Show x^4 - 22x^2 + 1 is irreducible in Q
Daniel C. Bastos
Proof. Suppose f(x) is reducible in Q[x]. Then there exists p/q in in
Q, such that f(p/q) = 0, where p/q is in lowest terms. By the
``rational zeros theorem,'' p divides a(0) = 1 and q divides a(n) = 1.
So p/q is 1 or -1 because the only divisors of 1 are 1 and -1, but
neither 1 nor -1 are roots of f, so there is no rational p/q such that
f(p/q) = 0; hence, f(x) is irreducible.
William Elliot
> Exercise. Show f(x) = x^4 - 22x^2 + 1 is irreducible over Q.
>
> Proof. Suppose f(x) is reducible in Q[x]. Then there exists p/q in in
> Q, such that f(p/q) = 0, where p/q is in lowest terms.
f(x) may be reducible and not have a rational root because f is of even
degree. For example, g(x) = (x^2 + 1)^2 is obviously reducible over Q
but there is no p/q in Q with g(p/q) = 0. The only roots to g are +-i.
> By the
> ``rational zeros theorem,'' p divides a(0) = 1 and q divides a(n) = 1.
> So p/q is 1 or -1 because the only divisors of 1 are 1 and -1, but
> neither 1 nor -1 are roots of f, so there is no rational p/q such that
> f(p/q) = 0; hence, f(x) is irreducible.
>
You've show f has no rational roots, thus no rational linear factor.
You haven't excluded that f can have two quadratic factors irreduciable
over Q.
For example, (x^2 - 2)(x^2 - 3) = x^4 - 5x^2 + 6 has no rational roots
and is reduciable over Q.
Paul Sperry
<dba...@yahoo.com.br> wrote:
f(x) can be reducible but not have any rational roots - it can be the
product of two quadratics in Q[x] which are irreducible in Q[x].
Note that you can factor f(x) in R[x] into linear factors (use the
quadratic formula). As you correctly noted f(x) has no rational roots
and thus no linear factors (in Q[x]). Show, by brute force, that no two
of the R[x] linear factors multiply to give a quadratic in Q[x].
--
Paul Sperry
Columbia, SC (USA)
Jim Heckman
On 27-Mar-2006, "Daniel C. Bastos" <dba...@yahoo.com.br>
wrote in message <20060328025512.0000778e@Saturn>:
No, that shows that there's no rational root, but being irreducible
is a stronger condition. For example, x^4 - 5x^2 + 6 =
(x^2 - 2)(x^2 - 3) is reducible over Q, but has no rational root.
Hint: Use the quadratic formula to solve for x^2.
--
Jim Heckman
William Elliot
>
x^2 = (22 +- sqr(22^2 - 4))/2 = 11 +- sqr(11^2 - 1) = 11 +- sqr 120
= 11 +- 2 sqr 30
What now? Why isn't f(x) the product of two quadratic polynomials?
Daniel C. Bastos
Paul Sperry wrote:
I'm not familiar with the brute force approach; can you explain? It
gives me the idea that I could suppose I have some factor g(x) and it
must be of degree 3, 2 or 1. So I could look at all possible
factorizations and show somehow that it won't happen in Q[x]; but I
don't really know how to work with this right now.
In any case, I've built a new argument here. By the quadratic formula,
x^2 = 11 +/- sqrt(22^2 - 4)/2
= 11 +/- 2 sqrt(30)
So x = +/- sqrt(11 +/- 2 sqrt(30)).
Now I can write f(x) as the product of 4 linear factors in R[x]. The
linear factors are irreducible polynomials in R[x]; so, the
factorization of f(x) is unique --- except for order and scalar
multiples. Therefore, if f(x) were to factor in Q[x], we would see
rational roots here, which we don't. So f(x) is irreducible in Q[x].
The idea, in other words, is: factorization by linear factors is
unique; so if we express some f(x) in R[x] and the linear factors come
with irrational numbers (x - i(1))(x - i(2)) ... (x - i(n)), where
i(k) is an irrational number, then f(x) is irreducible in Q[x]. Does
this work?
Paul Sperry
<dba...@yahoo.com.br> wrote:
> In article <280320062351454632%plsp...@sc.rr.com>,
> Paul Sperry wrote:
>
> > In article <20060328025512.0000778e@Saturn>, Daniel C. Bastos
> > <dba...@yahoo.com.br> wrote:
> >
> > > Exercise. Show f(x) = x^4 - 22x^2 + 1 is irreducible over Q.
