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{Gau\ss}}) Let {p(x)\in{\mathbb Z}[x].}

  1. Suppose that {p(x)=q(x)r(x)} with {q,r\in{\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,r\in{\mathbb Q}[x].} Then there is a rational {\lambda\ne0} such that, letting {\bar q=\lambda q} and {\bar r=\lambda^{-1}r,} then {\bar q,\bar r\in {\mathbb Z}[x].} 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,r\in{\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,} {0\le i\le m,} such that

\displaystyle t\nmid b_i

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

\displaystyle t\nmid 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\mid \sum_{s=0}^{i-1} b_s c_{i+j-s},

by minimality of {i.} Similarly,

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

by minimality of {j.}

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

2. Now suppose that {p=qr,} that {p\in{\mathbb Z}[x],} and that {q,r\in{\mathbb Q}[x].} By clearing up denominators, we may find a nonzero integer {n} such that the following principle {(*)} holds: {np=\hat q\hat r} for some {\hat q,\hat r\in{\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\mid 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<m} in place of {m.} This contradicts the minimality of {m.} \Box

For example,

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

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\mid 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{Gau\ss}}‘ 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^n\in{\mathbb Z}[x].} If there is a prime number {t} such that:

  1. {t\nmid a_n,}
  2. {t\mid a_i} for {0\le i<n,}
  3. {t^2\nmid a_0,}

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

Proof: By {\mbox{Gau\ss}}‘ 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<m,k.} Since {t\nmid a_n= b_m c_k,} then {t\nmid b_m} and {t\nmid c_k.} It follows that there are least integers {i} with {0\le i\le m} such that {t\nmid b_i,} and {j} with {0\le j\le k} such that {t\nmid c_j.} Since {t^2\nmid a_0=b_0c_0,} then either {t\nmid b_0} or {t\nmid c_0.} By exchanging the roles of {q,r,} we may assume that {t\nmid c_0,} so {j=0} in the paragraph above. On the other hand, {t\mid a_0,} so {t\mid b_0,} so that {i} as above is positive.

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

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

Note that {t\mid a_i,} because {i\le m<n.} Note also that

\displaystyle t \mid\sum_{s=0}^{i-1} b_s c_{i-s},

by minimality of {i.}

It follows that {t \mid b_ic_0,} which is a contradiction, since {t} is prime, {t\nmid b_i} and {t\nmid 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 {t\choose i} y^i,

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

Note that

\displaystyle t\mid{t\choose i}

for all {i} with {1\le i<t,} since

\displaystyle {t\choose i}=\frac{t!}{i!(t-i)!}

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

Since {{t\choose 1}=t} and {{t\choose 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. {t\nmid a_0,}
  2. {t\nmid a_i} for {0<i\le n,}
  3. {t^2\nmid 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^7\left(\frac{21}{x^7}+\frac{60}{x^5}-\frac{9}{x^2}+1\right)=21+60x^2-9x^5+x^7.

Typeset using LaTeX2WP. Here is a printable version of this post.

Advertisement

This entry was posted on Monday, April 20th, 2009 at 12:03 pm and is filed under 305: Abstract Algebra I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s

%d bloggers like this: