Irredicuble Polynomials

Section 16.1 Irredicuble Polynomials

Subsection Basic Irreducibility Tests

“Eliminate all other factors, and the one which remains must be the truth.”
―Sir Arthur Conan Doyle

Definition 16.1.

For an integral domain \(R\text{,}\) a polynomial \(p(x) \in R[x]\) is called irreducible if \(p(x)\) is not a unit and whenever \(p(x) = f(x) g(x)\text{,}\) either \(f(x)\) or \(g(x)\) is a unit. (It follows that such a \(p(x)\) also cannot be \(0\text{.}\))

In general, it can be very difficult to determine if a polynomial is irreducible. Here we cover some tricks that sometimes work. We will focus mostly on polynomials in \(\Q[x]\text{.}\)

Theorem 16.2.

Let \(F\) be a field and \(p(x) \in F[x]\text{.}\)

If \(p\) has degree one, it is irreducible.
If \(p\) has a root \(a \in F\) and \(\deg(p) \geq 2\text{,}\) then \(p(x)\) is not irreducible (since it factors as \(p = (x-a)q\) for some \(q\) of degree at least \(1\)).
If \(2 \leq \deg(p) \leq 3\text{,}\) then \(p(x)\) is irreducible if and only if \(p(x)\) has no roots.
Rational Root Test.
If \(p(x) \in \Q[x]\) and all the coefficients of \(p = a_n x^n + \cdots + a_0\) are integers and \(\frac{i}{j}\) is a root of \(p\) with \(i,j \in \Z\text{,}\) then \(j\) divides \(a_n\) and \(i\) divides \(a_0\text{.}\) More generally, the same holds with \(\Z\) replaced by any PID and \(\Q\) replaced by its field of fractions.

Remark 16.3.

Never, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever, ever try to use the converse of (2) or a version of (3) for polynomials of degree more than \(3\text{;}\) they are false.

Example 16.4.

Does \(x^4 + 8x^3 + 21x^2 + 20x + 13 \in \Q[x]\) have any roots? No. By the Rational Root Test the only possible roots are \(\pm 1\) and \(\pm 13\text{,}\) and a careful check rules these out. Is \(x^4 + 8x^3 + 21x^2 + 20x + 13 \in \Q[x]\) irreducible? No, it factors as \(x^2 + x + 1\) times \(x^2 + 7x + 13\)

Subsection Eisenstein’s Criterion

Definition 16.5. Content.

For a PID \(R\) and \(f \in R[x]\text{,}\) define the content of \(f\text{,}\) written \(\cont(f)\text{,}\) to be the gcd of the coefficients of \(f\text{.}\) Equivalently, \(\cont(f)\) is the generator of the principal ideal generated by the coefficients of \(f\text{.}\)

Theorem 16.6. Gauss’s Lemma: Part 1.

Let \(f(x), g(x) \in R[x]\) where \(R\) is a PID. Then \(\cont(fg) = \cont(f) \cont(g)\) (up to associates).

Proof.

We first show the following special case: If \(\cont(f) = 1\) and \(\cont(g) = 1\text{,}\) then \(\cont(fg) =1\text{.}\)

To see this, pick a prime \(p \in R\) and write \(\overline{f} \in (R/p)[x]\) for the result of modding out the coefficients of \(f\) by \(p\text{.}\) Then \(\ov{fg} = \ov{f}\ov{g}\) (since the map \(R[x] \to (R/p)[x]\) sending \(f\) to \(\ov{f}\) is a ring map). Since \(\cont(f) =1 = \cont(g)\text{,}\) we have that \(\ov{f} \ne 0\) and \(\ov{g} \ne 0\text{.}\) Since \((R/p)[x]\) is a domain, it follows that \(\ov{fg} \ne 0\text{.}\) This proves \(p\) does not divide all of the coefficients of \(fg\text{.}\) Since \(p\) was arbitrary, \(\cont(fg) = 1\text{.}\)

For the general case, let \(f = a h(x)\) and \(g = b l(x)\) where \(a =\cont(f)\) and \(b = \cont(g)\text{.}\) Note that \(\cont(h) = 1 = \cont(l)\) and \(fg = (ab) hl\text{.}\) By the case already proven, \(\cont(lh) =1\text{.}\) It follows \(\cont (abhl) =ab \cont(hl) = ab = \cont(f) \cont(g)\text{.}\)

Theorem 16.7. Gauss’s Lemma: Part 2.

Let \(R\) be a PID and let \(F\) be its field of fractions, and assume \(f(x) \in R[x]\) is a non-constant polynomial with coefficients in \(R\text{.}\) \(f(x)\) is irreducible in \(R[x]\) if and only if it is irreducible in \(F[x]\) and \(\cont(f) = 1\text{.}\)

Proof.

