Proving Q irreducility without using specific criteria for this [closed]
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Given a polynomial $p(x) in mathbb Q[x]$, we know in advance its possible roots in $mathbb Q$. Showing that none of these is a root of $p(x)$, concludes its irreducibility over $mathbb Q$?
abstract-algebra polynomials irreducible-polynomials
asked Jan 8 at 11:40
math_noobmath_noob
13
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closed as off-topic by Saad, Alexander Gruber♦ Jan 9 at 0:46
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$begingroup$
Given a polynomial $p(x) in mathbb Q[x]$, we know in advance its possible roots in $mathbb Q$. Showing that none of these is a root of $p(x)$, concludes its irreducibility over $mathbb Q$?
abstract-algebra polynomials irreducible-polynomials
asked Jan 8 at 11:40
math_noobmath_noob
13
$endgroup$
closed as off-topic by Saad, Alexander Gruber♦ Jan 9 at 0:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
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– lulu
Jan 8 at 11:42
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In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
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– lulu
Jan 8 at 11:44
add a comment |
$begingroup$
Given a polynomial $p(x) in mathbb Q[x]$, we know in advance its possible roots in $mathbb Q$. Showing that none of these is a root of $p(x)$, concludes its irreducibility over $mathbb Q$?
abstract-algebra polynomials irreducible-polynomials
asked Jan 8 at 11:40
math_noobmath_noob
13
$endgroup$
Given a polynomial $p(x) in mathbb Q[x]$, we know in advance its possible roots in $mathbb Q$. Showing that none of these is a root of $p(x)$, concludes its irreducibility over $mathbb Q$?
abstract-algebra polynomials irreducible-polynomials
abstract-algebra polynomials irreducible-polynomials
asked Jan 8 at 11:40
math_noobmath_noob
13
asked Jan 8 at 11:40
math_noobmath_noob
13
asked Jan 8 at 11:40
math_noobmath_noob
13
asked Jan 8 at 11:40
math_noobmath_noob
13
asked Jan 8 at 11:40
math_noobmath_noob
13
13
closed as off-topic by Saad, Alexander Gruber♦ Jan 9 at 0:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Alexander Gruber♦ Jan 9 at 0:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
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– lulu
Jan 8 at 11:42
$begingroup$
In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
$endgroup$
– lulu
Jan 8 at 11:44
add a comment |
$begingroup$
No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
$endgroup$
– lulu
Jan 8 at 11:42
$begingroup$
In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
$endgroup$
– lulu
Jan 8 at 11:44
$begingroup$
No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
$endgroup$
– lulu
Jan 8 at 11:42
$begingroup$
No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
$endgroup$
– lulu
Jan 8 at 11:42
$begingroup$
In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
$endgroup$
– lulu
Jan 8 at 11:44
$begingroup$
In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
$endgroup$
– lulu
Jan 8 at 11:44
add a comment |
1 Answer
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Counter-example:
$(x^2+1)(x^2+2)=x^4+3x^2+2;$ is not irreducible over $mathbf Q$, yet has no rational (nor real) root.
As to general methods, you have the following strategies:
- Let $alphainmathbf Q$ and $P(X)$ be a polynomial in $mathbf Q[X]$. Then $P(X)$ is irreducible if an only if $P(X+alpha)$ is.
It may happen that by a suitable choice of $alpha$, Eisenstein's criterion applies.
- One always may reduce the problem to the case a polynomial with integer coefficients. If for some $ninmathbf Z$, and $P(X)inmathbf Z[X]$, $overline P(X)$ is irreducible mod. $n$, then $P(x)$ is irreducible.
answered Jan 8 at 12:02
BernardBernard
119k639112
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Counter-example:
$(x^2+1)(x^2+2)=x^4+3x^2+2;$ is not irreducible over $mathbf Q$, yet has no rational (nor real) root.
As to general methods, you have the following strategies:
- Let $alphainmathbf Q$ and $P(X)$ be a polynomial in $mathbf Q[X]$. Then $P(X)$ is irreducible if an only if $P(X+alpha)$ is.
It may happen that by a suitable choice of $alpha$, Eisenstein's criterion applies.
- One always may reduce the problem to the case a polynomial with integer coefficients. If for some $ninmathbf Z$, and $P(X)inmathbf Z[X]$, $overline P(X)$ is irreducible mod. $n$, then $P(x)$ is irreducible.
answered Jan 8 at 12:02
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Counter-example:
$(x^2+1)(x^2+2)=x^4+3x^2+2;$ is not irreducible over $mathbf Q$, yet has no rational (nor real) root.
As to general methods, you have the following strategies:
- Let $alphainmathbf Q$ and $P(X)$ be a polynomial in $mathbf Q[X]$. Then $P(X)$ is irreducible if an only if $P(X+alpha)$ is.
It may happen that by a suitable choice of $alpha$, Eisenstein's criterion applies.
- One always may reduce the problem to the case a polynomial with integer coefficients. If for some $ninmathbf Z$, and $P(X)inmathbf Z[X]$, $overline P(X)$ is irreducible mod. $n$, then $P(x)$ is irreducible.
answered Jan 8 at 12:02
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Counter-example:
$(x^2+1)(x^2+2)=x^4+3x^2+2;$ is not irreducible over $mathbf Q$, yet has no rational (nor real) root.
As to general methods, you have the following strategies:
- Let $alphainmathbf Q$ and $P(X)$ be a polynomial in $mathbf Q[X]$. Then $P(X)$ is irreducible if an only if $P(X+alpha)$ is.
It may happen that by a suitable choice of $alpha$, Eisenstein's criterion applies.
- One always may reduce the problem to the case a polynomial with integer coefficients. If for some $ninmathbf Z$, and $P(X)inmathbf Z[X]$, $overline P(X)$ is irreducible mod. $n$, then $P(x)$ is irreducible.
answered Jan 8 at 12:02
BernardBernard
119k639112
$endgroup$
Counter-example:
$(x^2+1)(x^2+2)=x^4+3x^2+2;$ is not irreducible over $mathbf Q$, yet has no rational (nor real) root.
As to general methods, you have the following strategies:
- Let $alphainmathbf Q$ and $P(X)$ be a polynomial in $mathbf Q[X]$. Then $P(X)$ is irreducible if an only if $P(X+alpha)$ is.
It may happen that by a suitable choice of $alpha$, Eisenstein's criterion applies.
- One always may reduce the problem to the case a polynomial with integer coefficients. If for some $ninmathbf Z$, and $P(X)inmathbf Z[X]$, $overline P(X)$ is irreducible mod. $n$, then $P(x)$ is irreducible.
answered Jan 8 at 12:02
BernardBernard
119k639112
answered Jan 8 at 12:02
BernardBernard
119k639112
answered Jan 8 at 12:02
BernardBernard
119k639112
answered Jan 8 at 12:02
BernardBernard
119k639112
119k639112
add a comment |
add a comment |
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No. $(x^2+1)^2$ is reducible (clearly) but it has no rational roots.
$endgroup$
– lulu
Jan 8 at 11:42
$begingroup$
In general, it can be very hard to show that a polynomial over $mathbb Q$ is irreducible. There are certain tests like Eisenstein's criterion that work in certain cases, but there isn't any sort of silver bullet.
$endgroup$
– lulu
Jan 8 at 11:44