Fdtxjr

Writing in a Christian voice

How to get the n-th line after a grepped one?

A Ri-diddley-iley Riddle

How to generate binary array whose elements with values 1 are randomly drawn

Is there a hypothetical scenario that would make Earth uninhabitable for humans, but not for (the majority of) other animals?

How are passwords stolen from companies if they only store hashes?

Light propagating through a sound wave

Suggestions on how to spend Shaabath (constructively) alone

gerund and noun applications

HP P840 HDD RAID 5 many strange drive failures

Does multi-classing into Fighter give you heavy armor proficiency?

How might a highly intelligent aquatic species (mermaids) communicate underwater?

Help rendering a complicated sum/product formula

Does .bashrc contain syntax errors?

Tikz: place node leftmost of two nodes of different widths

Should I use acronyms in dialogues before telling the readers what it stands for in fiction?

How to define limit operations in general topological spaces? Are nets able to do this?

How does one measure the Fourier components of a signal?

Variable completely messes up echoed string

Why didn't Héctor fade away after this character died in the movie Coco?

What can I do if I am asked to learn different programming languages very frequently?

Synchronized implementation of a bank account in Java

Could Sinn Fein swing any Brexit vote in Parliament?

Print last inputted byte

Homogeneous elements of an ideal over a quotient ring

The definition of direct sums and subspaces?A graded ring $R$ is graded-local iff $R_0$ is a local ring?Finitely generated graded modules over $K[x]$Is a graded module over a graded ring zero when all of it's graded localizations at graded primes not containing the irrelevant ideal are zero?$M_1 oplus M_2$ is a cyclic $A$-module $iff rmAnn(M_1)+rmAnn(M_2)=A$Is the zero ideal of a graded ring considered homogeneous?$mathbbZ$-graded ring with no homogeneous prime ideals is isomorphic to Laurent polynomial ringIdeal of direct sum of rings.When a homogeneous ideal is written as a product of two ideals, then each of two ideals is homogeneous?Passing from a set of generators to a set of homogeneous generatorsPrime ideal of ring without gradation generating the prime homogeneous ideal contained in the same ring with gradation?

1

$begingroup$

Let $k$ be a field. Consider the ideals $I_1=(x),I_2=(y),J=(x^2,y)$ of $R=k[x,y]/(xy,y^2)$. Show that the homogeneous elements of $J$ are contained in $I_1cup I_2$ but that $Jnotsubset I_1$ and $Jnotsubset I_2$.

First of all, what is meant by "homogeneous elements of $J$"? The notion of homogeneous element makes sense in a graded module $M=oplus_i=-infty^inftyM_i$ over a graded ring $R=R_0oplus R_1oplusdots$. The ideal $J$ is an $R$-module, but how is the grading of $J$ and $R$ defined?

Also, whichever the definition of a homogeneous element is, I think the element $x+y$ must be homogeneous according to the definition. But it doesn't lie in the union $I_1cup I_2$...

abstract-algebra ring-theory commutative-algebra modules ideals

share|cite|improve this question

asked Mar 12 at 20:13

user419669user419669

25629

$endgroup$

add a comment | 

1

$begingroup$

Let $k$ be a field. Consider the ideals $I_1=(x),I_2=(y),J=(x^2,y)$ of $R=k[x,y]/(xy,y^2)$. Show that the homogeneous elements of $J$ are contained in $I_1cup I_2$ but that $Jnotsubset I_1$ and $Jnotsubset I_2$.

First of all, what is meant by "homogeneous elements of $J$"? The notion of homogeneous element makes sense in a graded module $M=oplus_i=-infty^inftyM_i$ over a graded ring $R=R_0oplus R_1oplusdots$. The ideal $J$ is an $R$-module, but how is the grading of $J$ and $R$ defined?

Also, whichever the definition of a homogeneous element is, I think the element $x+y$ must be homogeneous according to the definition. But it doesn't lie in the union $I_1cup I_2$...

abstract-algebra ring-theory commutative-algebra modules ideals

share|cite|improve this question

asked Mar 12 at 20:13

user419669user419669

25629

$endgroup$

add a comment | 

$begingroup$

Let $k$ be a field. Consider the ideals $I_1=(x),I_2=(y),J=(x^2,y)$ of $R=k[x,y]/(xy,y^2)$. Show that the homogeneous elements of $J$ are contained in $I_1cup I_2$ but that $Jnotsubset I_1$ and $Jnotsubset I_2$.

