s,x_2,x_3$ and $x_0$ is a root of a quadratic polynomial in terms of $p,q,r,s,x_1,x_2,x_3$. And finally, the four roots $X_1,X_2,X_3,X_4$ are polynomial expressions in terms of $p,q,r,s,x_0,x_1,x_2,x_3$.


Actually, we can do better, because so far we don't know yet what $n$ root we have to take of who, and it is also a stupid idea to solve for $x_3$ like we would do for any other cubic. The fact that each quotient group is cyclic of order $n_i$ means that each $x_i$ can be obtained from $K_{i+1}$ by adjoining some $n_i$th root of an element in $K_{i+1}$ :

If $\sigma_i$ is a generator of $G_{i+1}/G_i = Gal(K_i / K_{i+1}) = \Bbb Z/n_i \Bbb Z$, then if we let $y_i = \sum_{0 \le j < n_i} \sigma_i^j(x_i)\zeta_{n_i}^{-j} \in K_i$, we have $\sigma_i(y_i) = \zeta_n y_i$, and so $\sigma_i(y_i^{n_i}) = \zeta_{n_i}^{n_i} y_i^{n_i} = y_i^{n_i}$, hence $y_i^{n_i} \in K_{i+1}$. If $y_i$ is nonzero, we then have $\sigma(y_i) \neq y_i$ so $K_i = K_{i+1}(y_i)$.

Instead of computing the resolvant of the $x_i$ and trying to express things in terms of the $x_i$, we should use $y_i$ instead of $x_i$ everywhere, provided they are nonzero (if they are, pick another $x_i$) : we express things in terms of the $y_i$ and then the minimal polynomial of $y_i$ over $K_{i+1}$ is $T^{n_i} - y_i^{n_i}$ so we only need to express $z_i = y_i^n$ in terms of $p,q,r,s,y_j$ for $j > i$.

So in the end, the complicated formulas for the $z_i$ in terms of $p,q,r,s,y_j$ for $j>i$, and the complicated formulas for $X_1,X_2,X_3,X_4$ in terms of $p,q,r,s,y_j$ are what is needed to solve any quartic equation. Galois theory says that those formulas exist, it doesn't say that they are nice.


Since everything relies on the procedure to express things in terms of some $x$ and more symmetric terms, it is good to know that you can have an explicit formula using the Galois action when $x$ is the $n$th root of a more symmetric term :

suppose $Gal(L/K)$ is cyclic or oder $n$, $L = K(x)$ where $x^n \in L$, and $\sigma(x) = \zeta_nx$. Then we have a decomposition $L = \bigoplus_{0 \le k < n} (x^k). K$ where $\sigma$ acts on the $k$th component as multiplication-by-$\zeta_n^k$.

If $y \in L$, knowing the expression of $y$ in terms of $x$ and elements of $L$ is the same as knowing the coefficients of $y$ in this basis.

But in fact, we can express each projection as a polynomial in $\sigma$ : let $\pi_l(y) = \sum \zeta_n^{-kl} \sigma^k(y)$. We have $\sigma(\pi_l(y)) = \sum \zeta_n^{-kl} \sigma^{k+1}(y) = \zeta_n^l \sum \zeta_n^{-(k+1)l} \sigma^{k+1}(y) = \zeta_n^l \pi_l(y)$, which means that $\pi_l(y)$ is on the $l$-th component, and since $\pi_l(x^l) = \sum x^l = nx^l$ ; $\frac 1n \pi_l$ is the projection on the $l$-th component.

And so we have for all $y \in L$, $y = \frac 1n \sum \pi_l(y) = \sum \frac {\pi_l(y)}{nx^l}x^l$, where $\frac {\pi_l(y)}{nx^l} \in K$ is the expression we want.

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