Irreducibility of a simple polynomial
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 13 hours ago
JonHalesJonHales
520311
$endgroup$
add a comment |
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 13 hours ago
JonHalesJonHales
520311
$endgroup$
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago
add a comment |
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 13 hours ago
JonHalesJonHales
520311
$endgroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
abstract-algebra field-theory irreducible-polynomials
asked 13 hours ago
JonHalesJonHales
520311
asked 13 hours ago
JonHalesJonHales
520311
asked 13 hours ago
JonHalesJonHales
520311
asked 13 hours ago
JonHalesJonHales
520311
asked 13 hours ago
JonHalesJonHales
520311
520311
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago
add a comment |
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago
2
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 13 hours ago
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 13 hours ago
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 13 hours ago
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 13 hours ago
egregegreg
185k1486206
$endgroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 13 hours ago
egregegreg
185k1486206
answered 13 hours ago
egregegreg
185k1486206
answered 13 hours ago
egregegreg
185k1486206
answered 13 hours ago
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 13 hours ago
EurekaEureka
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 13 hours ago
EurekaEureka
25611
answered 13 hours ago
EurekaEureka
25611
25611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
add a comment |
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
13 hours ago
1
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
13 hours ago
add a comment |
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$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
13 hours ago