Additive group of local rings Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Krull's intersection theorem for commutative local not necessarily noetherian ringsIs the class of additive groups of rings axiomatizable?Characterization of non-commutative local rings of orders 64 and 128Local rings with simple radicalA question on local ringsIs every commutative group structure underlying at least one (unitary, commutative) ring structureProjecting solutions of Hermitian forms over local ringsautomorphisms of local rings vs local change of coordinatesQuotients of rings with finite free additive groupWhen is a zero dimensional local ring a chain ring?
Additive group of local rings
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Krull's intersection theorem for commutative local not necessarily noetherian ringsIs the class of additive groups of rings axiomatizable?Characterization of non-commutative local rings of orders 64 and 128Local rings with simple radicalA question on local ringsIs every commutative group structure underlying at least one (unitary, commutative) ring structureProjecting solutions of Hermitian forms over local ringsautomorphisms of local rings vs local change of coordinatesQuotients of rings with finite free additive groupWhen is a zero dimensional local ring a chain ring?
$begingroup$
Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?
ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings
asked 4 hours ago
Lisa_KLisa_K
604
$endgroup$
add a comment |
$begingroup$
Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?
ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings
asked 4 hours ago
Lisa_KLisa_K
604
$endgroup$
add a comment |
$begingroup$
Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?
ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings
asked 4 hours ago
Lisa_KLisa_K
604
$endgroup$
Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?
ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings
ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings
asked 4 hours ago
Lisa_KLisa_K
604
asked 4 hours ago
Lisa_KLisa_K
604
asked 4 hours ago
Lisa_KLisa_K
604
asked 4 hours ago
Lisa_KLisa_K
604
asked 4 hours ago
Lisa_KLisa_K
604
604
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbbZ/p^k mathbbZ times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbbZ/p^k mathbbZ $ - module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ (x,m) mid p $ divides $x $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.
answered 3 hours ago
MaxMax
6191619
$endgroup$
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f329877%2fadditive-group-of-local-rings%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbbZ/p^k mathbbZ times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbbZ/p^k mathbbZ $ - module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ (x,m) mid p $ divides $x $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.
answered 3 hours ago
MaxMax
6191619
$endgroup$
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
add a comment |
$begingroup$
Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbbZ/p^k mathbbZ times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbbZ/p^k mathbbZ $ - module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ (x,m) mid p $ divides $x $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.
answered 3 hours ago
MaxMax
6191619
$endgroup$
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
add a comment |
$begingroup$
Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbbZ/p^k mathbbZ times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbbZ/p^k mathbbZ $ - module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ (x,m) mid p $ divides $x $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.
answered 3 hours ago
MaxMax
6191619
$endgroup$
Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbbZ/p^k mathbbZ times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbbZ/p^k mathbbZ $ - module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ (x,m) mid p $ divides $x $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.
answered 3 hours ago
MaxMax
6191619
edited 3 hours ago
answered 3 hours ago
MaxMax
6191619
answered 3 hours ago
MaxMax
6191619
answered 3 hours ago
MaxMax
6191619
6191619
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
add a comment |
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
1
1
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
Precisely every nonzero such group.
$endgroup$
– YCor
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
@YCor : indeed, let me correct that
$endgroup$
– Max
3 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
$begingroup$
What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
$endgroup$
– LSpice
2 hours ago
1
1
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
$begingroup$
@LSpice : you can see it as $G/(mathbbZ/p^kmathbbZ)$ for instance; it comes from the structure theorem for finite abelian groups
$endgroup$
– Max
2 hours ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f329877%2fadditive-group-of-local-rings%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
