Lemma 2.16.1.
Let \(R\) be a nonzero commutative ring. If \(f\in R[x]\) is a monic polynomial, then if \(f\) factors as a product of two polynomials in \(R[x]\text{,}\) then it factors as a product of two monic polynomials.
Section 2.16 Factorization over integral domains
Gauss’s lemma is one important tool for answering factorization questions about polynomials \(f\in R[x]\) where \(R\) is a UFD. Roughly speaking, the main result there is that we investigate factorization in this context by comparing factorization in \(R[x]\) with factorization in \(F[x]\text{.}\)
In this section we add a couple of additional tools that can be used for arbitrary integral domains \(R\) (that need not be UFDs). First a simple lemma about factorization of monic polynomials in \(R[x]\) for an arbitrary nonzero commutative ring \(R\text{.}\)
Theorem 2.16.2. Reduction modulo \(I\).
Let \(R\) be an integral domain, and let \(f\) be a nonconstant monic polynomial in \(R[x]\text{.}\) If there is a proper ideal \(I\) of \(R\) such that the reduction \(f\boldmod I \in (R/I)[x]\) cannot be factored as a product of two polynomials of smaller degree, then \(f\) is irreducible in \(R[x]\text{.}\)
In the special case where \(I=P\) is a prime ideal of \(R\text{,}\) the condition that \(\overline{f}=f\boldmod P\in (R/P)[x]\) cannot be factored as a product of two polynomials of smaller degree is equivalent to \(\overline{f}\) being irreducible.
Proof.
Given \(f\) and \(I\) as in the statement, suppose by contradiction that \(f\) is reducible. In that case, since \(f\) is monic, we can write \(f(x)=g(x)h(x)\) for polynomials \(g,h\in R[x]\text{.}\) Since \(f\) is monic the products of the leading coefficients of \(g\) and \(h\) is \(1\text{.}\) It follows that these leading coefficients are units in \(R\text{,}\) and thus we factor them out of \(\)
Remark 2.16.3. Monic is important.
The condition that \(f\) be monic is important here. Note that \(f(x)=2x\) is irreducible modulo \(3\text{,}\) but is reducible in \(\Z[x]\text{.}\)
As another illuminating example in this regard, consider \(f(x)=3x^2+4x+1\text{.}\) We have \(f\boldmod 3=x+1\in \Z/3[x]\text{,}\) which clearly cannot be factored as a product of two polynomials of smaller degree. However we have \(f(x)=(3x+1)(x+1)\) in \(\Z[x]\text{.}\)
Remark 2.16.4. Nondefinitive irreducibility test.
It is important to note that Theorem 2.16.2 only gives us a partial or nondefinitive test for irreducibility. That is, with the setup of the theorem, if \(\overline{f}\) does not factor as a product as a polynomial of smaller degree, then \(f\) is irreducible. However, there is no useful converse to this statement. For example, it is not true that if \(\overline{f}\) is reducible in \(R/P[x]\) for all prime ideals then \(f\) is reducible in \(R[x]\text{;}\) nor is this true if we look at \(\overline{f}\in R/I[x]\) for all nontrivial ideals (prime or otherwise).
Indeed you will show in homework the the polynomial \(f(x)=x^4+1\in \Z[x]\) is irreducible, and yet \(\overline{f}\) is reducible in \(\Z/p\Z[x]\) for all prime integers \(p\text{.}\) By the way, in this case it is not true that \(f\) is reducible modulo all nonzero integers \(n\text{:}\) indeed, you can show that \(x^4+1\) is irreducible in \(\Z/4\Z\) in a straightforward manner.
Similarly, it is possible (but more difficult) to show that \(f(x)=x^4-72x^2+4\) is irreducible modulo \(n\) for all nonzero integers \(n\text{,}\) and yet \(f\) is irreducible in \(\Z[x]\text{.}\)
Example 2.16.5. Reduction mod \(I\text{:}\) cubic.
Decide whether \(f(x)=x^3+246x^2+608x+85\in \Q[x]\) is irreducible.
Solution.
