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Kernel of a polynomial ring homomorphism

Order of kernel of a homomorphism.The kernel of homomorphism of a local ring into a field is its maximal ideal?Homomorphism and kernel proof helpIdeal as kernel of a homomorphism in Gaußian integersKernel of a module homomorphismRing homomorphism of polynomial ringVisualising the kernel of a homomorphism and quotient groupsDescribe the Kernel of the map from this polynomial ringRing Homomorphisms, ideals and kernelDetermine kernel of localization map of ring

1

$begingroup$

Let $mathbbF$ be a field, and define a homomorphism $phi:mathbbF[x,y]rightarrow mathbbF[z]$ by:
$$f(x,y)mapsto f(z^a,z^b)$$
where $aneq bin mathbbN$.

My question is: For what $mathbbF,a$ and $b$, we have:
$$ker(phi)=langle x^b-y^arangle ?$$

Or rather this is a statement true for any polynomial ring; otherwise, is there any general technique to find the kernel of a map like this?

For example, let $phi:mathbbC[x,y]rightarrowmathbbC[z]$ defined by $f(x,y)mapsto f(z^2,z^3)$, then $ker(phi)=langle x^3-y^2rangle$.

abstract-algebra polynomials ring-theory field-theory polynomial-rings

share|cite|improve this question

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want to assume $gcd(a,b)=1$?
  $endgroup$
  – Daniel Schepler
  Mar 21 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler No, I am assuming nothing but $aneq b$. However, I am curious about whether $gcd(a,b)=1$ is sufficient for the statement to be true, and what happens when $gcd(a,b)neq 1$.
  $endgroup$
  – Wenze 'Sylvester' Zhang
  Mar 21 at 19:45
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  In general I think the kernel would be $langle x^b/d - y^a/d rangle$ where $d = gcd(a,b)$.
  $endgroup$
  – Daniel Schepler
  Mar 21 at 19:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler I agree with you, but I found it hard to prove $ker(phi)subseteq langle x^b/d-y^a/drangle$. Do you have a proof in mind?
  $endgroup$
  – Wenze 'Sylvester' Zhang
  Mar 21 at 23:55
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $mathbbF$ be a field, and define a homomorphism $phi:mathbbF[x,y]rightarrow mathbbF[z]$ by:
$$f(x,y)mapsto f(z^a,z^b)$$
where $aneq bin mathbbN$.

My question is: For what $mathbbF,a$ and $b$, we have:
$$ker(phi)=langle x^b-y^arangle ?$$

Or rather this is a statement true for any polynomial ring; otherwise, is there any general technique to find the kernel of a map like this?

For example, let $phi:mathbbC[x,y]rightarrowmathbbC[z]$ defined by $f(x,y)mapsto f(z^2,z^3)$, then $ker(phi)=langle x^3-y^2rangle$.

abstract-algebra polynomials ring-theory field-theory polynomial-rings

share|cite|improve this question

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want to assume $gcd(a,b)=1$?
  $endgroup$
  – Daniel Schepler
  Mar 21 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler No, I am assuming nothing but $aneq b$. However, I am curious about whether $gcd(a,b)=1$ is sufficient for the statement to be true, and what happens when $gcd(a,b)neq 1$.
  $endgroup$
  – Wenze 'Sylvester' Zhang
  Mar 21 at 19:45
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  In general I think the kernel would be $langle x^b/d - y^a/d rangle$ where $d = gcd(a,b)$.
  $endgroup$
  – Daniel Schepler
  Mar 21 at 19:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler I agree with you, but I found it hard to prove $ker(phi)subseteq langle x^b/d-y^a/drangle$. Do you have a proof in mind?
  $endgroup$
  – Wenze 'Sylvester' Zhang
  Mar 21 at 23:55
  

  
  
  
  

add a comment | 

$begingroup$

Let $mathbbF$ be a field, and define a homomorphism $phi:mathbbF[x,y]rightarrow mathbbF[z]$ by:
$$f(x,y)mapsto f(z^a,z^b)$$
where $aneq bin mathbbN$.

My question is: For what $mathbbF,a$ and $b$, we have:
$$ker(phi)=langle x^b-y^arangle ?$$

Or rather this is a statement true for any polynomial ring; otherwise, is there any general technique to find the kernel of a map like this?

For example, let $phi:mathbbC[x,y]rightarrowmathbbC[z]$ defined by $f(x,y)mapsto f(z^2,z^3)$, then $ker(phi)=langle x^3-y^2rangle$.

abstract-algebra polynomials ring-theory field-theory polynomial-rings

share|cite|improve this question

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

$endgroup$

share|cite|improve this question

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

share|cite|improve this question

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

edited Mar 22 at 1:38

darij grinberg

11.5k33168

edited Mar 22 at 1:38

darij grinberg

11.5k33168

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

asked Mar 21 at 19:31

Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

498

1 Answer
1

active

oldest

votes

3

$begingroup$

Observe that in general, if $d = gcd(a,b)$, then $x^b/d - y^a/d in ker(phi)$, so $langle x^b/d - y^a/d rangle subseteq ker(phi)$.

