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16. Eisenstein’s criterion
16.1 Eisenstein’s irreducibility criterion 16.2 Examples
1. Eisenstein’s irreducibility criterion
Let R be a commutative ring with 1, and suppose that R is a unique factorization domain. Let k be the field of fractions of R, and consider R as imbedded in k.
[1.0.1] Theorem: Let
f (x) = xN^ + aN − 1 xN^ −^1 + aN − 2 xN^ −^2 +... + a 2 x^2 + a 1 x + a 0
be a polynomial in R[x]. If p is a prime in R such that p divides every coefficient ai but p^2 does not divide a 0 , then f (x) is irreducible in R[x], and is irreducible in k[x].
Proof: Since f has coefficients in R, its content (in the sense of Gauss’ lemma) is in R. Since it is monic,
its content is 1. Thus, by Gauss’ lemma, if f (x) = g(x) · h(x) in k[x] we can adjust constants so that the content of both g and h is 1. In particular, we can suppose that both g and h have coefficients in R, and are monic.
Let g(x) = xm^ + bm− 1 xm−^1 + b 1 x + b 0
h(x) = xn^ + cm− 1 xm−^1 + c 1 x + c 0
Not both b 0 and c 0 can be divisible by p, since a 0 is not divisible by p^2. Without loss of generality, suppose that p|b 0. Suppose that p|bi for i in the range 0 ≤ i ≤ i 1 , and p does not divide bi 1. There is such an index i 1 , since g is monic. Then ai 1 = bi 1 c 0 + bi 1 − 1 c 1 +...
On the right-hand side, since p divides b 0 ,... , bi 1 − 1 , necessarily p divides all summands but possible the first. Since p divides neither bi 1 nor c 0 , and since R is a UFD, p cannot divide bi 1 c 0 , so cannot divide ai 1 , contradiction. Thus, after all, f does not factor. ///
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220 Eisenstein’s criterion
- Examples
[2.0.1] Example: For a rational prime p, and for any integer n > 1, not only does
xn^ − p = 0
not have a root in Q, but, in fact, the polynomial xn^ − p is irreducible in Q[x].
[2.0.2] Example: Let p be a prime number. Consider the pth^ cyclotomic polynomial
Φp(x) = xp−^1 + xp−^2 =... + x^2 + x + 1 =
xp^ − 1 x − 1
We claim that Φp(x) is irreducible in Q[x]. Although Φp(x) itself does not directly admit application of Eisenstein’s criterion, a minor variant of it does. That is, consider
f (x) = Φp(x + 1) =
(x + 1)p^ − 1 (x + 1) − 1
xp^ +
(p 1
xp−^1 +
(p 2
xp−^2 +... +
( (^) p p− 2
x^2 +
( (^) p p− 1
x x
= xp−^1 +
p 1
xp−^2 +
p 2
xp−^3 +... +
p p − 2
x +
p p − 1
All the lower coefficients are divisible by p, and the constant coefficient is exactly p, so is not divisible by p^2. Thus, Eisenstein’s criterion applies, and f is irreducible. Certainly if Φp(x) = g(x)h(x) then f (x) = Φp(x + 1) = g(x + 1)h(x + 1) gives a factorization of f. Thus, Φp has no proper factorization.
[2.0.3] Example: Let f (x) = x^2 + y^2 + z^2 in k[x, y, z] where k is not of characteristic 2. We make
identifications like k[x, y, z] = k[y, z][x]
via the natural isomorphisms. We want to show that y^2 + z^2 is divisible by some prime p in k[y, z], and not by p^2. It suffices to show that y^2 + z^2 is divisible by some prime p in k(z)[y], and not by p^2. Thus, it suffices to show that y^2 + z^2 is not a unit, and has no repeated factor, in k(z)[y]. Since it is of degree 2, it is certainly not a unit, so has some irreducible factor. To test for repeated factors, compute the gcd of this polynomial and its derivative, viewed as having coefficients in the field k(z): [1]
(y^2 + z^2 ) −
y 2
(2y) = z^2 = non-zero constant
Thus, y^2 + z^2 is a square-free non-unit in k(z)[y], so is divisible by some irreducible p in k[y, z] (Gauss’ lemma), so Eisenstein’s criterion applies to x^2 + y^2 + z^2 and p.
[2.0.4] Example: Let f (x) = x^2 + y^3 + z^5 in k[x, y, z] where k is not of characteristic dividing 30. We
want to show that y^3 + z^5 is divisible by some prime p in k[y, z], and not by p^2. It suffices to show that y^3 + z^5 is divisible by some prime p in k(z)[y], and not by p^2. Thus, it suffices to show that y^2 + z^2 is not a unit, and has no repeated factor, in k(z)[y]. Since it is of degree 2, it is certainly not a unit, so has some irreducible factor. To test for repeated factors, compute the gcd of this polynomial and its derivative, viewed as having coefficients in the field k(z): [2]
(y^2 + z^2 ) −
y 2
(2y) = z^2 = non-zero constant
[1] (^) It is here that the requirement that the characteristic not be 2 is visible.
[2] (^) It is here that the requirement that the characteristic not be 2 is visible.
