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Abstract Algebra 2 (MATH 4140/5140)
Solvability by Radicals
P
Let f = i ai xi be a monic polynomial of degree n ≥ 1. Informally, the phrase “the equation
f (x) = 0 is solvable by radicals” means that all roots α1 , . . . , αn of f can be obtained from the
coefficients an−1 , . . . , a0 of f by field operations (that is, by +, −, √
·, and −1 where −1 is only a
m
−1
(m ≥ 2).
partial operation, since 0 is undefined) and by extracting roots
We can give a precise definition, using field extensions and splitting fields.
Definitions 1–3.
1. A simple radical extension is a simple (algebraic) extension K ≤ K(α) such that α is a
root of a polynomial of the form xm − a ∈ K[x] with m ≥ 2 (i.e., α is an m-th root of
a ∈ K for some m and a).
2. A radical tower or a tower of simple radical extensions is a tower K0 ≤ K1 ≤ · · · ≤
Ki−1 ≤ Ki ≤ · · · ≤ Kr of fields such that for every i = 1, . . . , r,
• Ki is a simple radical extension of Ki−1 ;
or equivalently,
• Ki = Ki−1 (αi ) where αimi ∈ Ki−1 for some αi ∈ Ki and some integer mi ≥ 2.
3. An extension K ≤ L is called a radical extension if there exists a radical tower K =
K0 ≤ K1 ≤ · · · ≤ Ki−1 ≤ Ki ≤ · · · ≤ Kr = L.
It is easy to see from these definitions that the concept ‘β is contained in a radical extension
of a field K’ expresses precisely what the informal phrase ‘β can be obtained from elements of
K by field operations and extractions of roots’ suggests.
qp
√
√
5
Example 1. Construct a radical extension of Q which contains the number
1 + 3 7 − 4 2.
√
√
Remark. Recall that 3 2 (∈ √
R) and ε 3 √
2 are two roots of x3 − 1 ∈ Q[x] which yield different
3
simple radical extensions: Q( 2) 6= Q(ε 3 2). Similarly, if m > 2, then for different m-th roots
α and β of a ∈ K (in a splitting field L for xm − a over K), the simple radical extensions K(α)
√
and K(β) of K might be different subfields of L. Therefore, we do not use the notation K(√m a)
for K(α) or K(β) – unless there is a way to specify which m-th root of a is denoted by m a.
Definition 4. Let K be a field, and let f ∈ K[x] be a monic polynomial of degree n ≥ 1. We
say that the equation f (x) = 0 is solvable by radicals over K if K has a radical extension L
which contains a splitting field for f over K. We say that f (x) = 0 is solvable by radicals if
f (x) = 0 is solvable by radicals over the field generated by its coefficients.
Solving Quadratic and Cubic Equations by Radicals
Easy Facts. Let K be a field, let f =
g = a1n f (x − c) for some c ∈ K.
Pn
i=0
ai xi be a polynomial of degree n in K[x], and let
(1) L is a splitting field for f over K if and only if L is a splitting field for g over K, and
if the roots of f in L are α1 , . . . , αn , then the roots of g in L are α1 + c, . . . , αn + c.
1
2
(2) If char(K) - n and we choose c =
g = xn + bn−2 xn−2 + · · · + b0 .
an−1
,
nan
then g is a monic polynomial of the form
Therefore, to determine whether the equation f (x) = 0 is solvable by radicals, we may replace
it by the simpler equation g(x) = 0 as described in (2), provided char(K) - n. In particular,
• if char(K) 6= 2, it suffices to consider quadratic equations of the form x2 + p ∈ K[x];
• if char(K) 6= 3, it suffices to consider cubic equations of the form x3 + px + q ∈ K[x].
Corollary. If char(K) 6= 2, then a splitting field for a quadratic polynomial f ∈ K[x] over K
is a simple radical extension K(α) of K where α2 ∈ K. Hence the equation f (x) = 0 is solvable
by radicals over K.
Example 2. A splitting field for x2 + x + 1 ∈ Z2 [x] over Z2 is a simple extension L = Z2 (α)
of Z2 by a root α of x2 + x + 1 ∈ Z2 [x] (because the other root is 1 + α ∈ Z2 (α)). L cannot be
obtained from Z2 by adjoining a square root of an element of Z2 to Z2 . But α3 = (1 + α)3 = 1,
therefore the equation x2 + x + 1 = 0 is solvable by radicals over Z2 .
Theorem. If char(K) 6= 2, 3, then for every cubic polynomial f ∈ K[x], the equation f (x) = 0
is solvable by radicals over K.
Proof. We may assume that f = x3 +px+q ∈ K[x], q 6= 0. Let L be a splitting field for f over K.
Trick: We look for the roots α of f in the form
• α = u + v, where
• u, v are in some extension field of L, and
• uv = a ∈ K for some element a ∈ K that will be chosen later.
Since u, v satisfy the equations u+v = α and uv = a if and only if u, v are roots of the quadratic
polynomial (x − u)(x − v) = x2 − αx + a, such u, v will exists for every root α of f .
Choosing a: Since (u + v)3 = u3 + 3u2 v + 3uv 2 + v 3 = u3 + 3uv(u + v) + v 3 , we get that u + v
is a root of f if and only if
(∗)
0 = f (u + v) = (u + v)3 + p(u + v) + q = u3 + v 3 + (3uv + p)(u + v) + q.
Choose a = − p3 (∈ K); that is, we require that 3uv + p = 0.
Solve: With this choice the equality in (∗) becomes 0 = u3 + v 3 + q. Hence u3 , v 3 satisfy
p3
.
27
p3
Equivalently, u3 , v 3 are the two roots of the quadratic equation x2 + qx − 27
= 0, which we can
solve:
r
q
q 2 p3
u3 , v 3 = − ±
+ .
2
4
27
Let C denote one of the elements on the right hand side, which are not both 0, so let C 6= 0.
The roles of u, v are symmetric, so there is no loss of generality in assuming that u3 = C. Since
char(K) 6= 3, it can be shown (we will see this later) that C has three distinct cube roots (in a
splitting field for x3 − C over K(C)), say u1 , u2 , u3 . Since we required uv = − p3 to hold for u, v
when expressing a root α of f as α = u + v, the three roots of f are
p
αi = ui + vi = ui −
(i = 1, 2, 3).
3ui
This is Cardano’s formula.
u3 + v 3 = −q
and
u3 v 3 = −