## 305 -8. Irreducibility

In this lecture I want to present a couple of short results that are nevertheless very useful in practice when trying to show that a given polynomial in ${{mathbb Q}[x]}$ is irreducible. Of course, we may assume that the polynomial actually has integer coefficients. In this case, it turns out that analyzing whether the polynomial factors over ${{mathbb Z}[x]}$ suffices.

Lemma 1 (${mbox{Gauss}}$) Let ${p(x)in{mathbb Z}[x].}$

1. Suppose that ${p(x)=q(x)r(x)}$ with ${q,rin{mathbb Z}[x].}$ Let ${t}$ be a prime number. If ${t}$ divides all the coefficients of ${p,}$ then either it divides all the coefficients of ${q}$ or else it divides all the coefficients of ${r.}$
2. Suppose that ${p(x)=q(x)r(x)}$ with ${q,rin{mathbb Q}[x].}$ Then there is a rational ${lambdane0}$ such that, letting ${bar q=lambda q}$ and ${bar r=lambda^{-1}r,}$ then ${bar q,bar rin {mathbb Z}[z].}$ In other words, if ${p}$ is irreducible over ${{mathbb Z},}$ then it is irreducible over ${{mathbb Q}.}$

Proof: To fix notation, let’s say that

$displaystyle p(x)=a_0+a_1x+dots+a_n x^n,$

that

$displaystyle q(x)=b_0+b_1x+dots+b_m x^m,$

and that

$displaystyle r(x)=c_0+c_1x+dots+c_k x^k.$

1. Suppose first ${q,rin{mathbb Z}[x]}$ (i.e., all the ${b_i}$ and all the ${c_j}$ are integers), and let ${t}$ be a prime that divides all the coefficients of their product. Towards a contradiction, suppose that ${t}$ does not divide all the coefficients of ${q}$ and does not divide all the coefficients of ${r.}$ By the well-ordering principle there is then a least ${i,}$ ${0le ile m,}$ such that

$displaystyle tnot| b_i$

and, similarly, there is a least ${j,}$ ${0le jle k,}$ such that

$displaystyle tnot| c_j.$

Now compare the coefficient of ${x^{i+j}}$ in both sides of the equation

$displaystyle p(x)=q(x)r(x).$

We have

$displaystyle a_{i+j}=sum_{s=0}^{i+j}b_s c_{i+j-s},$

where, if necessary, we set ${b_s=0}$ if ${s>m,}$ and ${c_s=0}$ if ${s>k.}$

Note that

$displaystyle t|sum_{s=0}^{i-1} b_s c_{i+j-s},$

by minimality of ${i.}$ Similarly,

$displaystyle t|sum_{s=i+1}^{i+j} b_s c_{i+j-s},$

by minimality of ${j.}$

We are assuming that ${t|a_{i+j},}$ so we are forced to conclude that ${t| b_ic_j.}$ This is a contradiction, since ${t}$ is prime but ${tnot| b_i}$ and ${tnot| c_j.}$

2. Now suppose that ${p=qr,}$ that ${pin{mathbb Z}[x],}$ and that ${q,rin{mathbb Q}[x].}$ By clearing up denominators, we may find a nonzero integer ${n}$ such that the following principle ${(*)}$ holds: ${np=hat qhat r}$ for some ${hat q,hat rin{mathbb Z}[x],}$ with ${hat q}$ a rational multiple of ${q}$ and ${hat r}$ a rational multiple of ${r,}$ as we can take ${hat q=k_1 q}$ and ${hat r=k_2 r}$ for appropriate integers ${k_1,k_2}$ with ${n=k_1k_2.}$ By the well-ordering principle, there is a least positive integer ${m}$ such that ${(*)}$ holds with ${m}$ in the place of ${n.}$ We are done if we show that ${m=1.}$ Suppose otherwise. Then there is a prime number ${t}$ such that ${t|m.}$ Then ${t}$ divides all the coefficients of ${tp.}$ Then, by item 1, ${t}$ divides all the coefficients of ${hat q,}$ or it divides all the coefficients of ${hat r.}$ By dividing both sides of the equation by ${t,}$ we find that ${(*)}$ holds with ${m/t in place of ${m.}$ This contradicts the minimality of ${m.}$ $Box$

For example,

$displaystyle x^2-1=left (frac23 x+frac 23right )left (frac 32 x -frac 32right ).$

We can then take ${n=6,}$ to obtain

$displaystyle 6(x^2-1)=(3x+3)(2x-2).$

The prime ${t=2}$ divides all the coefficients of ${6(x^2-1)=6x^2-6,}$ so it must divide all the coefficients of one of the factors in the right hand side. A summary inspection reveals that, indeed, 2 divides the coefficients of ${2x-2.}$ Dividing both sides by 2 gives us the equation

