Section 3.5 Identifying irreducibles
In previous sections we have discussed the notion of unique factorization of elements into a product of irreducibles, but in no context other that the integers,Definition 3.5.1.
Let
and (i.e., is a common divisor)If
and then meaning any other common divisor divides making the greatest in terms of divisibility.
Remark 3.5.2.
It is an easy exercise that in an integral domain
Remark 3.5.3.
Also in the integers, unless someone asked you to compute the gcd of large numbers, your most likely method of finding a gcd was to factor both numbers, and determine the largest power of each prime dividing both numbers.
But as it turns out, factoring in
Exercise 3.5.1.
Greatest common divisors are guaranteed to exist in any UFD, but how they manifest themselves differs depending on whether the UFD has any more robust properties.
(a)
Let
Just as in \(\Z\text{,}\) show that if \(a = bq+r\text{,}\) then
where by \(\gcd(a,b)\) we mean any greatest common divisor.
(b)
Let
show that
Observe that if the ideal \((a,b) = (d)\text{,}\) then \(d\) is a common divisor of \(a,b\text{.}\) Moreover, recall that
so if \((d) = (a,b),\) then \(d = ax_0 + by_0\) for some \(x_0, y_0 \in R\text{.}\) Use that to show that \(d\) satisfies the other requirement of being a gcd.
(c)
Let
Theorem 3.5.4. Roots and linear factors.
Let
(sketch).
Apply the division algorithm in \(F[x],\) dividing \((x-a)\) into \(p(x),\) and use the evaluation homomorphism at \(a\) on the resulting equation.
Theorem 3.5.5. Rational Root Test.
Let
Proof.
Assume that \(\alpha\) is a root of \(p\text{,}\) and evaluate \(p\) at \(\alpha\) and clear the resulting denominators to produce the equation:
If we move one of the end terms to the other side of the equation, we acquire two equations
By inspection of those equations,
However as \(\gcd(r,s) = 1\text{,}\) we must have \(r\mid a_0\) and \(s\mid a_n.\)
Definition 3.5.6.
Let
Theorem 3.5.7. Gauss's lemma.
Let
Remark 3.5.8.
In some texts, Gauss's lemma is phrased as the product of two primitive polynomials is primitive which implies that for two polynomials
This version of Gauss's lemma is necessary to prove the one above.
Corollary 3.5.9.
Let
Proof.
If \(p\) is irreducible in \(R[x]\text{,}\) then it is necessarily primitive. If it were reducible in \(K[x],\) then \(p = FG\) for non-constant polynomials in \(K[x].\) By Gauss’s lemma, there exists \(\alpha, \beta\in K^\times\) so that
Since \(\deg f, \deg g \ge 1,\) this would imply \(p\) is reducible in \(R[x]\text{,}\) a contradiction.
Conversely, suppose that \(p\) is primitive and irreducible in \(K[x].\) If it were reducible in \(R[x],\) then \(p = fg\) for \(f,g \in R[x]\) with \(\deg f, \deg g \ge 1\) since \(p\) is primitive. Then neither \(f\) nor \(g\) are units in \(K[x]\text{,}\) so the factorization shows \(p\) is reducible in \(K[x]\text{,}\) a contradiction.
Example 3.5.10.
Consider the polynomial
Theorem 3.5.11. Reduction criterion.
Let
Proof.
The proof is by contradiction. Suppose that \(\overline p\) cannot be factored into the product of two non-constant polynomials, but \(p\) is reducible in \(R[x]\text{,}\) say \(p = fg.\) Since \(p\) is monic, the leading coefficients of \(f,g\) must be units, which forces \(\deg f, \deg g \ge 1\) to provide a witness to reducibility. Set
As we noted, \(a_m, b_n \in R^\times.\)
Consider the reduction modulo \(I:\)
We note that
Since \(a_m, b_n \in R^\times\) and \(I\) is a proper ideal, we see \(a_m + I \ne 0+ I\) and \(b_n+I \ne 0+I,\) so \(\deg \overline f, \deg \overline g \ge 1\) which gives a factorization of \(\overline p\) into two non-constant polynomials, a contradiction.
Example 3.5.12.
Let's try the reduction criterion on the polynomial
As a first guess, let
Let
Example 3.5.13.
Let
Let
Since the degree is 3,
We soon give a different proof that this polynomial is irreducible using Eisenstein's criterion.
Theorem 3.5.14. Eisenstein's criterion.
Let
Then
Proof.
We proceed by contradiction and assume that \(p\) is reducible in \(R[x]\text{.}\) Since \(p\) is primitive, this means
say
Recalling that \(\pi\) is a prime element, then since \(\pi \mid a_0=b_0c_0\text{,}\) it must divide one of \(b_0\) or \(c_0\text{,}\) but \(\pi^2 \nmid a_0\) says it can divide precisely one, say \(\pi\mid b_0\) and \(\pi \nmid c_0.\) Also since \(\pi \nmid a_n = b_rc_s\text{,}\) we see \(\pi \nmid b_r\) and \(\pi \nmid c_s.\)
Since \(\pi \mid c_0\text{,}\) but \(\pi \nmid c_s\text{,}\) let \(\ell\) be the smallest index so that \(\pi \nmid c_\ell.\) So
Since \(\ell \lt n\text{,}\) we know that \(\pi \mid a_\ell,\) and by the choice of \(\ell\text{,}\) \(\pi \mid c_0, \dots, c_{\ell-1}\text{,}\) which implies \(\pi\mid b_0c_\ell\text{.}\) But \(\pi\) prime then says \(\pi \mid b_0\) or \(\pi \mid c_\ell\text{,}\) a contradiction.
Example 3.5.15.
There exist irreducible polynomials of all degrees
Let \(p\) be a prime in \(\Z.\) Then by Eisenstein, \(x^n \pm p\) is irreducible in \(\Z[x].\)
Example 3.5.16.
Show that
It is irreducible by Eisenstein with \(p=3.\)
Example 3.5.17.
Show that
We looked at this polynomial when considering the reduction criterion. So here \(R=\Q[x]\) is our UFD and we write \(p\) as an element of \(R[y]\) as
Moreover, \(\pi = (x-1) \in R=\Q[x]\) is an irreducible. We check that Eisenstein's criterion applies, and we are done.
Example 3.5.18.
Let
is irreducible in
Let \(g(x) = f(x+1)\text{.}\) Then \(f\) is irreducible iff \(g\) is. We see that
which is irreducible in \(\Z[x]\) using Eisenstein with the prime \(p\) since \(p \mid \binom{p}{k}\) for \(1\le k \le p-1\) and \(\binom{p}{p-1}=p.\)