Since \(f\in \Z[x]\) and is primitive, irreducibility of \(f\) in \(\Q[x]\) is equivalent to irreducibility of \(f\) in \(\Z[x]\text{.}\) We investigate the latter by examining \(f\boldmod p\) for various prime integers \(p\text{.}\)
We have \(f\boldmod 2=x^3+1\in \F_2[x]\text{,}\) which is reducible: $x^3+1=(x+1)(x^2+x+1)$ in \(\F_2[x]\text{.}\)
We have \(f\boldmod 3=x^3+2x+1\in \F_3[x]\text{,}\) which is irreducible. Indeed, since the degree is 3, it suffices to check that this polynomial has no roots in \(\F_3\text{,}\) which is easy to verify.
Since \(f\boldmod 3\) is irreducible, we conclude that \(f\) is irreducible in \(\Z[x]\text{,}\) and thus also in \(\Q[x]\text{.}\)
Example 2.16.6. Reduction mod \(I\text{:}\) quartic.
Solution.
Again, by Gauss’s lemma and its consequences, \(f\) is irreducible in \(\Q[x]\) if and only if \(f\) is irreducible in \(\Z[x]\text{.}\)
We have \(\overline{f}=f\boldmod 2=x^4+x+1\in \F_2[x]\text{.}\) We claim \(\overline{f}\) in \(\F_2[x]\text{.}\) It is easily verified that it has no roots, so if it factored, we would have \(x^4+x+1=(x^2+ax+b)(x^2+cx+d)\) in \(\F_2[x]\text{.}\) Expanding the product out and comparing coefficients with \(\overline{f}\text{,}\) we would have
\begin{align*}
a+c \amp = 0\\
ac+b+d \amp =0\\
ad+bc \amp = 1 \\
bd \amp =1\text{.}
\end{align*}
Using the very simple arithmetic of \(\F_2\text{,}\) we see that the last equation implies \(b=d=1\text{.}\) But then the third equation becomes
\begin{align*}
b(a+c) \amp =a+c=1\text{.}
\end{align*}
But this contradicts the first equation, which implies \(a+c=0\text{.}\) We conclude there is no such factorization, and that \(\overline{f}\) is irreducible over \(\F_2\text{.}\) It follows that \(f\) is irreducible in \(\Z[x]\text{,}\) and hence also in \(\Q[x]\text{.}\)
The following irreducibilty criterion is strictly speaking a corollary of Theorem 2.16.2, but it is famous enough to warrant the theorem designation.
Theorem 2.16.7. Eisenstein’s criterion.
Let \(R\) be an integral domain, and let \(f(x)=\sum_{i=0}^{n}a_ix^i\) be a nonconstant monic polynomial of \(R[x]\text{.}\) If \(P\) is a prime ideal of \(R\) satisfying
-
\(a_0\notin P^2=PP\text{,}\)
then \(f\) is irreducible in \(R[x]\text{.}\)
Proof.
Suppose by contradiction that \(f\) is reducible. Since \(f\) is monic, we may assume that \(f(x)=g(x)h(x)\text{,}\) where \(g(x)=\sum_{k=0}^mb_kx^k\text{,}\) \(h(x)=\sum_{k=0}^{n-m}c_kx^k\) are both monic and \(m, n-m< n\text{.}\) It follows that modulo \(P\) we have
\begin{align*}
\overline{f}(x) \amp =x^n=\overline{g}(x)\overline{h}(x)\text{,}
\end{align*}
where \(\deg \overline{g}=m\text{,}\) \(\deg \overline{h}=n-m\text{.}\) It is not difficult to see that since \(R/P\) is an integral domain, we must have \(\overline{b_0}=\overline{c_0}=0\text{:}\) indeed, if \(\overline{b_0}\ne 0\text{,}\) then \(\overline{g}\overline{h}\) has a nonzero term of degree \(n-m< n\text{;}\) if \(\overline{c_0}\ne 0\text{,}\) then \(\overline{f}\overline{g}\) has a nonzero term of degree \(m< n\text{.}\) It follows that \(b_0\in P\) and \(c_0\in P\text{,}\) and thus \(a_0=b_0c_0\in P^2\text{,}\) contradicting our assumption! Thus \(f\) is irreducible.
