irreducible polynomials
Please help me to prove that these polynomials are irreducible over $mathbb{Q}$:
1) $x^4+5x^3+7x^2-6$
2) $x^4-4x^3+8x^2-8x+9$
Unfortunately , it takes infinite amount of time to prove that given polynomials are irreducible according to Eisenstein's criterion. So could you recommend and show other ways, that requires less time , because such exercises will come in final exam where I should solve a problems in a short time.
abstract-algebra polynomials
edited Dec 23 at 20:06
André 3000
12.4k22042
asked Dec 23 at 20:03
Nicko
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Please help me to prove that these polynomials are irreducible over $mathbb{Q}$:
1) $x^4+5x^3+7x^2-6$
2) $x^4-4x^3+8x^2-8x+9$
Unfortunately , it takes infinite amount of time to prove that given polynomials are irreducible according to Eisenstein's criterion. So could you recommend and show other ways, that requires less time , because such exercises will come in final exam where I should solve a problems in a short time.
abstract-algebra polynomials
edited Dec 23 at 20:06
André 3000
12.4k22042
asked Dec 23 at 20:03
Nicko
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Please help me to prove that these polynomials are irreducible over $mathbb{Q}$:
1) $x^4+5x^3+7x^2-6$
2) $x^4-4x^3+8x^2-8x+9$
Unfortunately , it takes infinite amount of time to prove that given polynomials are irreducible according to Eisenstein's criterion. So could you recommend and show other ways, that requires less time , because such exercises will come in final exam where I should solve a problems in a short time.
abstract-algebra polynomials
edited Dec 23 at 20:06
André 3000
12.4k22042
asked Dec 23 at 20:03
Nicko
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please help me to prove that these polynomials are irreducible over $mathbb{Q}$:
1) $x^4+5x^3+7x^2-6$
2) $x^4-4x^3+8x^2-8x+9$
Unfortunately , it takes infinite amount of time to prove that given polynomials are irreducible according to Eisenstein's criterion. So could you recommend and show other ways, that requires less time , because such exercises will come in final exam where I should solve a problems in a short time.
abstract-algebra polynomials
abstract-algebra polynomials
edited Dec 23 at 20:06
André 3000
12.4k22042
asked Dec 23 at 20:03
Nicko
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 23 at 20:06
André 3000
12.4k22042
asked Dec 23 at 20:03
Nicko
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 23 at 20:06
André 3000
12.4k22042
edited Dec 23 at 20:06
André 3000
12.4k22042
edited Dec 23 at 20:06
André 3000
12.4k22042
12.4k22042
asked Dec 23 at 20:03
Nicko
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 23 at 20:03
Nicko
192
asked Dec 23 at 20:03
Nicko
192
192
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nicko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3 Answers
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The first polynomial is not irreducible since we have
$$
x^4+5x^3+7x^2-6=
(x^2 + 3x + 3)(x^2 + 2x - 2).
$$
For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
add a comment |
Actually Eisenstein is very quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.
Hint $ $ for $2):, bmodcolor{#c00}2!:, f(x) equiv x^{ 4}!+!1equiv (x!+!1)^{4} $ is a prime power.
So Eisenstein works on $,g(x) = f(x!-!1) equiv x^4 $ by
$,g(0) = f(-1)equiv 1!+!9equiv 2pmod{!4}$
Remark $ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $,p,,$ powers of a prime, viz. $,equiv x^n,,$ and products of primes always factor uniquely. The same works for its shift $,(x-c)^n,,$ so we seek primes $,p,$ such that, mod $,p,,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $,p,$ that can yield such powers are those dividing the discriminant (here by Alpha = $,color{#c00}2^9cdot 3cdot 5^2)$. Indeed, if $,fequiv a (x-c)^n,, n> 1,$ then $,f,$ and $,f',$ have a common root $,xequiv c,,$ hence their resultant $, R(f,f')equiv 0.,$ But this is, up to sign, the discriminant of $,f,$ (presumed monic).
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
add a comment |
Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$
or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$
Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.
answered Dec 23 at 20:11
greedoid
37.7k114794
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
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votes
The first polynomial is not irreducible since we have
$$
x^4+5x^3+7x^2-6=
(x^2 + 3x + 3)(x^2 + 2x - 2).
$$
For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
add a comment |
The first polynomial is not irreducible since we have
$$
x^4+5x^3+7x^2-6=
(x^2 + 3x + 3)(x^2 + 2x - 2).
$$
For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
add a comment |
The first polynomial is not irreducible since we have
$$
x^4+5x^3+7x^2-6=
(x^2 + 3x + 3)(x^2 + 2x - 2).
$$
For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
The first polynomial is not irreducible since we have
$$
x^4+5x^3+7x^2-6=
(x^2 + 3x + 3)(x^2 + 2x - 2).
$$
For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
answered Dec 23 at 20:08
Dietrich Burde
77.4k64386
77.4k64386
add a comment |
add a comment |
Actually Eisenstein is very quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.
