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Do even degree polynomials have even degree factors (no conjugates)?
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An even degree polynomial is a polynomial which has terms of only even degree, for example $3x^6 + x^4 + 2x^2 + 5$.
Let $p$ and $q$ be two non-zero polynomials such that both don't have a term with the same degree (for example, $p = x^3 + x + 1$, $q = x^4 + 3x^2 + 5$ is not allowed since both have a term of degree 0). A conjugate factor pair is a pair of form $p + q$ and $p - q$.
Assume all factorisation is over integers.
Now suppose there is a even polynomial which doesn't have a conjugate factor pair in it's list of factors. Will all factors be even degree polynomials too?
I really have no idea how to even approach this problem. Only thing I tried was taking examples. But I wasn't successful in finding a counter example.
My inspiration behind asking this was that, if you remove the conjugate pair restriction, there there are examples like $(x-1)(x+1) = x^2 - 1$ or $(x^2 + 1 + x)(x^2 + 1 - x) = x^4 + x^2 + 1$.
abstract-algebra number-theory polynomials factoring
asked Aug 25 at 11:40
Confuse
507513
add a comment |Â
up vote
8
down vote
favorite
An even degree polynomial is a polynomial which has terms of only even degree, for example $3x^6 + x^4 + 2x^2 + 5$.
Let $p$ and $q$ be two non-zero polynomials such that both don't have a term with the same degree (for example, $p = x^3 + x + 1$, $q = x^4 + 3x^2 + 5$ is not allowed since both have a term of degree 0). A conjugate factor pair is a pair of form $p + q$ and $p - q$.
Assume all factorisation is over integers.
Now suppose there is a even polynomial which doesn't have a conjugate factor pair in it's list of factors. Will all factors be even degree polynomials too?
I really have no idea how to even approach this problem. Only thing I tried was taking examples. But I wasn't successful in finding a counter example.
My inspiration behind asking this was that, if you remove the conjugate pair restriction, there there are examples like $(x-1)(x+1) = x^2 - 1$ or $(x^2 + 1 + x)(x^2 + 1 - x) = x^4 + x^2 + 1$.
abstract-algebra number-theory polynomials factoring
asked Aug 25 at 11:40
Confuse
507513
Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
An even degree polynomial is a polynomial which has terms of only even degree, for example $3x^6 + x^4 + 2x^2 + 5$.
Let $p$ and $q$ be two non-zero polynomials such that both don't have a term with the same degree (for example, $p = x^3 + x + 1$, $q = x^4 + 3x^2 + 5$ is not allowed since both have a term of degree 0). A conjugate factor pair is a pair of form $p + q$ and $p - q$.
Assume all factorisation is over integers.
Now suppose there is a even polynomial which doesn't have a conjugate factor pair in it's list of factors. Will all factors be even degree polynomials too?
I really have no idea how to even approach this problem. Only thing I tried was taking examples. But I wasn't successful in finding a counter example.
My inspiration behind asking this was that, if you remove the conjugate pair restriction, there there are examples like $(x-1)(x+1) = x^2 - 1$ or $(x^2 + 1 + x)(x^2 + 1 - x) = x^4 + x^2 + 1$.
abstract-algebra number-theory polynomials factoring
asked Aug 25 at 11:40
Confuse
507513
An even degree polynomial is a polynomial which has terms of only even degree, for example $3x^6 + x^4 + 2x^2 + 5$.
Let $p$ and $q$ be two non-zero polynomials such that both don't have a term with the same degree (for example, $p = x^3 + x + 1$, $q = x^4 + 3x^2 + 5$ is not allowed since both have a term of degree 0). A conjugate factor pair is a pair of form $p + q$ and $p - q$.
Assume all factorisation is over integers.
Now suppose there is a even polynomial which doesn't have a conjugate factor pair in it's list of factors. Will all factors be even degree polynomials too?
I really have no idea how to even approach this problem. Only thing I tried was taking examples. But I wasn't successful in finding a counter example.
My inspiration behind asking this was that, if you remove the conjugate pair restriction, there there are examples like $(x-1)(x+1) = x^2 - 1$ or $(x^2 + 1 + x)(x^2 + 1 - x) = x^4 + x^2 + 1$.
abstract-algebra number-theory polynomials factoring
asked Aug 25 at 11:40
Confuse
507513
edited Aug 25 at 13:05
asked Aug 25 at 11:40
Confuse
507513
asked Aug 25 at 11:40
Confuse
507513
asked Aug 25 at 11:40
Confuse
507513
507513
Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15
add a comment |Â
Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15
Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
accepted
Let $R:=BbbZ[X^2]subsetBbbZ[X]$ be the subring of 'even degree' polynomials. Let $Pin R$ factor in $BbbZ[X]$ as
$$P=prod_i=1^nQ_i,$$
where the $Q_i$ are irreducible. Because $Pin R$ we have $P(X)=P(-X)$ and so
$$prod_i=1^nQ_i(-X)=P(-X)=P(X)=prod_i=1^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=fracQ_i+Q_j2qquadtext and qquad G:=fracQ_i-Q_j2,$$
we have $F,GinBbbZ[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_iin R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
answered Aug 25 at 13:09
Servaes
1
add a comment |Â
up vote
3
down vote
Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.
