Mixing
Is $mathbb{Z}[x]/langle 4,x^2+x+1 rangle$ a field?
$begingroup$
I know that $mathbb{Z}[x]/langle4, x^2+x+1rangle$ is isomorphic to $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle $. Since $(Bbb Z/4Bbb Z)[X]$ is not a PID, irreducibility/reducibility of $x^2+x+1$ is not enough to say whether or not this is a field. My idea is to prove that this ring is not a domain. For example, $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle$ is not a domain if $2$ is not in the ideal $langle x^2+x+1rangle$ but I can't prove this.
abstract-algebra ring-theory
edited Feb 2 at 16:48
Shaun
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
$endgroup$
add a comment |
$begingroup$
I know that $mathbb{Z}[x]/langle4, x^2+x+1rangle$ is isomorphic to $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle $. Since $(Bbb Z/4Bbb Z)[X]$ is not a PID, irreducibility/reducibility of $x^2+x+1$ is not enough to say whether or not this is a field. My idea is to prove that this ring is not a domain. For example, $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle$ is not a domain if $2$ is not in the ideal $langle x^2+x+1rangle$ but I can't prove this.
abstract-algebra ring-theory
edited Feb 2 at 16:48
Shaun
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
$endgroup$
add a comment |
$begingroup$
I know that $mathbb{Z}[x]/langle4, x^2+x+1rangle$ is isomorphic to $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle $. Since $(Bbb Z/4Bbb Z)[X]$ is not a PID, irreducibility/reducibility of $x^2+x+1$ is not enough to say whether or not this is a field. My idea is to prove that this ring is not a domain. For example, $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle$ is not a domain if $2$ is not in the ideal $langle x^2+x+1rangle$ but I can't prove this.
abstract-algebra ring-theory
edited Feb 2 at 16:48
Shaun
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
$endgroup$
I know that $mathbb{Z}[x]/langle4, x^2+x+1rangle$ is isomorphic to $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle $. Since $(Bbb Z/4Bbb Z)[X]$ is not a PID, irreducibility/reducibility of $x^2+x+1$ is not enough to say whether or not this is a field. My idea is to prove that this ring is not a domain. For example, $(mathbb{Z}/4mathbb{Z})[X]/langle x^2+x+1rangle$ is not a domain if $2$ is not in the ideal $langle x^2+x+1rangle$ but I can't prove this.
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Feb 2 at 16:48
Shaun
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
edited Feb 2 at 16:48
Shaun
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
edited Feb 2 at 16:48
Shaun
10.5k113687
edited Feb 2 at 16:48
Shaun
10.5k113687
edited Feb 2 at 16:48
Shaun
10.5k113687
10.5k113687
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
asked Feb 2 at 12:52
Andreas DahlbergAndreas Dahlberg
806
806
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint $ $ Let $,h = x^2!+!x!+!1. $ If $,2,$ is a unit: $,2 f = 1 + 4g + hh',$ in $Bbb Z[x],$ $overset{bmod 2}Longrightarrow,hmid 1,$ in $,Bbb Z_2[x] Rightarrow!Leftarrow$
And if $,2=0,$ then $,2 = 4g + hh',$ so $,2mid h',$ hence $,1 = 2g+ h(h'/2),$ $overset{bmod 2}Longrightarrow, hmid 1,$ in $Bbb Z_2[x] Rightarrow!Leftarrow$
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
$endgroup$
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
|
show 1 more comment
$begingroup$
Hint: $2$ is in $(x^2+x+1){bf Z}/4{bf Z}[x]$ if and only if for some $kin {mathbf Z}$ such that $kequiv 2 pmod 4$, $k$ is in $(x^2+x+1){bf Z}[x]$.