Suppose \(f = gh\) with \(g, h \in F[x]\text{.}\) We can find an element \(M\) of \(R\) such that \(M \cdot g(x) \in R[x]\text{.}\) (We can take \(M\) to be the product of all the denominators of the coefficients of \(g\text{,}\) for example; this is called “clearing denominators’’.) Similarly, there is an \(N \in R\) such that \(N \cdot h(x) \in R[x]\text{.}\) Further, write \(M \cdot g(x) = a \cdot l(x)\) where \(a = \cont(M g(x)) \in R\text{,}\) \(l(x) \in R[x]\text{,}\) and \(\cont(l) = 1,\) and similarly \(N \cdot g(x) = b \cdot m(x)\) where \(b = \cont(N h(x)) \in R\text{,}\) \(m(x) \in R[x]\text{,}\) and \(\cont(m) = 1\text{.}\) We have

\begin{equation*} abl(x)m(x) = MN f(x) \end{equation*}

Using Gauss’s Lemma: Part 1 we have \(\cont(lm) = \cont(l) \cont(m) = 1\) and hence

\begin{equation*} \cont(ab l(x) m(x)) = ab \cont(l(x) m(x)) = ab. \end{equation*}

We also have \(\cont(MNf(x)) = MN\cont(f) = MN\) (since we’ve already proven that \(\cont(f) = 1\)).

So \(ab = MN\) and thus by dividing we arrive at

\begin{equation*} l(x) m(x) = f(x). \end{equation*}

But \(f(x)\) is irreducible in \(R[x]\text{.}\) It follows that either \(l\) or \(m\) must be a unit in \(R[x]\) (i.e., a unit in \(R\)) and it follows that either \(g(x)\) or \(h(x)\) is a unit in \(F[x]\text{.}\) This proves \(f(x)\) is irreducible in \(F[x]\text{.}\)

(\(\Leftarrow\)) Say \(f(x) \in R[x]\) irreducible in \(F[x]\) and \(\cont(f) = 1\text{.}\) If \(f = q(x) r(x)\) with \(q,r \in R[x]\) then, since \(f\) is irreducible in \(F[x]\text{,}\) one of \(q\) or \(r\) must be a constant. But since \(\cont(f) = 1\text{,}\) this constant must be \(1\text{.}\)

Example 16.8. Irreducibility and Gauss.

Let’s prove \(f = x^4 + 7x^3 + 16 x^2 + 124 x + 23 \in \Q[x]\) is irreducible. By Gauss’s Lemma, we just need to show it is irreducible in \(\Z[x]\text{.}\)

If \(f\) did factor in \(\Z[x]\) as a product of monic polynomials, then \(\ov{f} \in \Z/2[x]\) would also factor in this way. Now \(\ov{f} = x^4 + x^3 + 1 \in \Z/2[x]\text{.}\) This has no roots (the only possibilities are \(0\) or \(1\)) and so if it did factor it would have to factor as a product of two quadratic, irreducible polynomials. The only such polynomial in \(\Z/2[x]\) is \(q = x^2+ x + 1\) (the others all have roots). But \(q^2 \ne \ov{f}\text{.}\) This proves \(\ov{f}\) is irreducible in \((\Z/2)[x]\) and hence \(f\) must be irreducible in \(\Z[x]\text{.}\)

Theorem 16.9. Eisenstein’s Criterion.

Let \(R\) be a PID with field of fractions \(F\) and assume \(f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 \in R[x]\) is a polynomial. Suppose there is a prime element \(p \in R\) such that \(p \nmid a_n\text{,}\) \(p \mid a_i\) for all \(0 \leq i \leq n-1\text{,}\) and \(p^2 \nmid a_0\text{.}\) Then \(f\) is irreducible in \(F[x]\text{.}\)

Proof.

We have \(f(x) = \cont(f) \cdot l(x)\) with \(l(x) \in R[x]\) and \(\cont(l) = 1\text{.}\) Since \(p\) does not divide the leading coefficient of \(f\text{,}\) we have \(p \nmid \cont(f)\text{.}\) It therefore follows that the three assumptions involving the coefficients of \(f(x)\) are also satisfied by the coefficients \(l(x)\text{.}\) Moreover, \(f\) is irreducible in \(F[x]\) if and only if \(l(x)\) is irreducible in \(F[x]\text{.}\) We may therefore assume without loss of generality that \(\cont(f) = 1\text{.}\)

By Gauss’s Lemma we just need to prove it is irreducible in \(R[x]\text{.}\) Suppose \(f(x) = g(x)h(x)\) with \(g(x), h(x) \in R[x]\text{,}\) where

\begin{equation*} g(x) = b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0 \end{equation*}

and

\begin{equation*} h(x) = c_l x^l + c_{l-1} x^{l-1} + \cdots + c_0 \end{equation*}

where \(m+l = n\) and \(b_m c_l = a_n\text{.}\)

Upon modding out by \(p\) we get

\begin{equation*} \ov{f} = \ov{g} \ov{h} \in (R/p)[x]. \end{equation*}

The assumptions on \(f\) give that

\begin{equation*} \ov{f} = \ov{a_n} x^n \end{equation*}

with \(\ov{a_n} \ne 0\text{.}\) Since \(R/p\) is a domain, it follows that

\begin{equation*} \ov{g} = \ov{b_m} x^m \, \text{ and } \, \ov{h} = \ov{c_l} x^l \end{equation*}

On the other hand, we also have

\begin{equation*} a_0 = b_0c_0, \end{equation*}

and since \(p^2 \nmid a_0\text{,}\) we have that \(p \nmid b_0\) or \(p \nmid c_0\text{.}\) So \(\ov{b_0} \ne 0\) or \(\ov{c_0} \ne 0\text{.}\) This can only occur if \(m = 0\) or \(l = 0\text{.}\)

We have shown that if \(f\) factors in \(R[x]\) as \(f = g \cdot h\text{,}\) then at least one of \(g\) of \(h\) must be a constant polynomial. Since \(\cont(f) = 1\text{,}\) this factor must be a unit. So \(f\) is irreducible in \(R[x]\text{.}\)

Example 16.10.

