Check whether quotient ring is a field or an integral domainTrue/False about ring and integral domainIs the...
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Check whether quotient ring is a field or an integral domain
True/False about ring and integral domainIs the ring $K=mathbb R[t] /( t^3+1)$ integral domain and/or a field? What's $[K: mathbb R]$?Quotient Field of an Integral DomainWhat is the quotient ring $Bbb Z[x]/(2x-3)$?check whether $mathbb{Z}[x]/(x-3)$ is a field or not?Proving that a quotient ring is not an integral domainOnto homomorphism keeps integral domain and field?how to check whether it is a Galois extension.The ring S is integral domain or field?The cardinality of a integral domain and its quotient field.
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$begingroup$
Given the quotient ring $R:=mathbb{C}[x,y,z]/(xy-z^2)$, I've to check whether it's a field or an integral domain.
I'd say that $R$ is a field and an integral domain because $(xy-z^²)$ is irreducible in $mathbb{C}[x,y,z]$ (Eisenstein with $x in mathbb{C}[x,y][z]$).
However I know that $R$ is not a field. I don't know what I've done wrong. Is my application of Eisenstein criteria false? How can I show that $R$ is not a field, but an integral domain?
abstract-algebra ring-theory commutative-algebra
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
Given the quotient ring $R:=mathbb{C}[x,y,z]/(xy-z^2)$, I've to check whether it's a field or an integral domain.
I'd say that $R$ is a field and an integral domain because $(xy-z^²)$ is irreducible in $mathbb{C}[x,y,z]$ (Eisenstein with $x in mathbb{C}[x,y][z]$).
However I know that $R$ is not a field. I don't know what I've done wrong. Is my application of Eisenstein criteria false? How can I show that $R$ is not a field, but an integral domain?
abstract-algebra ring-theory commutative-algebra
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
$endgroup$
$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
1
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago
|
show 1 more comment
$begingroup$
Given the quotient ring $R:=mathbb{C}[x,y,z]/(xy-z^2)$, I've to check whether it's a field or an integral domain.
I'd say that $R$ is a field and an integral domain because $(xy-z^²)$ is irreducible in $mathbb{C}[x,y,z]$ (Eisenstein with $x in mathbb{C}[x,y][z]$).
However I know that $R$ is not a field. I don't know what I've done wrong. Is my application of Eisenstein criteria false? How can I show that $R$ is not a field, but an integral domain?
abstract-algebra ring-theory commutative-algebra
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
$endgroup$
Given the quotient ring $R:=mathbb{C}[x,y,z]/(xy-z^2)$, I've to check whether it's a field or an integral domain.
I'd say that $R$ is a field and an integral domain because $(xy-z^²)$ is irreducible in $mathbb{C}[x,y,z]$ (Eisenstein with $x in mathbb{C}[x,y][z]$).
However I know that $R$ is not a field. I don't know what I've done wrong. Is my application of Eisenstein criteria false? How can I show that $R$ is not a field, but an integral domain?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
edited 6 hours ago
user26857
39.8k12 gold badges44 silver badges86 bronze badges
39.8k12 gold badges44 silver badges86 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
asked 8 hours ago
ThesinusThesinus
3042 silver badges10 bronze badges
3042 silver badges10 bronze badges
$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
1
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago
|
show 1 more comment
$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
1
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago
$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
1
1
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$mathbf C[X,Y,Z]$ is a U.F.D., so ideals generated by irreducible elements have height $1$. Can these ideals be maximal?
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
$endgroup$
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
add a comment |
$begingroup$
A quotient ring is a field if and only if the original ideal is maximal. Use the fact that $mathbb{C}[x,y,z]$ has natural gradings by degree; does $(xy-z^2)$ contain any elements of degree $1$?
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$mathbf C[X,Y,Z]$ is a U.F.D., so ideals generated by irreducible elements have height $1$. Can these ideals be maximal?
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
$endgroup$
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
add a comment |
$begingroup$
Hint:
$mathbf C[X,Y,Z]$ is a U.F.D., so ideals generated by irreducible elements have height $1$. Can these ideals be maximal?
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
$endgroup$
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
add a comment |
$begingroup$
Hint:
$mathbf C[X,Y,Z]$ is a U.F.D., so ideals generated by irreducible elements have height $1$. Can these ideals be maximal?
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
$endgroup$
Hint:
$mathbf C[X,Y,Z]$ is a U.F.D., so ideals generated by irreducible elements have height $1$. Can these ideals be maximal?
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
answered 8 hours ago
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
129k7 gold badges43 silver badges122 bronze badges
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
add a comment |
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Why can't they be maximal? Have maximal ideals height greater than one?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
$begingroup$
For any field, $F[X,Y,Z]$ has Krull dimension $3$, and is catenary, so maximal ideals have height $3$.
$endgroup$
– Bernard
7 hours ago
add a comment |
$begingroup$
A quotient ring is a field if and only if the original ideal is maximal. Use the fact that $mathbb{C}[x,y,z]$ has natural gradings by degree; does $(xy-z^2)$ contain any elements of degree $1$?
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
$endgroup$
add a comment |
$begingroup$
A quotient ring is a field if and only if the original ideal is maximal. Use the fact that $mathbb{C}[x,y,z]$ has natural gradings by degree; does $(xy-z^2)$ contain any elements of degree $1$?
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
$endgroup$
add a comment |
$begingroup$
A quotient ring is a field if and only if the original ideal is maximal. Use the fact that $mathbb{C}[x,y,z]$ has natural gradings by degree; does $(xy-z^2)$ contain any elements of degree $1$?
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
$endgroup$
A quotient ring is a field if and only if the original ideal is maximal. Use the fact that $mathbb{C}[x,y,z]$ has natural gradings by degree; does $(xy-z^2)$ contain any elements of degree $1$?
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
answered 8 hours ago
TomGrubbTomGrubb
11.3k1 gold badge16 silver badges39 bronze badges
11.3k1 gold badge16 silver badges39 bronze badges
add a comment |
add a comment |
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$begingroup$
$z$ is also irreducible, but $Bbb C[x,y,z] / (z) simeq Bbb C[x,y]$ is not a field (since $x$ has no inverse, for example).
$endgroup$
– Ruben
8 hours ago
1
$begingroup$
In your $R$ the element $bar x$ is not invertible. If $bar xbar f=bar 1$ then $xf-1in(xy-z^2)$, hence $xf-1=(xy-z^2)g$. Now set $x=y=z=0$.
$endgroup$
– user26857
8 hours ago
$begingroup$
Why do I have to set x=y=z=0 in order to show that x has no inverse??
$endgroup$
– Thesinus
8 hours ago
$begingroup$
Is it because of the contradiction $-1 =0$.Could I also set $x=y=z=5$, for example?
$endgroup$
– Thesinus
8 hours ago
$begingroup$
@Thesinus I'm not sure that x=y=z=5 leads to a contradiction.
$endgroup$
– user26857
6 hours ago