> > >
> > > Proof. Suppose f(x) is reducible in Q[x]. Then there exists p/q in in
> > > Q, such that f(p/q) = 0, where p/q is in lowest terms. By the
> > > ``rational zeros theorem,'' p divides a(0) = 1 and q divides a(n) = 1.
> > > So p/q is 1 or -1 because the only divisors of 1 are 1 and -1, but
> > > neither 1 nor -1 are roots of f, so there is no rational p/q such that
> > > f(p/q) = 0; hence, f(x) is irreducible.
> >
> > f(x) can be reducible but not have any rational roots - it can be the
> > product of two quadratics in Q[x] which are irreducible in Q[x].
> >
> > Note that you can factor f(x) in R[x] into linear factors (use the
> > quadratic formula). As you correctly noted f(x) has no rational roots
> > and thus no linear factors (in Q[x]). Show, by brute force, that no two
> > of the R[x] linear factors multiply to give a quadratic in Q[x].
>
> I'm not familiar with the brute force approach; can you explain?
"Brute force" is just an informal term for an approach which is not
particularly sophisticated - you just grit your teeth, roll up your
sleeves and do it.
> It
> gives me the idea that I could suppose I have some factor g(x) and it
> must be of degree 3, 2 or 1. So I could look at all possible
> factorizations and show somehow that it won't happen in Q[x]; but I
> don't really know how to work with this right now.
>
> In any case, I've built a new argument here. By the quadratic formula,
>
> x^2 = 11 +/- sqrt(22^2 - 4)/2
> = 11 +/- 2 sqrt(30)
>
> So x = +/- sqrt(11 +/- 2 sqrt(30)).
>
> Now I can write f(x) as the product of 4 linear factors in R[x]. The
> linear factors are irreducible polynomials in R[x]; so, the
> factorization of f(x) is unique --- except for order and scalar
> multiples. Therefore, if f(x) were to factor in Q[x], we would see
> rational roots here, which we don't. So f(x) is irreducible in Q[x].
>
> The idea, in other words, is: factorization by linear factors is
> unique; so if we express some f(x) in R[x] and the linear factors come
> with irrational numbers (x - i(1))(x - i(2)) ... (x - i(n)), where
> i(k) is an irrational number, then f(x) is irreducible in Q[x]. Does
> this work?
First notice that x^2 - 2 = (x + sqrt(2))(x - sqrt(2)) is in Q[x] but
neither factor is in Q[x]. Consider x^4 - 4 = (x^2 + 2)(x^2 - 2).
You were OK until you asserted that two polynomials with irrational
coefficients couldn't multiply and get a product with rational
coefficients.
Since f(x) has no rational roots it has no factors in Q[x] of degree 1
or 3. So, the only remaining possibility is that there is a factor in
Q[x] of degree 2. That factor must be a product of two of the
irreducible R[x] factors.
The "brute force" comes in because I don't know any way to show that
the product of two of those R[x] factors is not in Q[x] other than just
trying all necessary products. It _could_ happen that such a product
_is_ in Q[x] just like (x + sqrt(2))(x - sqrt(2)).
Daniel C. Bastos
Paul Sperry wrote:
I see; irrationals are not closed under multiplication, so, as you
show, two irrationals may yield a rational one. The four roots are
irrational, so it's irreducible to linear factors. Take the linear
factors and multiply them in all possible ways.
A = x - sqrt{11 + 2 sqrt(30)}
B = x + sqrt{11 + 2 sqrt(30)}
C = x - sqrt{11 - 2 sqrt(30)}
D = x + sqrt{11 - 2 sqrt(30)}
AB = x^2 - [11 + 2 sqrt(30)]
AC = x^2 (-sqrt{11 + 2 sqrt(30)} - sqrt{11 - 2 sqrt(30)})x + 1
AD = x^2 (-sqrt{11 + 2 sqrt(30)} + sqrt{11 - 2 sqrt(30)})x + 1
BC = x^2 (-sqrt{11 - 2 sqrt(30)} + sqrt{11 + 2 sqrt(30)})x + 1
BD = x^2 (sqrt{11 - 2 sqrt(30)} + sqrt{11 + 2 sqrt(30)})x + 1
CD = x^2 + 241 - 44 sqrt{30}
In each possible square factor, not all of the coefficients are
rationals, so it can't be reducible in Q[x]. If I don't miss any
possible factor, I guess I'm done.
