Where is wrong with this fake proof that Gaussian integer is a field?
$begingroup$
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 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
$endgroup$
|
show 1 more comment
$begingroup$
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 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
$endgroup$
2
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
2
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
4
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
1
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53
|
show 1 more comment
$begingroup$
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 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
$endgroup$
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
fake-proofs gaussian-integers
asked Aug 29 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
asked Aug 29 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
asked Aug 29 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
asked Aug 29 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
asked Aug 29 '18 at 2:37
Math.StackExchangeMath.StackExchange
2,412921
2,412921
2
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
2
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
4
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
1
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53
|
show 1 more comment
2
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
2
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
4
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
1
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53
2
2
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
2
2
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
4
4
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
1
1
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
"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 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
$endgroup$
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
add a comment |
$begingroup$
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:31
Al JebrAl Jebr
4,42543478
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"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 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
$endgroup$
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
add a comment |
$begingroup$
"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 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
$endgroup$
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
add a comment |
$begingroup$
"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 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
$endgroup$
"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 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
answered Aug 29 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
answered Aug 29 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
answered Aug 29 '18 at 2:55
JoshuaZJoshuaZ
1,1681010
1,1681010
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
add a comment |
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
16
16
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
$begingroup$
I believe a counterexample in this particular case is $(x^2+1) subsetneq (x^2+3,2) subsetneq mathbb Z[x].$
$endgroup$
– ktoi
Aug 29 '18 at 2:58
3
3
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
$begingroup$
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.
$endgroup$
– aginensky
Aug 29 '18 at 15:27
add a comment |
$begingroup$
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
$endgroup$
add a comment |
$begingroup$
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
$endgroup$
add a comment |
$begingroup$
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
$endgroup$
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $geq 2$. $dim mathbbZ[X] = 2$.
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
answered Aug 29 '18 at 3:36
qu binggangqu binggang
37817
37817
add a comment |
add a comment |
$begingroup$
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 '18 at 20:31
Al JebrAl Jebr
4,42543478
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:31
Al JebrAl Jebr
4,42543478
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:31
Al JebrAl Jebr
4,42543478
$endgroup$
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 '18 at 20:31
Al JebrAl Jebr
4,42543478
answered Sep 5 '18 at 20:31
Al JebrAl Jebr
4,42543478
answered Sep 5 '18 at 20:31
Al JebrAl Jebr
4,42543478
answered Sep 5 '18 at 20:31
Al JebrAl Jebr
4,42543478
4,42543478
add a comment |
add a comment |
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2
$begingroup$
$x^2+1$ is an irreducible element in $mathbbZ[X]$ but the ideal $(x^2+1)$ is not maximal. Note that $mathbbZ[X]$ is not a PID.
$endgroup$
– Shahab
Aug 29 '18 at 2:59
2
$begingroup$
Irreducible element generates a prime ideal, but not always a maximal ideal.
$endgroup$
– Cave Johnson
Aug 29 '18 at 3:04
4
$begingroup$
@CaveJohnson: An irreducible element doesn't even always generate a prime ideal!
$endgroup$
– Eric Wofsey
Aug 29 '18 at 3:10
1
$begingroup$
@EricWofsey: in UFD, irreducible elements generate prime ideals. $mathbbZ[X]$ is a UFD.
$endgroup$
– qu binggang
Aug 29 '18 at 3:26
$begingroup$
Oh no! First fake news, now fake proofs...
$endgroup$
– Henrik
Aug 29 '18 at 14:53