First of all, what is meant by "homogeneous elements of $J$"? The notion of homogeneous element makes sense in a graded module $M=oplus_i=-infty^inftyM_i$ over a graded ring $R=R_0oplus R_1oplusdots$. The ideal $J$ is an $R$-module, but how is the grading of $J$ and $R$ defined?

Also, whichever the definition of a homogeneous element is, I think the element $x+y$ must be homogeneous according to the definition. But it doesn't lie in the union $I_1cup I_2$...

abstract-algebra ring-theory commutative-algebra modules ideals

share|cite|improve this question

asked Mar 12 at 20:13

user419669user419669

25629

$endgroup$

share|cite|improve this question

asked Mar 12 at 20:13

user419669user419669

25629

share|cite|improve this question

asked Mar 12 at 20:13

user419669user419669

25629

asked Mar 12 at 20:13

user419669user419669

25629

asked Mar 12 at 20:13

user419669user419669

25629

1 Answer
1

active

oldest

votes

1

$begingroup$

The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups
$$A_n:=bigoplus_i=0^nx^iy^n-ik,$$
give the grading on $k[x,y]=bigoplus_ngeq0A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that
$$B_0=k,qquad
B_1=overlinexkoplusoverlineyk
qquadtext and qquad
B_n=overlinex^nk text for n>1.$$
Then $R=bigoplus_ngeq0B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.

As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.

share|cite|improve this answer

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I found out I don't quite understand the grading of $k[x,y]$. By your definition, $A_1=kxoplus ky$, $A_2=kx^2oplus kxyoplus y^2$. They are both abelian groups But the definition of the graded ring says that one must have $R_iR_jsubset R_i+j$. How is multiplication of elements of $A_1$ by elements of $A_2$ defined? The elements of $A_1$ are of the form $(a_1x,a_2y)$, and those of $A_2$ are of the form $(b_1x^2,b_2xy,b_3y^2)$; how to multiply them? Also, how does the grading of $R$ give a grading on its submodule?
  $endgroup$
  – user419669
  Mar 12 at 21:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These groups are still subsets of their respective rings; their elements are multiplied as elements of the ring. So writing more conventionally, two elements of $A_1$ and $A_2$ are multiplied as begineqnarray* (a_1x,a_2y)cdot(b_1x^2,b_2xy,b_3y^2)&=&(a_1x+a_2y)(b_1x^2+b_2xy+b-3y^2)\ &=&a_1b_1x^3+(a_1b_2+a_2b_1)x^2y+(a_1b_3+a_2b_2)xy^2+a_2b_3y^3\ &=&(a_1b_1,a_1b_2+a_2b_1,a_1b_3+a_2b_2,a_2b_3)in A_3. endeqnarray*
  $endgroup$
  – Servaes
  Mar 12 at 23:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For some more context: The grading on $k[x,y]$ is by degree; an element of $k[x,y]$ is homogeneous of degree $n$ if it is a linear combinations of monomials of degree $n$. For example the polynomials $$x^3+3xy^2+4y^3qquadtext and qquad x^7+5x^3y^4+2xy^6,$$ are homogeneous of degree $3$ and $7$, respectively. But the polynomial $x^2+3y$ is not homogeneous at all, because the individual monomials do not have the same degree. The subgroup $A_n$ is then the subset of all elements of $k[x,y]$ of degree $n$, including $0$. This makes clear that $A_iA_jsubset A_ij$.
  $endgroup$
  – Servaes
  Mar 12 at 23:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second description is how I imagined grading on $k[x,y]$. The notation with direct sums (like $kxoplus kx$) is confusing because formally speaking, this is not a subgroup of the ring $k[x,y]$ (because the ring does not contain any ordered pairs, it contains elements of different nature). Is there a way to formalize the statement that $A_i$ is a subgroup of $k[x,y]$? And how does the grading on $R$ give a grading on $J$? Can I think of homogeneous elements on $J$ as homogeneous polynomials in the residue classes of $x,y$?
  $endgroup$
  – user419669
  Mar 12 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The direct sums are subgroups of $k[x,y]$, in fact they are even subspaces ($k$-vector spaces). If you distinguish between internal and external direct sums, perhaps read this. It helps not to think of a direct sum as a product of sets with an operation. But if you prefer, you could write $A_1=xk+yk$ though the notation gets uglier if you can't write $A_n=bigoplus_i=0^nx^iy^n-ik$.
  $endgroup$
  – Servaes
  Mar 13 at 0:06
  