answered Aug 25 at 13:16
Marco
1,55917
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
add a comment |Â
up vote
0
down vote
$$ x ^ 2 = x cdot x $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
answered Aug 25 at 20:36
Henning Makholm
230k16296527
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Let $R:=BbbZ[X^2]subsetBbbZ[X]$ be the subring of 'even degree' polynomials. Let $Pin R$ factor in $BbbZ[X]$ as
$$P=prod_i=1^nQ_i,$$
where the $Q_i$ are irreducible. Because $Pin R$ we have $P(X)=P(-X)$ and so
$$prod_i=1^nQ_i(-X)=P(-X)=P(X)=prod_i=1^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=fracQ_i+Q_j2qquadtext and qquad G:=fracQ_i-Q_j2,$$
we have $F,GinBbbZ[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_iin R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
answered Aug 25 at 13:09
Servaes
1
add a comment |Â
up vote
8
down vote
accepted
Let $R:=BbbZ[X^2]subsetBbbZ[X]$ be the subring of 'even degree' polynomials. Let $Pin R$ factor in $BbbZ[X]$ as
$$P=prod_i=1^nQ_i,$$
where the $Q_i$ are irreducible. Because $Pin R$ we have $P(X)=P(-X)$ and so
$$prod_i=1^nQ_i(-X)=P(-X)=P(X)=prod_i=1^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=fracQ_i+Q_j2qquadtext and qquad G:=fracQ_i-Q_j2,$$
we have $F,GinBbbZ[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_iin R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
answered Aug 25 at 13:09
Servaes
1
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Let $R:=BbbZ[X^2]subsetBbbZ[X]$ be the subring of 'even degree' polynomials. Let $Pin R$ factor in $BbbZ[X]$ as
$$P=prod_i=1^nQ_i,$$
where the $Q_i$ are irreducible. Because $Pin R$ we have $P(X)=P(-X)$ and so
$$prod_i=1^nQ_i(-X)=P(-X)=P(X)=prod_i=1^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=fracQ_i+Q_j2qquadtext and qquad G:=fracQ_i-Q_j2,$$
we have $F,GinBbbZ[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_iin R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
answered Aug 25 at 13:09
Servaes
1
Let $R:=BbbZ[X^2]subsetBbbZ[X]$ be the subring of 'even degree' polynomials. Let $Pin R$ factor in $BbbZ[X]$ as
$$P=prod_i=1^nQ_i,$$
where the $Q_i$ are irreducible. Because $Pin R$ we have $P(X)=P(-X)$ and so
$$prod_i=1^nQ_i(-X)=P(-X)=P(X)=prod_i=1^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=fracQ_i+Q_j2qquadtext and qquad G:=fracQ_i-Q_j2,$$
we have $F,GinBbbZ[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_iin R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
answered Aug 25 at 13:09
Servaes
1
answered Aug 25 at 13:09
Servaes
1
answered Aug 25 at 13:09
Servaes
1
answered Aug 25 at 13:09
Servaes
1
1
add a comment |Â
add a comment |Â
up vote
3
down vote
Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.
answered Aug 25 at 13:16
Marco
1,55917
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
add a comment |Â
up vote
3
down vote
Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.
answered Aug 25 at 13:16
Marco
1,55917
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.
answered Aug 25 at 13:16
Marco
1,55917
Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.
answered Aug 25 at 13:16
Marco
1,55917
answered Aug 25 at 13:16
Marco
1,55917
answered Aug 25 at 13:16
Marco
1,55917
answered Aug 25 at 13:16
Marco
1,55917
1,55917
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
add a comment |Â
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
But what if $p(-x)$ and $p(x)$ have some common factors?
– Confuse
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
I take $p(x)$ to be irreducible.
– Marco
Aug 25 at 13:20
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
Oh right. I missed that. Thank you.
– Confuse
Aug 25 at 13:21
add a comment |Â
up vote
0
down vote
$$ x ^ 2 = x cdot x $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
answered Aug 25 at 20:36
Henning Makholm
230k16296527
add a comment |Â
up vote
0
down vote
$$ x ^ 2 = x cdot x $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
answered Aug 25 at 20:36
Henning Makholm
230k16296527
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$ x ^ 2 = x cdot x $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
answered Aug 25 at 20:36
Henning Makholm
230k16296527
$$ x ^ 2 = x cdot x $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
answered Aug 25 at 20:36
Henning Makholm
230k16296527
answered Aug 25 at 20:36
Henning Makholm
230k16296527
answered Aug 25 at 20:36
Henning Makholm
230k16296527
answered Aug 25 at 20:36
Henning Makholm
230k16296527
230k16296527
add a comment |Â
add a comment |Â
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Every polynomial $P$ with $P(0)=0$ has the conjugate factor pair $P+0$ and $P-0$, so such a polynomial must have nonzero constant term.
– Servaes
Aug 25 at 12:57
Thanks. I'll edit that in.
– Confuse
Aug 25 at 13:01
It also depends a bit on your convention for $deg0$; some consider $deg0=0$ but often $deg0=-infty$ is more practical. In this case my previous comment does not hold, and in stead every polynomial has a conjugate factor pair, making the question void.
– Servaes
Aug 25 at 13:15