William Elliot
Newsgroups: alt.algebra.help
Subject: Re: Show x^4 - 22x^2 + 1 is irreducible in Q
Paul Sperry wrote:
> <dba...@yahoo.com.br> wrote:
>
> > Exercise. Show f(x) = x^4 - 22x^2 + 1 is irreducible over Q.
> >
> > Proof. Suppose f(x) is reducible in Q[x]. Then there exists p/q in in
> > Q, such that f(p/q) = 0, where p/q is in lowest terms. By the
> > ``rational zeros theorem,'' p divides a(0) = 1 and q divides a(n) = 1.
> > So p/q is 1 or -1 because the only divisors of 1 are 1 and -1, but
> > neither 1 nor -1 are roots of f, so there is no rational p/q such that
> > f(p/q) = 0; hence, f(x) is irreducible.
>
> f(x) can be reducible but not have any rational roots - it can be the
> product of two quadratics in Q[x] which are irreducible in Q[x].
>
> Note that you can factor f(x) in R[x] into linear factors (use the
> quadratic formula). As you correctly noted f(x) has no rational roots
> and thus no linear factors (in Q[x]). Show, by brute force, that no two
> of the R[x] linear factors multiply to give a quadratic in Q[x].
| I'm not familiar with the brute force approach; can you explain?
It's called "taking the bull by the horns".
Have you bull fighting in Brazil?
In US, policeman have been called "the bull".
Assume for rational a,b, r,s,
f(x) = (x^2 + ax + b)(x^2 + rx + s)
Notice b /= 0 /= s and either a /= 0 or r /= 0
Othewise if a and r both zero, then
f(x) = x^4 + (b + s)x^2 + bs
b + 1/b = -22; b^2 + 22b + 1 = 0; b not rational
x^4 - 22x^2 + 1 =
x^4 + (a + r)x^3 + (b + ar + s)x^2 + (as + br)x + bs
r = -a /= 0; s = b; b^2 = 1
b - a^2 + b = -22
a^2 - 2b = 22; b = +-1
a = sqr 24 or a = sqr 20; a not rational
> In any case, I've built a new argument here.
> By the quadratic formula,
> x^2 = 11 +/- sqrt(22^2 - 4)/2
> = 11 +/- 2 sqrt(30)
As I previously commented about that approch, show it's any good.
----
Paul Sperry
<dba...@yahoo.com.br> wrote:
I haven't checked your algebra but you certainly have the right idea.
You could have stopped with AB, AC and AD since for example for BC the
_other_ factor is AD.
Paul Sperry
William Elliot <ma...@hevanet.remove.com> wrote:
> From: Daniel C. Bastos <dba...@yahoo.com.br>
> Newsgroups: alt.algebra.help
> Subject: Re: Show x^4 - 22x^2 + 1 is irreducible in Q
>
> Paul Sperry wrote:
> > <dba...@yahoo.com.br> wrote:
> >
> > > Exercise. Show f(x) = x^4 - 22x^2 + 1 is irreducible over Q.
> > >
> > > Proof. Suppose f(x) is reducible in Q[x]. Then there exists p/q in in
> > > Q, such that f(p/q) = 0, where p/q is in lowest terms. By the
> > > ``rational zeros theorem,'' p divides a(0) = 1 and q divides a(n) = 1.
> > > So p/q is 1 or -1 because the only divisors of 1 are 1 and -1, but
> > > neither 1 nor -1 are roots of f, so there is no rational p/q such that
> > > f(p/q) = 0; hence, f(x) is irreducible.
> >
> > f(x) can be reducible but not have any rational roots - it can be the
> > product of two quadratics in Q[x] which are irreducible in Q[x].
> >
> > Note that you can factor f(x) in R[x] into linear factors (use the
> > quadratic formula). As you correctly noted f(x) has no rational roots
> > and thus no linear factors (in Q[x]). Show, by brute force, that no two
> > of the R[x] linear factors multiply to give a quadratic in Q[x].
>
> | I'm not familiar with the brute force approach; can you explain?
You have the quotes screwed up. _I_ was the one who made the "brute
force" comment and subsequently explained it.
[...]