  

  
  
  
  

 | 
show 3 more comments

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3145614%2fhomogeneous-elements-of-an-ideal-over-a-quotient-ring%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups
$$A_n:=bigoplus_i=0^nx^iy^n-ik,$$
give the grading on $k[x,y]=bigoplus_ngeq0A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that
$$B_0=k,qquad
B_1=overlinexkoplusoverlineyk
qquadtext and qquad
B_n=overlinex^nk text for n>1.$$
Then $R=bigoplus_ngeq0B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.

As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.

share|cite|improve this answer

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I found out I don't quite understand the grading of $k[x,y]$. By your definition, $A_1=kxoplus ky$, $A_2=kx^2oplus kxyoplus y^2$. They are both abelian groups But the definition of the graded ring says that one must have $R_iR_jsubset R_i+j$. How is multiplication of elements of $A_1$ by elements of $A_2$ defined? The elements of $A_1$ are of the form $(a_1x,a_2y)$, and those of $A_2$ are of the form $(b_1x^2,b_2xy,b_3y^2)$; how to multiply them? Also, how does the grading of $R$ give a grading on its submodule?
  $endgroup$
  – user419669
  Mar 12 at 21:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These groups are still subsets of their respective rings; their elements are multiplied as elements of the ring. So writing more conventionally, two elements of $A_1$ and $A_2$ are multiplied as begineqnarray* (a_1x,a_2y)cdot(b_1x^2,b_2xy,b_3y^2)&=&(a_1x+a_2y)(b_1x^2+b_2xy+b-3y^2)\ &=&a_1b_1x^3+(a_1b_2+a_2b_1)x^2y+(a_1b_3+a_2b_2)xy^2+a_2b_3y^3\ &=&(a_1b_1,a_1b_2+a_2b_1,a_1b_3+a_2b_2,a_2b_3)in A_3. endeqnarray*
  $endgroup$
  – Servaes
  Mar 12 at 23:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For some more context: The grading on $k[x,y]$ is by degree; an element of $k[x,y]$ is homogeneous of degree $n$ if it is a linear combinations of monomials of degree $n$. For example the polynomials $$x^3+3xy^2+4y^3qquadtext and qquad x^7+5x^3y^4+2xy^6,$$ are homogeneous of degree $3$ and $7$, respectively. But the polynomial $x^2+3y$ is not homogeneous at all, because the individual monomials do not have the same degree. The subgroup $A_n$ is then the subset of all elements of $k[x,y]$ of degree $n$, including $0$. This makes clear that $A_iA_jsubset A_ij$.
  $endgroup$
  – Servaes
  Mar 12 at 23:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second description is how I imagined grading on $k[x,y]$. The notation with direct sums (like $kxoplus kx$) is confusing because formally speaking, this is not a subgroup of the ring $k[x,y]$ (because the ring does not contain any ordered pairs, it contains elements of different nature). Is there a way to formalize the statement that $A_i$ is a subgroup of $k[x,y]$? And how does the grading on $R$ give a grading on $J$? Can I think of homogeneous elements on $J$ as homogeneous polynomials in the residue classes of $x,y$?
  $endgroup$
  – user419669
  Mar 12 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The direct sums are subgroups of $k[x,y]$, in fact they are even subspaces ($k$-vector spaces). If you distinguish between internal and external direct sums, perhaps read this. It helps not to think of a direct sum as a product of sets with an operation. But if you prefer, you could write $A_1=xk+yk$ though the notation gets uglier if you can't write $A_n=bigoplus_i=0^nx^iy^n-ik$.
  $endgroup$
  – Servaes
  Mar 13 at 0:06
  
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups
$$A_n:=bigoplus_i=0^nx^iy^n-ik,$$
give the grading on $k[x,y]=bigoplus_ngeq0A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that
$$B_0=k,qquad
B_1=overlinexkoplusoverlineyk
qquadtext and qquad
B_n=overlinex^nk text for n>1.$$
Then $R=bigoplus_ngeq0B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.

