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 $(x-r_1)(x-r_2)(x-r_3)(x-r_4) = x^4 + ax^3 + bx^2 + cx + d$ . The goal is to exhibit the field extension $\C(r_1,r_2,r_3,r_4)/\C(a,b,c,d)$ 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 $\C(r_1,r_2,r_3,r_4)$ and $F$ for $\C(a,b,c,d)$ . The Galois group $\Gal(N/F)$ 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

• $A_4$ is the alternating group in $S_4$ , consisting of the even permutations.
• $V_4 = \{1, (12)(34), (13)(24), (14)(23)\}$ is the Klein four-group.
• $\Z/2$ 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:

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: \begin{eqnarray*} \Delta^2 & = & 256 d^3 - d^2 (27 a^4 - 144 a^2 b + 128 b^2 + 192 a c) - \\ & & c^2 (27c^2 - 18 abc + 4a^3 c + 4 b^3 - a^2 b^2) - \\ & & 2d (abc (9a^2 - 40 b) - 2b^3 (a^2-4b) - 3c^2 (a^2-24b)). \end{eqnarray*} 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: \begin{eqnarray*} t_1 & := & (r_1+r_2)(r_3+r_4) \\ t_2 & := & (r_1+r_3)(r_2+r_4) \\ t_3 & := & (r_1+r_4)(r_2+r_3) \end{eqnarray*} 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 $$ \Phi(x) := (x-t_1) (x-t_2) (x-t_3) $$ 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 \begin{equation}\label{resolvent} \Phi(x) = x^3 - 2b x^2 + (b^2+ac-4d) x + (c^2+a^2d - abc). \end{equation}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 \begin{equation}\label{step} (x-(r_1+r_2)) (x-(r_3+r_4)) = x^2 + a x + t_1 \in L[x] \end{equation}and therefore equals $(-a + \sqrt{a^2 - 4 t_1})/2$ . Hence $M = L[r_1+r_2] = L[(-a + \sqrt{a^2 - 4 t_1})/2]$ 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 \begin{equation}\label{final} (x-r_1)(x-r_2) = x^2 - (r_1+r_2) x + r_1 r_2. \end{equation}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 $$ (x - r_1 r_2) (x - r_3 r_4) = x^2 - \frac{(t_2 + t_3 - t_1)}{2} x + d $$ and, upon solving for $r_1 r_2$ with the quadratic formula, yields \begin{eqnarray} \label{post} r_1 r_2 & = & \frac{(t_2 + t_3 - t_1) + \sqrt{(t_2 + t_3 - t_1)^2 - 16d}}{4} \\ \label{post2} r_3 r_4 & = & \frac{(t_2 + t_3 - t_1) - \sqrt{(t_2 + t_3 - t_1)^2 - 16d}}{4} \end{eqnarray}We can then use this expression, combined with Equation (), 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 ()) using the cubic formula,
2. From Equation (), obtain \begin{eqnarray*} r_1+r_2 & = & \frac{(-a + \sqrt{a^2 - 4 t_1})}{2} \\ r_3+r_4 & = & \frac{(-a - \sqrt{a^2 - 4 t_1})}{2} \end{eqnarray*}
3. Using Equation (), write \begin{eqnarray*} r_1 & = & \frac{(r_1+r_2) + \sqrt{(r_1+r_2)^2 - 4 (r_1 r_2)}}{2} \\ r_2 & = & \frac{(r_1+r_2) - \sqrt{(r_1+r_2)^2 - 4 (r_1 r_2)}}{2} \\ r_3 & = & \frac{(r_3+r_4) + \sqrt{(r_3+r_4)^2 - 4 (r_3 r_4)}}{2} \\ r_4 & = & \frac{(r_3+r_4) - \sqrt{(r_3+r_4)^2 - 4 (r_3 r_4)}}{2} \\ \end{eqnarray*}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 () and ().
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!