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
Hint $ $ Let $,h = x^2!+!x!+!1. $ If $,2,$ is a unit: $,2 f = 1 + 4g + hh',$ in $Bbb Z[x],$ $overset{bmod 2}Longrightarrow,hmid 1,$ in $,Bbb Z_2[x] Rightarrow!Leftarrow$
And if $,2=0,$ then $,2 = 4g + hh',$ so $,2mid h',$ hence $,1 = 2g+ h(h'/2),$ $overset{bmod 2}Longrightarrow, hmid 1,$ in $Bbb Z_2[x] Rightarrow!Leftarrow$
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
$endgroup$
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
|
show 1 more comment
$begingroup$
Hint $ $ Let $,h = x^2!+!x!+!1. $ If $,2,$ is a unit: $,2 f = 1 + 4g + hh',$ in $Bbb Z[x],$ $overset{bmod 2}Longrightarrow,hmid 1,$ in $,Bbb Z_2[x] Rightarrow!Leftarrow$
And if $,2=0,$ then $,2 = 4g + hh',$ so $,2mid h',$ hence $,1 = 2g+ h(h'/2),$ $overset{bmod 2}Longrightarrow, hmid 1,$ in $Bbb Z_2[x] Rightarrow!Leftarrow$
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
$endgroup$
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
|
show 1 more comment
$begingroup$
Hint $ $ Let $,h = x^2!+!x!+!1. $ If $,2,$ is a unit: $,2 f = 1 + 4g + hh',$ in $Bbb Z[x],$ $overset{bmod 2}Longrightarrow,hmid 1,$ in $,Bbb Z_2[x] Rightarrow!Leftarrow$
And if $,2=0,$ then $,2 = 4g + hh',$ so $,2mid h',$ hence $,1 = 2g+ h(h'/2),$ $overset{bmod 2}Longrightarrow, hmid 1,$ in $Bbb Z_2[x] Rightarrow!Leftarrow$
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
$endgroup$
Hint $ $ Let $,h = x^2!+!x!+!1. $ If $,2,$ is a unit: $,2 f = 1 + 4g + hh',$ in $Bbb Z[x],$ $overset{bmod 2}Longrightarrow,hmid 1,$ in $,Bbb Z_2[x] Rightarrow!Leftarrow$
And if $,2=0,$ then $,2 = 4g + hh',$ so $,2mid h',$ hence $,1 = 2g+ h(h'/2),$ $overset{bmod 2}Longrightarrow, hmid 1,$ in $Bbb Z_2[x] Rightarrow!Leftarrow$
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
edited Feb 2 at 14:06
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
answered Feb 2 at 13:12
Bill DubuqueBill Dubuque
214k29197657
214k29197657
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
|
show 1 more comment
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
$begingroup$
Doesn’t our ring plainly contain $Bbb Z/4Bbb Z$ as a subring? Therefore not a field. Am I missing something?
$endgroup$
– Lubin
Feb 2 at 17:42
1
1
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
@Lubin I did it more directly in case the OP does not yet know the isomorphism theorems etc needed to rigorously deduce that. Presumably if they did then they wouldn't have asked the question.
$endgroup$
– Bill Dubuque
Feb 2 at 17:56
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
Grudgingly, I must agree.
$endgroup$
– Lubin
Feb 2 at 18:01
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
@Lubin: It is not so clear. It admits a homomorphism from ${bf Z}/4{bf Z}$, but it could be (a priori) not injective.
$endgroup$
– tomasz
Feb 4 at 14:29
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
$begingroup$
I’m sorry @tomasz, I think it is clear: any time $R$ is a ring and you take $S=R[x]/(f(x))$ where $f$ is monic, then $S$ is free over $R$ of rank equal to the degree of $f$, and the map $Rto S$ is an injection.
$endgroup$
– Lubin
Feb 4 at 20:36
|
show 1 more comment
$begingroup$
Hint: $2$ is in $(x^2+x+1){bf Z}/4{bf Z}[x]$ if and only if for some $kin {mathbf Z}$ such that $kequiv 2 pmod 4$, $k$ is in $(x^2+x+1){bf Z}[x]$.
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
$endgroup$
add a comment |
$begingroup$
Hint: $2$ is in $(x^2+x+1){bf Z}/4{bf Z}[x]$ if and only if for some $kin {mathbf Z}$ such that $kequiv 2 pmod 4$, $k$ is in $(x^2+x+1){bf Z}[x]$.
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
$endgroup$
add a comment |
$begingroup$
Hint: $2$ is in $(x^2+x+1){bf Z}/4{bf Z}[x]$ if and only if for some $kin {mathbf Z}$ such that $kequiv 2 pmod 4$, $k$ is in $(x^2+x+1){bf Z}[x]$.
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
$endgroup$
Hint: $2$ is in $(x^2+x+1){bf Z}/4{bf Z}[x]$ if and only if for some $kin {mathbf Z}$ such that $kequiv 2 pmod 4$, $k$ is in $(x^2+x+1){bf Z}[x]$.
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
answered Feb 2 at 13:12
tomasztomasz
24.1k23482
24.1k23482
add a comment |
add a comment |
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