$displaystyle 3(x^2-1)=(3x+3)(x-1).$

The prime ${t=3}$ divides all the coefficients of ${3(x^2-1)=3x^2-3,}$ so it must divide all the coefficients of one of the factors in the right hand side, in this case, ${3x+3.}$ Dividing by 3 shows that

$displaystyle x^2-1=(x+1)(x-1).$

As another example, consider ${p(x)=x^3+3x+1.}$ If this polynomial factors over ${{mathbb Z}[x],}$ one of its factors must be linear, and monic, say ${x-r.}$ We also must have ${r|1,}$ so ${r=pm1.}$ Since ${r=1}$ and ${r=-1}$ are not roots of ${p(x),}$ we conclude that ${p}$ is irreducible over ${{mathbb Z}[x].}$ It follows that ${p(x)}$ is irreducible over ${{mathbb Q}[x].}$ ${mbox{Gauss}}$‘ lemma provides us with the following criterion that allows us to show in many cases irreducibility over ${{mathbb Q}[x]:}$

Theorem 2 (Eisenstein) Suppose that ${p(x)=a_0+a_1x+dots+a_nx^nin{mathbb Z}[x].}$ If there is a prime number ${t}$ such that:

1. ${tnot| a_n,}$
2. ${t|a_i}$ for ${0le i
3. ${t^2not| a_0,}$

then ${p}$ is irreducible over ${{mathbb Q}[x].}$

Proof: By ${mbox{Gauss}}$‘ lemma it suffices to show that ${p}$ is irreducible over ${{mathbb Z}[x].}$ Suppose otherwise, and let

$displaystyle q(x)=b_0+b_1x+dots+b_m x^m$

and

$displaystyle r(x)=c_0+c_1x+dots+c_k x^k$

be polynomials in ${{mathbb Z}[x]}$ with ${p=qr}$ and ${0

Since ${tnot|a_n= b_m c_k,}$ then ${tnot|b_m}$ and ${tnot|c_k.}$ It follows that there are least integers ${i}$ with ${0le ile m}$ such that ${tnot|b_i,}$ and ${j}$ with ${0le jle k}$ such that ${tnot|c_j.}$

Since ${t^2not| a_0=b_0c_0,}$ then either ${tnot|b0}$ or ${tnot|c_0.}$ by exchanging the roles of ${q,r,}$ we may assume that ${tnot|c_0,}$ so ${j=0}$ in the paragraph above. On the other hand, ${t|a_0,}$ so ${t|b_0,}$ so ${i>0}$ in the paragraph above.

Now compare the coefficient of ${x^i}$ in both sides of the equation ${p=qr.}$ We have

$displaystyle a_i=sum_{s=0}^i b_s c_{i-s}.$

Note that ${t|a_i,}$ because ${ile m0.}$ Note also that

$displaystyle t |sum_{s=0}^{i-1} b_s c_{i-s},$

by minimality of ${i.}$

It follows that ${t | b_ic_0,}$ which is a contradiction, since ${t}$ is prime, ${tnot|b_i}$ and ${tnot|c_0.}$ $Box$

For example, ${p(x)=x^7-9x^5+60x^2+21}$ is irreducible over ${{mathbb Q}[x],}$ as witnessed by the prime ${t=3.}$

As a more significant example, suppose that ${t}$ is prime, and let’s check that

$displaystyle p(x)=1+x+dots+x^{t-2}+x^{t-1}$

is irreducible over ${{mathbb Q}[x].}$ This does not follow directly from Eisenstein’s criterion, since no prime divides 1.

We know that ${p(x)(x-1)=x^t-1.}$ Let ${y=x-1,}$ so ${y+1=x}$ and

$displaystyle yp(y+1)=sum_{i=1}^t {tchoose i} y^i,$

or ${p(y+1)=sum_{i=1}^t {tchoose i} y^{i-1}.}$

Note that

$displaystyle t|{tchoose i}$

for all ${i}$ with ${1le i since

$displaystyle {tchoose i}=frac{t!}{i!(t-i)!}$

is an integer and ${t}$ is a factor of the numerator but not of the denominator.

Since ${{tchoose 1}=t}$ and ${{tchoose t}=1,}$ Eisenstein’s criterion shows that ${p(y+1)}$ is irreducible over ${{mathbb Q}[x],}$ and therefore so is ${p(x).}$

Corollary 3 If in Theorem 2 the conditions on ${t}$ are instead:

1. ${tnot| a_0,}$
2. ${t|a_i}$ for ${0
3. ${t^2not| a_n,}$

then again ${p(x)}$ irreducible over ${{mathbb Q}[x].}$

Proof: Let ${q(x)=x^n p(1/x).}$ Then Eisenstein’s theorem applies to ${q.}$ $Box$

For example, if ${p(x)=21x^7+60 x^5-9 x^2 +1,}$ then

$displaystyle q(x)=x^7left(frac{21}{x^7}+frac{60}{x^5}-frac{9}{x^2}+1right)=21+60x^2-9x^5+x^7.$

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