Hint $ $ for $2):, bmodcolor{#c00}2!:, f(x) equiv x^{ 4}!+!1equiv (x!+!1)^{4} $ is a prime power.
So Eisenstein works on $,g(x) = f(x!-!1) equiv x^4 $ by
$,g(0) = f(-1)equiv 1!+!9equiv 2pmod{!4}$
Remark $ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $,p,,$ powers of a prime, viz. $,equiv x^n,,$ and products of primes always factor uniquely. The same works for its shift $,(x-c)^n,,$ so we seek primes $,p,$ such that, mod $,p,,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $,p,$ that can yield such powers are those dividing the discriminant (here by Alpha = $,color{#c00}2^9cdot 3cdot 5^2)$. Indeed, if $,fequiv a (x-c)^n,, n> 1,$ then $,f,$ and $,f',$ have a common root $,xequiv c,,$ hence their resultant $, R(f,f')equiv 0.,$ But this is, up to sign, the discriminant of $,f,$ (presumed monic).
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
add a comment |
Actually Eisenstein is very quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.
Hint $ $ for $2):, bmodcolor{#c00}2!:, f(x) equiv x^{ 4}!+!1equiv (x!+!1)^{4} $ is a prime power.
So Eisenstein works on $,g(x) = f(x!-!1) equiv x^4 $ by
$,g(0) = f(-1)equiv 1!+!9equiv 2pmod{!4}$
Remark $ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $,p,,$ powers of a prime, viz. $,equiv x^n,,$ and products of primes always factor uniquely. The same works for its shift $,(x-c)^n,,$ so we seek primes $,p,$ such that, mod $,p,,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $,p,$ that can yield such powers are those dividing the discriminant (here by Alpha = $,color{#c00}2^9cdot 3cdot 5^2)$. Indeed, if $,fequiv a (x-c)^n,, n> 1,$ then $,f,$ and $,f',$ have a common root $,xequiv c,,$ hence their resultant $, R(f,f')equiv 0.,$ But this is, up to sign, the discriminant of $,f,$ (presumed monic).
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
add a comment |
Actually Eisenstein is very quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.
Hint $ $ for $2):, bmodcolor{#c00}2!:, f(x) equiv x^{ 4}!+!1equiv (x!+!1)^{4} $ is a prime power.
So Eisenstein works on $,g(x) = f(x!-!1) equiv x^4 $ by
$,g(0) = f(-1)equiv 1!+!9equiv 2pmod{!4}$
Remark $ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $,p,,$ powers of a prime, viz. $,equiv x^n,,$ and products of primes always factor uniquely. The same works for its shift $,(x-c)^n,,$ so we seek primes $,p,$ such that, mod $,p,,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $,p,$ that can yield such powers are those dividing the discriminant (here by Alpha = $,color{#c00}2^9cdot 3cdot 5^2)$. Indeed, if $,fequiv a (x-c)^n,, n> 1,$ then $,f,$ and $,f',$ have a common root $,xequiv c,,$ hence their resultant $, R(f,f')equiv 0.,$ But this is, up to sign, the discriminant of $,f,$ (presumed monic).
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
Actually Eisenstein is very quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.
Hint $ $ for $2):, bmodcolor{#c00}2!:, f(x) equiv x^{ 4}!+!1equiv (x!+!1)^{4} $ is a prime power.
So Eisenstein works on $,g(x) = f(x!-!1) equiv x^4 $ by
$,g(0) = f(-1)equiv 1!+!9equiv 2pmod{!4}$
Remark $ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $,p,,$ powers of a prime, viz. $,equiv x^n,,$ and products of primes always factor uniquely. The same works for its shift $,(x-c)^n,,$ so we seek primes $,p,$ such that, mod $,p,,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $,p,$ that can yield such powers are those dividing the discriminant (here by Alpha = $,color{#c00}2^9cdot 3cdot 5^2)$. Indeed, if $,fequiv a (x-c)^n,, n> 1,$ then $,f,$ and $,f',$ have a common root $,xequiv c,,$ hence their resultant $, R(f,f')equiv 0.,$ But this is, up to sign, the discriminant of $,f,$ (presumed monic).
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
edited Dec 23 at 20:48
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
answered Dec 23 at 20:16
Bill Dubuque
208k29190628
208k29190628
add a comment |
add a comment |
Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$
or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$
Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.
answered Dec 23 at 20:11
greedoid
37.7k114794
add a comment |
Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$
or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$
Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.
answered Dec 23 at 20:11
greedoid
37.7k114794
add a comment |
Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$
or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$
Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.
answered Dec 23 at 20:11
greedoid
37.7k114794
Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$
or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$
Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.
answered Dec 23 at 20:11
greedoid
37.7k114794
answered Dec 23 at 20:11
greedoid
37.7k114794
answered Dec 23 at 20:11
greedoid
37.7k114794
answered Dec 23 at 20:11
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
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