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Where is wrong with this fake proof that Gaussian integer is a field?
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The Gaussian integer $mathbbZ[i]$ is an Euclidean domain that is not a field, since there is no inverse of $2$. So, where is wrong with the following proof?
Fake proof
First, note that $mathbbZ[X]$ is a integral domain. Since $x^2+1$
is an irreducible element in $mathbbZ[X]$, the ideal $(x^2+1)$ is
maximal, which implies $mathbbZ[i]simeqmathbbZ[X]/(x^2+1)$ is a
field.
fake-proofs gaussian-integers
asked Aug 29 at 2:37
Math.StackExchange
2,302919
 |Â
show 1 more comment
up vote
18
down vote
favorite
The Gaussian integer $mathbbZ[i]$ is an Euclidean domain that is not a field, since there is no inverse of $2$. So, where is wrong with the following proof?
Fake proof
First, note that $mathbbZ[X]$ is a integral domain. Since $x^2+1$
is an irreducible element in $mathbbZ[X]$, the ideal $(x^2+1)$ is
maximal, which implies $mathbbZ[i]simeqmathbbZ[X]/(x^2+1)$ is a
field.
fake-proofs gaussian-integers
asked Aug 29 at 2:37
Math.StackExchange
2,302919
2
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
2
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
4
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
1
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53
 |Â
show 1 more comment
up vote
18
down vote
favorite
up vote
18
down vote
favorite
The Gaussian integer $mathbbZ[i]$ is an Euclidean domain that is not a field, since there is no inverse of $2$. So, where is wrong with the following proof?
Fake proof
First, note that $mathbbZ[X]$ is a integral domain. Since $x^2+1$
is an irreducible element in $mathbbZ[X]$, the ideal $(x^2+1)$ is
maximal, which implies $mathbbZ[i]simeqmathbbZ[X]/(x^2+1)$ is a
field.
fake-proofs gaussian-integers
asked Aug 29 at 2:37
Math.StackExchange
2,302919
The Gaussian integer $mathbbZ[i]$ is an Euclidean domain that is not a field, since there is no inverse of $2$. So, where is wrong with the following proof?
Fake proof
First, note that $mathbbZ[X]$ is a integral domain. Since $x^2+1$
is an irreducible element in $mathbbZ[X]$, the ideal $(x^2+1)$ is
maximal, which implies $mathbbZ[i]simeqmathbbZ[X]/(x^2+1)$ is a
field.
fake-proofs gaussian-integers
asked Aug 29 at 2:37
Math.StackExchange
2,302919
asked Aug 29 at 2:37
Math.StackExchange
2,302919
asked Aug 29 at 2:37
Math.StackExchange
2,302919
asked Aug 29 at 2:37
Math.StackExchange
2,302919
2,302919
2
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
2
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
4
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
1
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53
 |Â
show 1 more comment
2
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
2
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
4
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
1
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53
2
2
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
2
2
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
4
4
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
1
1
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
25
down vote
accepted
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
answered Aug 29 at 2:55
JoshuaZ
95179
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
add a comment |Â
up vote
5
down vote
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 at 3:36
qu binggang
1287
add a comment |Â
up vote
0
down vote
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
answered Sep 5 at 20:31
Al Jebr
4,01943071
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
25
down vote
accepted
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
answered Aug 29 at 2:55
JoshuaZ
95179
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
add a comment |Â
up vote
25
down vote
accepted
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
answered Aug 29 at 2:55
JoshuaZ
95179
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
add a comment |Â
up vote
25
down vote
accepted
up vote
25
down vote
accepted
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
answered Aug 29 at 2:55
JoshuaZ
95179
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
answered Aug 29 at 2:55
JoshuaZ
95179
answered Aug 29 at 2:55
JoshuaZ
95179
answered Aug 29 at 2:55
JoshuaZ
95179
answered Aug 29 at 2:55
JoshuaZ
95179
95179
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
add a comment |Â
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
16
16
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
– ktoi
Aug 29 at 2:58
3
3
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring.
– aginensky
Aug 29 at 15:27
add a comment |Â
up vote
5
down vote
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 at 3:36
qu binggang
1287
add a comment |Â
up vote
5
down vote
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 at 3:36
qu binggang
1287
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 at 3:36
qu binggang
1287
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 at 3:36
qu binggang
1287
answered Aug 29 at 3:36
qu binggang
1287
answered Aug 29 at 3:36
qu binggang
1287
answered Aug 29 at 3:36
qu binggang
1287
1287
add a comment |Â
add a comment |Â
up vote
0
down vote
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
answered Sep 5 at 20:31
Al Jebr
4,01943071
add a comment |Â
up vote
0
down vote
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
answered Sep 5 at 20:31
Al Jebr
4,01943071
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
answered Sep 5 at 20:31
Al Jebr
4,01943071
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
answered Sep 5 at 20:31
Al Jebr
4,01943071
answered Sep 5 at 20:31
Al Jebr
4,01943071
answered Sep 5 at 20:31
Al Jebr
4,01943071
answered Sep 5 at 20:31
Al Jebr
4,01943071
4,01943071
add a comment |Â
add a comment |Â
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2
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
– Shahab
Aug 29 at 2:59
2
Irreducible element generates a prime ideal, but not always a maximal ideal.
– Cave Johnson
Aug 29 at 3:04
4
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
– Eric Wofsey
Aug 29 at 3:10
1
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
– qu binggang
Aug 29 at 3:26
Oh no! First fake news, now fake proofs...
– Henrik
Aug 29 at 14:53