As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.

share|cite|improve this answer

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I found out I don't quite understand the grading of $k[x,y]$. By your definition, $A_1=kxoplus ky$, $A_2=kx^2oplus kxyoplus y^2$. They are both abelian groups But the definition of the graded ring says that one must have $R_iR_jsubset R_i+j$. How is multiplication of elements of $A_1$ by elements of $A_2$ defined? The elements of $A_1$ are of the form $(a_1x,a_2y)$, and those of $A_2$ are of the form $(b_1x^2,b_2xy,b_3y^2)$; how to multiply them? Also, how does the grading of $R$ give a grading on its submodule?
  $endgroup$
  – user419669
  Mar 12 at 21:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These groups are still subsets of their respective rings; their elements are multiplied as elements of the ring. So writing more conventionally, two elements of $A_1$ and $A_2$ are multiplied as begineqnarray* (a_1x,a_2y)cdot(b_1x^2,b_2xy,b_3y^2)&=&(a_1x+a_2y)(b_1x^2+b_2xy+b-3y^2)\ &=&a_1b_1x^3+(a_1b_2+a_2b_1)x^2y+(a_1b_3+a_2b_2)xy^2+a_2b_3y^3\ &=&(a_1b_1,a_1b_2+a_2b_1,a_1b_3+a_2b_2,a_2b_3)in A_3. endeqnarray*
  $endgroup$
  – Servaes
  Mar 12 at 23:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For some more context: The grading on $k[x,y]$ is by degree; an element of $k[x,y]$ is homogeneous of degree $n$ if it is a linear combinations of monomials of degree $n$. For example the polynomials $$x^3+3xy^2+4y^3qquadtext and qquad x^7+5x^3y^4+2xy^6,$$ are homogeneous of degree $3$ and $7$, respectively. But the polynomial $x^2+3y$ is not homogeneous at all, because the individual monomials do not have the same degree. The subgroup $A_n$ is then the subset of all elements of $k[x,y]$ of degree $n$, including $0$. This makes clear that $A_iA_jsubset A_ij$.
  $endgroup$
  – Servaes
  Mar 12 at 23:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second description is how I imagined grading on $k[x,y]$. The notation with direct sums (like $kxoplus kx$) is confusing because formally speaking, this is not a subgroup of the ring $k[x,y]$ (because the ring does not contain any ordered pairs, it contains elements of different nature). Is there a way to formalize the statement that $A_i$ is a subgroup of $k[x,y]$? And how does the grading on $R$ give a grading on $J$? Can I think of homogeneous elements on $J$ as homogeneous polynomials in the residue classes of $x,y$?
  $endgroup$
  – user419669
  Mar 12 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The direct sums are subgroups of $k[x,y]$, in fact they are even subspaces ($k$-vector spaces). If you distinguish between internal and external direct sums, perhaps read this. It helps not to think of a direct sum as a product of sets with an operation. But if you prefer, you could write $A_1=xk+yk$ though the notation gets uglier if you can't write $A_n=bigoplus_i=0^nx^iy^n-ik$.
  $endgroup$
  – Servaes
  Mar 13 at 0:06
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups
$$A_n:=bigoplus_i=0^nx^iy^n-ik,$$
give the grading on $k[x,y]=bigoplus_ngeq0A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that
$$B_0=k,qquad
B_1=overlinexkoplusoverlineyk
qquadtext and qquad
B_n=overlinex^nk text for n>1.$$
Then $R=bigoplus_ngeq0B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.

As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.

share|cite|improve this answer

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

$endgroup$

share|cite|improve this answer

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

answered Mar 12 at 20:33

ServaesServaes

28.4k34099

answered Mar 12 at 20:33

ServaesServaes

28.4k34099