Irreducibility of a simple polynomial Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraShow $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine whether the polynomial $x^2-12$ in $mathbb Z[x]$ satisfies an Eisenstein criterion for irreducibility over $mathbb Q$Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $mathbb Q[x]$Constructibility of roots of a polynomialEisenstein's criterion for polynomials in Z mod pProving irreducibility; What is this method and what is the logic behind it?Irreducibility of special cyclotomic polynomial.Irreducibility of a Polynomial after a substitutionIrreducibility of Non-monic Quartic Polynomials in Q[x]Irreducible monic polynomial in $mathbbQ[x]$
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Irreducibility of a simple polynomial
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraShow $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine whether the polynomial $x^2-12$ in $mathbb Z[x]$ satisfies an Eisenstein criterion for irreducibility over $mathbb Q$Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $mathbb Q[x]$Constructibility of roots of a polynomialEisenstein's criterion for polynomials in Z mod pProving irreducibility; What is this method and what is the logic behind it?Irreducibility of special cyclotomic polynomial.Irreducibility of a Polynomial after a substitutionIrreducibility of Non-monic Quartic Polynomials in Q[x]Irreducible monic polynomial in $mathbbQ[x]$
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbbQ$.
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 $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked Mar 26 at 20:53
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 $mathbbQ$.
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 $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked Mar 26 at 20:53
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
Mar 26 at 21:08
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 $mathbbQ$.
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 $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked Mar 26 at 20:53
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 $mathbbQ$.
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 $mathbbQ(sqrtai)$ 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 Mar 26 at 20:53
JonHalesJonHales
520311
asked Mar 26 at 20:53
JonHalesJonHales
520311
asked Mar 26 at 20:53
JonHalesJonHales
520311
asked Mar 26 at 20:53
JonHalesJonHales
520311
asked Mar 26 at 20:53
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
Mar 26 at 21:08
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
Mar 26 at 21:08
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
Mar 26 at 21:08
$begingroup$
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
Mar 26 at 21:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.
answered Mar 26 at 21:28
egregegreg
186k1486209
$endgroup$
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:
$$2a-b^2=0$$
$$b=sqrt2a$$
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 Mar 26 at 21:23
EurekaEureka
1,189116
$endgroup$
$begingroup$
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.
answered Mar 26 at 21:28
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.
answered Mar 26 at 21:28
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.
answered Mar 26 at 21:28
egregegreg
186k1486209
$endgroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.
answered Mar 26 at 21:28
egregegreg
186k1486209
answered Mar 26 at 21:28
egregegreg
186k1486209
answered Mar 26 at 21:28
egregegreg
186k1486209
answered Mar 26 at 21:28
egregegreg
186k1486209
186k1486209
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:
$$2a-b^2=0$$
$$b=sqrt2a$$
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 Mar 26 at 21:23
EurekaEureka
1,189116
$endgroup$
$begingroup$
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
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:
$$2a-b^2=0$$
$$b=sqrt2a$$
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 Mar 26 at 21:23
EurekaEureka
1,189116
$endgroup$
$begingroup$
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
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:
$$2a-b^2=0$$
$$b=sqrt2a$$
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 Mar 26 at 21:23
EurekaEureka
1,189116
$endgroup$
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:
$$2a-b^2=0$$
$$b=sqrt2a$$
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 Mar 26 at 21:23
EurekaEureka
1,189116
answered Mar 26 at 21:23
EurekaEureka
1,189116
answered Mar 26 at 21:23
EurekaEureka
1,189116
answered Mar 26 at 21:23
EurekaEureka
1,189116
1,189116
$begingroup$
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
$begingroup$
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
Mar 26 at 21:27
$begingroup$
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
Mar 26 at 21:27
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
Mar 26 at 21:30
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
Mar 26 at 21:40
$begingroup$
@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
Mar 26 at 21:40
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
Mar 26 at 21:42
add a comment |
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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
Mar 26 at 21:08