The correspondence between maximal ideals in an algebra and it's unitalization
$begingroup$
Let $A_+$ denote the unitalization of a $mathbb{C}$-algebra $A$ ( which is $A oplus mathbb{C}$ endowed with well-know multiplication rule. I know that the map $Omega(A_+) to Omega(A)$, $J mapsto J cap A$ where $Omega$ means the maximal ideals and $A$ is considered as it's image in $A_+$, is a bijection (almost may be we should add $A$ to the image). I am asking for some hint. I know it is most likely pretty obvious but I can't see it for some reason.
abstract-algebra ideals maximal-and-prime-ideals algebras
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
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add a comment |
$begingroup$
Let $A_+$ denote the unitalization of a $mathbb{C}$-algebra $A$ ( which is $A oplus mathbb{C}$ endowed with well-know multiplication rule. I know that the map $Omega(A_+) to Omega(A)$, $J mapsto J cap A$ where $Omega$ means the maximal ideals and $A$ is considered as it's image in $A_+$, is a bijection (almost may be we should add $A$ to the image). I am asking for some hint. I know it is most likely pretty obvious but I can't see it for some reason.
abstract-algebra ideals maximal-and-prime-ideals algebras
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
$endgroup$
add a comment |
$begingroup$
Let $A_+$ denote the unitalization of a $mathbb{C}$-algebra $A$ ( which is $A oplus mathbb{C}$ endowed with well-know multiplication rule. I know that the map $Omega(A_+) to Omega(A)$, $J mapsto J cap A$ where $Omega$ means the maximal ideals and $A$ is considered as it's image in $A_+$, is a bijection (almost may be we should add $A$ to the image). I am asking for some hint. I know it is most likely pretty obvious but I can't see it for some reason.
abstract-algebra ideals maximal-and-prime-ideals algebras
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
$endgroup$
Let $A_+$ denote the unitalization of a $mathbb{C}$-algebra $A$ ( which is $A oplus mathbb{C}$ endowed with well-know multiplication rule. I know that the map $Omega(A_+) to Omega(A)$, $J mapsto J cap A$ where $Omega$ means the maximal ideals and $A$ is considered as it's image in $A_+$, is a bijection (almost may be we should add $A$ to the image). I am asking for some hint. I know it is most likely pretty obvious but I can't see it for some reason.
abstract-algebra ideals maximal-and-prime-ideals algebras
abstract-algebra ideals maximal-and-prime-ideals algebras
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
asked Dec 22 '18 at 20:32
VladislavVladislav
50728
50728
add a comment |
add a comment |
1 Answer
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The map you speak of seems not always to be a bijection. You can take a $mathbb C$ algebra with no maximal ideals, and embed it into its unitization (which must have maximal ideals.)
Or you can take something like $mathbb Ctimesmathbb C$ with trivial multiplication to be a $mathbb C$ algebra, and that has infinitely many maximal ideals (any $1$ dimensional subspace works.) but when you take the unitization, the original rng is the unique maximal ideal of the unitization.
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The map you speak of seems not always to be a bijection. You can take a $mathbb C$ algebra with no maximal ideals, and embed it into its unitization (which must have maximal ideals.)
Or you can take something like $mathbb Ctimesmathbb C$ with trivial multiplication to be a $mathbb C$ algebra, and that has infinitely many maximal ideals (any $1$ dimensional subspace works.) but when you take the unitization, the original rng is the unique maximal ideal of the unitization.
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
add a comment |
$begingroup$
The map you speak of seems not always to be a bijection. You can take a $mathbb C$ algebra with no maximal ideals, and embed it into its unitization (which must have maximal ideals.)
Or you can take something like $mathbb Ctimesmathbb C$ with trivial multiplication to be a $mathbb C$ algebra, and that has infinitely many maximal ideals (any $1$ dimensional subspace works.) but when you take the unitization, the original rng is the unique maximal ideal of the unitization.
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
add a comment |
$begingroup$
The map you speak of seems not always to be a bijection. You can take a $mathbb C$ algebra with no maximal ideals, and embed it into its unitization (which must have maximal ideals.)
Or you can take something like $mathbb Ctimesmathbb C$ with trivial multiplication to be a $mathbb C$ algebra, and that has infinitely many maximal ideals (any $1$ dimensional subspace works.) but when you take the unitization, the original rng is the unique maximal ideal of the unitization.
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
$endgroup$
The map you speak of seems not always to be a bijection. You can take a $mathbb C$ algebra with no maximal ideals, and embed it into its unitization (which must have maximal ideals.)
Or you can take something like $mathbb Ctimesmathbb C$ with trivial multiplication to be a $mathbb C$ algebra, and that has infinitely many maximal ideals (any $1$ dimensional subspace works.) but when you take the unitization, the original rng is the unique maximal ideal of the unitization.
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
answered Dec 23 '18 at 17:54
rschwiebrschwieb
107k12102251
107k12102251
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
add a comment |
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
what if we say those are modular maximal ideals?
$endgroup$
– Vladislav
Dec 24 '18 at 6:44
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
$begingroup$
@Vladislav The first example is one in which the modular maximal ideals are not in bijection...
$endgroup$
– rschwieb
Dec 24 '18 at 11:09
add a comment |
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