Proving polynomial over Q is irreducible
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Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$
I would be very much thankful if someone could help me with this one.
abstract-algebra elementary-number-theory polynomials
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
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add a comment |
$begingroup$
Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$
I would be very much thankful if someone could help me with this one.
abstract-algebra elementary-number-theory polynomials
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
$endgroup$
$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
4
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53
add a comment |
$begingroup$
Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$
I would be very much thankful if someone could help me with this one.
abstract-algebra elementary-number-theory polynomials
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
$endgroup$
Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$
I would be very much thankful if someone could help me with this one.
abstract-algebra elementary-number-theory polynomials
abstract-algebra elementary-number-theory polynomials
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
edited Dec 9 '18 at 10:11
greedoid
42.7k1153105
42.7k1153105
asked Dec 9 '18 at 9:38
user615503
$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
4
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53
add a comment |
$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
4
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53
$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
4
4
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53
add a comment |
2 Answers
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Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
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add a comment |
$begingroup$
A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
$endgroup$
add a comment |
$begingroup$
Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
$endgroup$
add a comment |
$begingroup$
Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
$endgroup$
Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
answered Dec 9 '18 at 10:00
greedoidgreedoid
42.7k1153105
42.7k1153105
add a comment |
add a comment |
$begingroup$
A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
$endgroup$
add a comment |
$begingroup$
A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
$endgroup$
add a comment |
$begingroup$
A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
$endgroup$
A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
answered Dec 9 '18 at 12:02
YiFanYiFan
3,9741627
3,9741627
add a comment |
add a comment |
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$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42
4
$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53