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How can I prove that $(x^3-y^2)$ is a radical ideal in $mathbb{F}_2[x,y]$?
In an algebraically closed field, it's easy to verify that $(x^3-y^2)$ is a prime ideal, hence a radical ideal. However, $mathbb{F}_2$ is not algebraically closed. So how can I prove this?
polynomials ideals finite-fields
asked Nov 21 '18 at 8:10
Hugo
1579
add a comment |
In an algebraically closed field, it's easy to verify that $(x^3-y^2)$ is a prime ideal, hence a radical ideal. However, $mathbb{F}_2$ is not algebraically closed. So how can I prove this?
polynomials ideals finite-fields
asked Nov 21 '18 at 8:10
Hugo
1579
Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
3
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
1
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30
add a comment |
In an algebraically closed field, it's easy to verify that $(x^3-y^2)$ is a prime ideal, hence a radical ideal. However, $mathbb{F}_2$ is not algebraically closed. So how can I prove this?
polynomials ideals finite-fields
asked Nov 21 '18 at 8:10
Hugo
1579
In an algebraically closed field, it's easy to verify that $(x^3-y^2)$ is a prime ideal, hence a radical ideal. However, $mathbb{F}_2$ is not algebraically closed. So how can I prove this?
polynomials ideals finite-fields
polynomials ideals finite-fields
asked Nov 21 '18 at 8:10
Hugo
1579
asked Nov 21 '18 at 8:10
Hugo
1579
edited Nov 21 '18 at 10:19
asked Nov 21 '18 at 8:10
Hugo
1579
asked Nov 21 '18 at 8:10
Hugo
1579
asked Nov 21 '18 at 8:10
Hugo
1579
1579
Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
3
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
1
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30
add a comment |
Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
3
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
1
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30
Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
3
3
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
1
1
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30
add a comment |
1 Answer
1
active
oldest
votes
I was reasoning along the following lines but now am not so sure. RTP that if $p^r in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
add a comment |
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1 Answer
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active
oldest
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I was reasoning along the following lines but now am not so sure. RTP that if $p^r in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
add a comment |
I was reasoning along the following lines but now am not so sure. RTP that if $p^r in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
add a comment |
I was reasoning along the following lines but now am not so sure. RTP that if $p^r in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
I was reasoning along the following lines but now am not so sure. RTP that if $p^r in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
answered Nov 21 '18 at 10:17
Richard Martin
1,61118
1,61118
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
add a comment |
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I'm not convinced by this myself. It shows that a radical of x^3-y^2 has to be x^3-y^2, which isn't quite what is wanted, but I think it is on the right lines.
– Richard Martin
Nov 21 '18 at 10:29
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
I think this is the definition for radical ideal.
– Hugo
Nov 22 '18 at 5:45
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
But I find elements in $(x^3-y^2)$ can have even powers for $x$.
– Hugo
Nov 22 '18 at 5:52
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
Exactly, that's why I think the above isn't right. As you say, $x^6+y^4$ is in there.
– Richard Martin
Nov 22 '18 at 11:06
add a comment |
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Did you try the definition?
– Math_QED
Nov 21 '18 at 8:47
Yes but I found it's difficult to calculate this.
– Hugo
Nov 21 '18 at 8:51
3
But $mathbb{F}_2[x,y]$ is a unique-factorization domain and $x^3-y^2$ is an irreducible element, so shouldn't $langle x^3-y^2rangle$ be a prime ideal (whence it is radical)?
– Batominovski
Nov 21 '18 at 10:16
1
You don't need to assume that the base field is algebraically closed here.
– Pedro Tamaroff♦
Nov 21 '18 at 10:30