To show the reverse inclusion, notice that by treating $mathbbF[x,y]$ as $mathbbF[y][x]$, using the division algorithm we can write any $f in mathbbF[x,y]$ as $q(x,y) (x^b/d - y^a/d) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $phi(f) = phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 le m < b/d$ maps to a distinct monomial $z^am + bn$ (see below for a proof), so if $f in ker(phi)$ then $phi(r) = 0$ implies $r = 0$.

So, all we have left to show is the easy purely number theoretic result:

The map $mathbbN_0 times mathbbN_0 to mathbbN_0, (m, n) mapsto am + bn$, is injective when restricted to $[0, b/d) times mathbbN_0$.

To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $gcd(a/d, b/d) = 1$, this implies that $m equiv m' pmodb/d$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.

(Note that this argument assumes that $0 notin mathbbN$ in your convention, so that $a ne 0$ and $b ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)

share|cite|improve this answer

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

$endgroup$

add a comment | 

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3

$begingroup$

Observe that in general, if $d = gcd(a,b)$, then $x^b/d - y^a/d in ker(phi)$, so $langle x^b/d - y^a/d rangle subseteq ker(phi)$.

To show the reverse inclusion, notice that by treating $mathbbF[x,y]$ as $mathbbF[y][x]$, using the division algorithm we can write any $f in mathbbF[x,y]$ as $q(x,y) (x^b/d - y^a/d) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $phi(f) = phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 le m < b/d$ maps to a distinct monomial $z^am + bn$ (see below for a proof), so if $f in ker(phi)$ then $phi(r) = 0$ implies $r = 0$.

So, all we have left to show is the easy purely number theoretic result:

The map $mathbbN_0 times mathbbN_0 to mathbbN_0, (m, n) mapsto am + bn$, is injective when restricted to $[0, b/d) times mathbbN_0$.

To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $gcd(a/d, b/d) = 1$, this implies that $m equiv m' pmodb/d$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.

(Note that this argument assumes that $0 notin mathbbN$ in your convention, so that $a ne 0$ and $b ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)

share|cite|improve this answer

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

$endgroup$

add a comment | 

3

$begingroup$

Observe that in general, if $d = gcd(a,b)$, then $x^b/d - y^a/d in ker(phi)$, so $langle x^b/d - y^a/d rangle subseteq ker(phi)$.

To show the reverse inclusion, notice that by treating $mathbbF[x,y]$ as $mathbbF[y][x]$, using the division algorithm we can write any $f in mathbbF[x,y]$ as $q(x,y) (x^b/d - y^a/d) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $phi(f) = phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 le m < b/d$ maps to a distinct monomial $z^am + bn$ (see below for a proof), so if $f in ker(phi)$ then $phi(r) = 0$ implies $r = 0$.

So, all we have left to show is the easy purely number theoretic result:

The map $mathbbN_0 times mathbbN_0 to mathbbN_0, (m, n) mapsto am + bn$, is injective when restricted to $[0, b/d) times mathbbN_0$.

To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $gcd(a/d, b/d) = 1$, this implies that $m equiv m' pmodb/d$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.

(Note that this argument assumes that $0 notin mathbbN$ in your convention, so that $a ne 0$ and $b ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)

share|cite|improve this answer

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

$endgroup$

add a comment | 

$begingroup$

Observe that in general, if $d = gcd(a,b)$, then $x^b/d - y^a/d in ker(phi)$, so $langle x^b/d - y^a/d rangle subseteq ker(phi)$.

To show the reverse inclusion, notice that by treating $mathbbF[x,y]$ as $mathbbF[y][x]$, using the division algorithm we can write any $f in mathbbF[x,y]$ as $q(x,y) (x^b/d - y^a/d) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $phi(f) = phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 le m < b/d$ maps to a distinct monomial $z^am + bn$ (see below for a proof), so if $f in ker(phi)$ then $phi(r) = 0$ implies $r = 0$.

So, all we have left to show is the easy purely number theoretic result:

The map $mathbbN_0 times mathbbN_0 to mathbbN_0, (m, n) mapsto am + bn$, is injective when restricted to $[0, b/d) times mathbbN_0$.

To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $gcd(a/d, b/d) = 1$, this implies that $m equiv m' pmodb/d$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.

(Note that this argument assumes that $0 notin mathbbN$ in your convention, so that $a ne 0$ and $b ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)

share|cite|improve this answer

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

$endgroup$

share|cite|improve this answer

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

edited Mar 22 at 1:40

darij grinberg

11.5k33168

edited Mar 22 at 1:40

darij grinberg

11.5k33168

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821

answered Mar 22 at 0:10

Daniel ScheplerDaniel Schepler

9,2741821