Show the following polynomial is Irreducible over the given ring
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Studying for a qualifying exam and this practice problem flat out has me stumped. I wish to show that the polynomial $(y+8)^2x^3 - x^2 + (y+7)(y+8) - y - 12$ is irreducible over $mathbb{Q}[x, y]$. My thought was to use Eisenstein's for $mathbb{Q}[x][y]$ and $mathbb{Q}[y][x]$, however both variations haven't yielded a solution. For example, in $mathbb{Q}[y][x]$, the prime ideal would need to contain $-1$, but this cannot happen. Writing this polynomial as a polynomial in
$mathbb{Q}[x][y]$ yields the polynomial $(x^3+1)y^2 + (16x^3 + 14)y + 64x^3 - x^2 + 44$, but haven't been able to find a prime ideal to satisfy Eisenstein's. Any idea would be appreciated.
ring-theory irreducible-polynomials polynomial-rings
asked Dec 21 '18 at 1:54
Travis62Travis62
714
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add a comment |
$begingroup$
Studying for a qualifying exam and this practice problem flat out has me stumped. I wish to show that the polynomial $(y+8)^2x^3 - x^2 + (y+7)(y+8) - y - 12$ is irreducible over $mathbb{Q}[x, y]$. My thought was to use Eisenstein's for $mathbb{Q}[x][y]$ and $mathbb{Q}[y][x]$, however both variations haven't yielded a solution. For example, in $mathbb{Q}[y][x]$, the prime ideal would need to contain $-1$, but this cannot happen. Writing this polynomial as a polynomial in
$mathbb{Q}[x][y]$ yields the polynomial $(x^3+1)y^2 + (16x^3 + 14)y + 64x^3 - x^2 + 44$, but haven't been able to find a prime ideal to satisfy Eisenstein's. Any idea would be appreciated.
ring-theory irreducible-polynomials polynomial-rings
asked Dec 21 '18 at 1:54
Travis62Travis62
714
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Figured it out, in case anyone is working on it.
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– Travis62
Dec 21 '18 at 19:38
1
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Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29
add a comment |
$begingroup$
Studying for a qualifying exam and this practice problem flat out has me stumped. I wish to show that the polynomial $(y+8)^2x^3 - x^2 + (y+7)(y+8) - y - 12$ is irreducible over $mathbb{Q}[x, y]$. My thought was to use Eisenstein's for $mathbb{Q}[x][y]$ and $mathbb{Q}[y][x]$, however both variations haven't yielded a solution. For example, in $mathbb{Q}[y][x]$, the prime ideal would need to contain $-1$, but this cannot happen. Writing this polynomial as a polynomial in
$mathbb{Q}[x][y]$ yields the polynomial $(x^3+1)y^2 + (16x^3 + 14)y + 64x^3 - x^2 + 44$, but haven't been able to find a prime ideal to satisfy Eisenstein's. Any idea would be appreciated.
ring-theory irreducible-polynomials polynomial-rings
asked Dec 21 '18 at 1:54
Travis62Travis62
714
$endgroup$
Studying for a qualifying exam and this practice problem flat out has me stumped. I wish to show that the polynomial $(y+8)^2x^3 - x^2 + (y+7)(y+8) - y - 12$ is irreducible over $mathbb{Q}[x, y]$. My thought was to use Eisenstein's for $mathbb{Q}[x][y]$ and $mathbb{Q}[y][x]$, however both variations haven't yielded a solution. For example, in $mathbb{Q}[y][x]$, the prime ideal would need to contain $-1$, but this cannot happen. Writing this polynomial as a polynomial in
$mathbb{Q}[x][y]$ yields the polynomial $(x^3+1)y^2 + (16x^3 + 14)y + 64x^3 - x^2 + 44$, but haven't been able to find a prime ideal to satisfy Eisenstein's. Any idea would be appreciated.
ring-theory irreducible-polynomials polynomial-rings
ring-theory irreducible-polynomials polynomial-rings
asked Dec 21 '18 at 1:54
Travis62Travis62
714
asked Dec 21 '18 at 1:54
Travis62Travis62
714
asked Dec 21 '18 at 1:54
Travis62Travis62
714
asked Dec 21 '18 at 1:54
Travis62Travis62
714
asked Dec 21 '18 at 1:54
Travis62Travis62
714
714
$begingroup$
Figured it out, in case anyone is working on it.
$endgroup$
– Travis62
Dec 21 '18 at 19:38
1
$begingroup$
Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29
add a comment |
$begingroup$
Figured it out, in case anyone is working on it.
$endgroup$
– Travis62
Dec 21 '18 at 19:38
1
$begingroup$
Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29
$begingroup$
Figured it out, in case anyone is working on it.
$endgroup$
– Travis62
Dec 21 '18 at 19:38
$begingroup$
Figured it out, in case anyone is working on it.
$endgroup$
– Travis62
Dec 21 '18 at 19:38
1
1
$begingroup$
Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29
$begingroup$
Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29
add a comment |
1 Answer
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Consider the homomorphism $y to -7$ from $mathbb{Q}[x, y] to mathbb{Q}[x]$. Then the image is a proper prime ideal of $mathbb{Q}[x]$ and one can show then that the preimage must also be a prime ideal in $mathbb{Q}[x, y]$.
answered Dec 27 '18 at 16:06
Travis62Travis62
714
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Consider the homomorphism $y to -7$ from $mathbb{Q}[x, y] to mathbb{Q}[x]$. Then the image is a proper prime ideal of $mathbb{Q}[x]$ and one can show then that the preimage must also be a prime ideal in $mathbb{Q}[x, y]$.
answered Dec 27 '18 at 16:06
Travis62Travis62
714
$endgroup$
add a comment |
$begingroup$
Consider the homomorphism $y to -7$ from $mathbb{Q}[x, y] to mathbb{Q}[x]$. Then the image is a proper prime ideal of $mathbb{Q}[x]$ and one can show then that the preimage must also be a prime ideal in $mathbb{Q}[x, y]$.
answered Dec 27 '18 at 16:06
Travis62Travis62
714
$endgroup$
add a comment |
$begingroup$
Consider the homomorphism $y to -7$ from $mathbb{Q}[x, y] to mathbb{Q}[x]$. Then the image is a proper prime ideal of $mathbb{Q}[x]$ and one can show then that the preimage must also be a prime ideal in $mathbb{Q}[x, y]$.
answered Dec 27 '18 at 16:06
Travis62Travis62
714
$endgroup$
Consider the homomorphism $y to -7$ from $mathbb{Q}[x, y] to mathbb{Q}[x]$. Then the image is a proper prime ideal of $mathbb{Q}[x]$ and one can show then that the preimage must also be a prime ideal in $mathbb{Q}[x, y]$.
answered Dec 27 '18 at 16:06
Travis62Travis62
714
answered Dec 27 '18 at 16:06
Travis62Travis62
714
answered Dec 27 '18 at 16:06
Travis62Travis62
714
answered Dec 27 '18 at 16:06
Travis62Travis62
714
714
add a comment |
add a comment |
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$begingroup$
Figured it out, in case anyone is working on it.
$endgroup$
– Travis62
Dec 21 '18 at 19:38
1
$begingroup$
Might I suggest that you post an answer to your own question? This would remove it from the "unanswered" tab.
$endgroup$
– Pierre-Guy Plamondon
Dec 23 '18 at 16:29