A New Proof of Abel-Ruffini Theorem
Abstract.
The Abel-Ruffini theorem shows that the general quintic equation is not solvable in radicals. Reference [4] gives a proof based on a theorem of Kronecker, but we notice a mistake in it. We give a corrected version of this mistake and provide a new proof of the Abel-Ruffini theorem.
Key words and phrases:
Abel–Ruffini Theorem, field theory2010 Mathematics Subject Classification:
Primary 12F15, 12E051. Introduction
Abel-Ruffini theorem plays a vital role in solving algebraic equations in history. Notions like the Galois Theory and solvable group originated here. For a long time, people attempted to find the expression of the root by using the coefficient of radicals and finite field operations but failed. It was not until 1824 [1] that Niels Henrik Abel, a Norwegian mathematician, discovered the proof of the theorem. Galois’s proof published in 1846 [2] after his death. Vladimir Arnold found a topological proof of this theorem in 1963 [3].
A proof based on a theorem of Kronecker was given in the reference [4], and it was only using basic algebraic knowledge. Unfortunately, there is a mistake that exists in the proof. Our work points out the mistake and gives a fixed version by Theorem 2. Then we use a different way to provide a new proof of Theorem 3 and use it proves the Abel-Ruffini theorem.
2. A Mistake [4]
The mistake in [4] is derived from a paragraph on pages 123 to 124:
“Also, with each substituted radical of our series, which still does not allow division of , we will also substitute at the same time the complex conjugate radical. Though this may be superfluous, it can certainly do no harm.”
We will use the field theory to prove that the assertion is wrong. Below we provide some definitions of conventions.
Definition 1 (radical extension).
A field extension is called a radical extension if there is a prime , is irreducible over , and .
Definition 2 (radical tower).
A field extension is said to be a radical tower over if there is a series of intermediate fields
such that for each , , where is a prime, and is irreducible over .
For convenience, when , we call is a radical tower.
Definition 3 (solvable in radicals).
We call is solvable in radicals if all the roots of belong to , where is a radical tower.
The following is an example to illustrate the mistake in [4].
We set and let . Because is irreducible over and all its roots are , we have
So we get and . If , we can get or from
When or , set is a root of , because
we have
and
That means , it is impossible. So , and is irreducible over .
3. Some Lemmas
In the following, fields are the subset of , and polynomials are monic.
Lemma 1.
Suppose , are irreducible over , is a prime, set be a root of and be a root of , if is not irreducible over , we have .
Proof.
We set , , thus
If is not irreducible over , then
and
According to and are both prime numbers, we have . ∎
Lemma 2.
Suppose , are irreducible over , and are both prime numbers, let all the roots of be , and be a root of . If is not irreducible over , we have can be factored into linear factors over .
Proof.
If is not irreducible over , has the form
Since is irreducible over , then
,
,
.
According to are all the roots of , we have
Since all the roots of are the roots of , we have
Here the is a positive integer. According to the above results, thus
Because of and are both prime numbers, , we have
That means can be factored into linear factors over . ∎
Lemma 3 (Abel).
Suppose is a prime, the equation is not irreducible over , only when there is a , and .
Proof.
See [4]. ∎
Lemma 4.
Suppose is a prime, , and . If is not irreducible over , we get can be factored into linear factors over . That is to say, if is a root of , the extension is a th radical extension or a trivial extension.
Proof.
By Lemma 3, if is not irreducible over , there is a belongs , . So we get
That means can be factored into linear factors over . Thus is a trivial extension.
If is irreducible over , by Definition 1, is a th radical extension. ∎
Lemma 5.
Suppose is an integer, is a field, and for any prime , the th root belongs to . We set , , then we can find a radical tower , and .
Proof.
We can find primes , , and the extension
By Lemma 4, the extension
is a trivial extension or a radical extension, so that we can find a radical tower , and . ∎
4. Proof of Abel-Ruffini Theorem
Theorem 1.
Suppose is a prime, and is a field. We have the extension
such that for each , , where is a prime, and is irreducible over .
Proof.
Let us use mathematical induction to prove this proposition.
When , due to is a root of . Thus is a radical extension or a trivial extension.
Set is a prime, and . Suppose for any prime , the proposition is true. We will prove when , the proposition is true. We set . By Lemma 5 and , we can get is a radical tower. We set , , and . Then we let
so we have
and
Thus we get
By Lemma 5, we can get , and by radical tower. Since we have
Thus
That means we get . So we can find a radical tower , and . That means when , the proposition is true. ∎
Theorem 2.
Suppose is irreducible over and has a prime degree . If is solvable in radicals, we can find a radical tower , , for any , , and is irreducible over , but is not irreducible over . In there, is a radical extension.
Proof.
According to is solvable in radicals, we can find a radical tower
such that for each , where is a prime, and is irreducible over .
Suppose . By Theorem 1, we have a radical tower
By Lemma 4, we add to , and ignore the trivial extensions. We can get a new radical tower
such that for each , , where is a prime, and is irreducible over .
By Lemma 4 and , if we ignore the trivial extensions in
such that for each , . We can get is a radical tower, , for any , , and . We set the new radical tower is
such that for each , , , where is a prime, and is irreducible over .
Because of all the roots of belong to . By Lemma 4, we can find a , and is irreducible over , but is not irreducible over .
If for any , , we get and . Else we consider this extension
If is irreducible over , we set and . If is not irreducible over , we set and . ∎
Remark 1.
Theorem 3 (Kronecker).
Suppose is irreducible over and has a prime degree . If is solvable in radicals and has an imaginary root, then it only has one real root.
Proof.
By Theorem 2, there is a radical tower , and . We have is irreducible over , but is not irreducible over . In there, is a root of , and is a prime number. By Lemma 1, . By Lemma 4, is not irreducible over . By Lemma 2, can be factored into linear factors over . So we have
In there, for any , we have , , and .
If has an imaginary root, we set it is . Because of , we can find another imaginary root , and have . Hence
(1) |
We have two cases:
CASE I. When , by equation 1, we have
(2) |
Since is irreducible over , and . We can change the in equation 2 to for any . Thus for any , the following formula holds
(3) |
CASE II. When , we set . If is not irreducible over , this will give us the situation of Case I. If is irreducible over , we get is irreducible over . Otherwise, by Lemma 4, we have . That means is not irreducible over , which against our premise. By equation 1, we have
(4) |
Since is irreducible over , and . We can change the in equation 4 to for any . Thus for any , the following formula holds
(5) |
Remark 2.
We provide an example that a solvable quintic equation only has real roots. We set and let . As proved earlier, is solvable in radicals and only has real roots.
Theorem 4 (Abel-Ruffini).
The general quintic equations not solvable in radicals.
Proof.
Consider , it has exactly three real roots. By Eisenstein’s criterion, is irreducible over . According to the Theorem 3, the equation is not solvable in radicals. ∎
References
- [1] Abel, N. H. (1824). Mémoire sur les équations algébrique: où on démontre l’impossiblité de la résolution de l’equation générale du cinquième dégré. Librarian, Faculty of Science, University of Oslo.
- [2] Galois, E. (1846). Sur les conditions de résolubilité des équations par radicaux. Journal de mathématiques pures et appliquées, 11, pp. 417-444.
- [3] Alekseev, V. B. (2004). Abel’s Theorem in Problems and Solutions: Based on the lectures of Professor VI Arnold. Springer Science & Business Media.
- [4] Dörrie, H. (2013). 100 great problems of elementary mathematics. Courier Corporationr, pp. 116-127.