It’s great when Eisenstein’s criterion can be used, but its scope of application is unfortunately somewhat narrow, as the following somewhat artificial examples illustrate.
Example 2.16.8. Eisenstein.
Show that the given \(f\in R[x]\) is irreducible.
-
\(\displaystyle f(x)=x^{17}-30x^2+600x-180\in \Z[x]\)
-
\(\displaystyle f(x)=x^4+26x^2+(6+4i)\in \Z[i][x]\)
Solution.
-
\(f\) is Eisenstein with respect to the prime ideal \((5)\text{,}\) thus irreducible. Note that \(f\) is not Eisenstein with respect to \(2\) or \(3\text{.}\)
-
\(f\) is Eisenstein with respect to the prime ideal \((3+2i)\) in \(\Z[i]\text{.}\) Note that \((2)\) is not prime in \(\Z[i]\text{,}\) and \(f\) is not Eisenstein with respect to the prime ideal \((1+i)\) lying over \((2)\text{,}\) since \(6+4i=2(3+2i)\in (1+i)^2\text{.}\)
Example 2.16.9. Eisenstein: \(n\)-th roots in UFD.
Let \(R\) be a UFD, and let \(a\in R-R^*\) satisfy \(v_p(a)\leq 1\) for all irreducible elements \(p\in R\text{.}\) Show that \(f(x)=x^n-a\) is irreducible in \(R[x]\) for any positive integer \(n\text{.}\)
Solution.
In general, for any element \(b\in R\) and irreducible element \(p\text{,}\) setting \(P=(p)\text{,}\) we have
\begin{align*}
a\in P^n \amp \iff p^n\mid a\\
\amp \iff v_p(a)\geq n\text{.}
\end{align*}
Since \(a\) is not a unit and \(v_p(a)\leq 1\) for all irreducible elements \(p\text{,}\) there is some irreducible element \(p\) such that \(v_p(a)=1\text{.}\) It follows that \(a\in P-P^2\text{,}\) and thus that \(f(x)=x^n-a\) is Eisenstein with respect to \((p)\text{,}\) making it irreducible.
Occasionally we can make indirect use of Eisenstein’s criterion, as the next result illustrates.
Theorem 2.16.10. Cyclotomic polynomial: \(p\) prime.
Let \(p\) be a prime integer and let \(\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots +x+1\text{.}\)
-
Let \(\zeta=e^{2\pi i/p}=\cos(2\pi/p)+\sin(2\pi/p)i\text{,}\) and let \(\Q[\zeta]=\{g(\zeta)\in \C\mid g\in \Q[x]\}\text{,}\) the subring of \(\C\) generated over \(\Q\) by \(\zeta\text{.}\) We have\begin{align*} \Q[x]/(\Phi_p(x)) \amp \cong \Q[\zeta]\text{.} \end{align*}In particular, \(\Q[\zeta]\) is a subfield of \(\C\) of dimension \(p-1\) as a \(\Q\)-vector space.
Remark 2.16.11. Primitive roots of unity.
In general, given positive integer \(n\text{,}\) the complex number \(\zeta_n=e^{2\pi/n}=\cos(2\pi/n)+\sin(2\pi/n)i\) is called a primitive \(n\)-th root of unity. This is because as an element of \(\C^*\text{,}\) we have \(\ord \zeta_n=n\text{.}\) As with prime integers the set
\begin{align*}
\Q[\zeta_n] \amp = \{g(\zeta_n)\in \C\mid g\in \Q[x]\}
\end{align*}
is in fact a subfield of \(\C\) that we call the \(n\)-th cyclotomic extension of \(\Q\text{.}\) Since it is a \(\Q\)-algebra, it has a well-defined dimension as a \(\Q\)-vector space. For \(n=p\text{,}\) the result above tells us that the \(p\)-th cyclotomic extension has \(\Q\)-dimension \(p-1\text{.}\) What do you think the dimension is for general \(n\text{?}\) You guessed it: \(\phi(n)\text{.}\) This will be one of the big results of the field theory portion of this course.