For instance, \(5x^{11} - 30 x^4 +60 x^3 +360 x- 30\) is irreducible in \(\Q[x]\) thanks to Eisenstein’s Criterion applied with \(p = 2\text{.}\) (Note that it isn’t irreducible in \(\Z[x]\) since it does not have unit content.)

Proposition 16.11.

For any prime \(p\text{,}\) the \(p\)-th cyclotomic polynomial

\begin{equation*} f(x) = x^{p-1} + x^{p-2} + \cdots + x^2 + x+1 \in \Z[x] \end{equation*}

is irreducible in \(\Q[x]\text{.}\)

Proof.

Consider the ring isomorphism \(\phi: \Q[x] \xra{\cong} \Q[y]\) given by \(\phi(h(x)) = h(y+1)\text{.}\) I claim that

\begin{equation*} \phi(f(x)) = y^{p-1} + p y^{p-2} + {p \choose 2} y^{p-3} +{p \choose 3} y^{p-4} +\cdots + {p \choose p-1} y + p. \end{equation*}

To see this, we note that \(f(x) (x-1) = x^p - 1\) and by the binomial theorem we have

\begin{equation*} \phi(x^p-1) = (y+1)^p - 1 = y^p + p y^{p-1} + {p \choose 2} y^{p-2} + \cdots + py. \end{equation*}

Since \(\phi(x^p-1) = \phi(f(x)) \phi(x-1) = \phi(f(x)) y\text{,}\) the claim follows.

By Eisenstein’s Criterion, \(\phi(f(x))\) is irreducible in \(\Q[y]\) and, since \(\phi\) is an isomorphism, it follows that \(f(x)\) is irreducible in \(\Q[x]\text{.}\)

Proposition 16.12.

Assume \(R\) is an integral domain and \(f(x) \in R[x]\) is a monic polynomial with coefficients in \(R\text{.}\) If there is a prime element \(p \in R\) such that \(\ov{f} \in (R/p)[x]\) is irreducible in \((R/p)[x]\text{,}\) then \(f\) is irreducible in \(R[x]\text{.}\)

Proof.

We prove the contrapositive. Suppose \(f\) is reducible in \(R[x]\text{.}\) Then we have \(f(x) = g(x) h(x)\) for some monic, non-constant polynomials \(g\) and \(h\) in \(R[x]\text{.}\) It follows that \(\ov{f} = \ov{gh} = \ov{g} \ov{h}\) holds in \((R/p)[x]\text{.}\) But since \(g\) and \(h\) are monic, non-constant polynomials, \(\ov{g}\) and \(\ov{h}\) are both non-constant polynomials in \((R/p)[x]\) and hence \(\ov{f}\) is reducible in \((R/p)[x]\text{.}\)

Example 16.13.

The assumption that \(f\) be monic in this Proposition is necessary. Consider for instance \(f(x) = (3x+1)(x+1) = 3 x^2 + 4x + 1 \in \Z[x]\text{.}\) Modding out by \(3\) yields a linear polynomial in \((\Z/3)[x]\) which is thus irreducible. But \(f\) isn’t irreducible in \(\Z[x]\text{.}\) (The Proposition CITEX can be generalized to non-monic polynomials by adding the assumption that \(p\) does not divided the leading coefficient of \(f\text{.}\))

Proposition 16.14.

Let \(F\) be a field and \(G\) a finite subgroup of the multiplicative group \(F^\times\text{.}\) Then \(G\) is cyclic.

Proof.

Let \(F\) be a field and \(G\) a finite subgroup of order \(n\) the multiplicative group \(F^\times\text{.}\) Let \(k\) denote the LCM of all orders of elements in \(G\text{.}\) Notice that as \(g^n=1\) for all \(g\in G\) we have \(k\leq n\text{.}\)

However, as \(g^k=1\) for all \(g\in G\) we know that \(g\) is the root of the polynomial \(f(x)=x^k-1\) in \(F[x]\text{.}\) By the Factor Theorem there are thus at most \(k\) roots of \(f\text{,}\) but there are \(n\) distinct roots. Thus \(k=n\text{.}\)

Thus the LCM of all orders of elements in \(G\) is \(n\text{.}\) Notice that as \(G\) is abelian every Sylow \(p\)-subgroup of \(G\) is normal, and thus there is only one of each. This means that \(G\) can be written as a direct product of cyclic groups of relatively prime order; hence \(G\) is itself cyclic.

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