Mixing
Quotient of two ideals equality
$begingroup$
I was following a sequence of ring isomorphisms, sorry about that but I cannot figure out why in the last step we have:
$ (x^2+ 1, 1+x) / (1+x) = ( 2) $ in the ring $ mathbb{Z} [X] / (1+x) $
ring-theory
asked Jan 23 at 23:52
PsylexPsylex
1068
$endgroup$
add a comment |
$begingroup$
I was following a sequence of ring isomorphisms, sorry about that but I cannot figure out why in the last step we have:
$ (x^2+ 1, 1+x) / (1+x) = ( 2) $ in the ring $ mathbb{Z} [X] / (1+x) $
ring-theory
asked Jan 23 at 23:52
PsylexPsylex
1068
$endgroup$
add a comment |
$begingroup$
I was following a sequence of ring isomorphisms, sorry about that but I cannot figure out why in the last step we have:
$ (x^2+ 1, 1+x) / (1+x) = ( 2) $ in the ring $ mathbb{Z} [X] / (1+x) $
ring-theory
asked Jan 23 at 23:52
PsylexPsylex
1068
$endgroup$
I was following a sequence of ring isomorphisms, sorry about that but I cannot figure out why in the last step we have:
$ (x^2+ 1, 1+x) / (1+x) = ( 2) $ in the ring $ mathbb{Z} [X] / (1+x) $
ring-theory
ring-theory
asked Jan 23 at 23:52
PsylexPsylex
1068
asked Jan 23 at 23:52
PsylexPsylex
1068
asked Jan 23 at 23:52
PsylexPsylex
1068
asked Jan 23 at 23:52
PsylexPsylex
1068
asked Jan 23 at 23:52
PsylexPsylex
1068
1068
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
We have that:
$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$
Thus in $mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = langle 1+x rangle$
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
$endgroup$
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have that:
$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$
Thus in $mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = langle 1+x rangle$
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
$endgroup$
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
add a comment |
$begingroup$
We have that:
$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$
Thus in $mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = langle 1+x rangle$
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
$endgroup$
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
add a comment |
$begingroup$
We have that:
$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$
Thus in $mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = langle 1+x rangle$
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
$endgroup$
We have that:
$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$
Thus in $mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = langle 1+x rangle$
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
answered Jan 23 at 23:55
Stefan4024Stefan4024
30.6k63479
30.6k63479
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
add a comment |
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
$begingroup$
Thanks. So when i consider the map $ Phi: mathbb{Z} [X] / (X+1) rightarrow mathbb{Z} : : $newline $ pi (P) mapsto pi ( P(-1)) $ then we would get $ Phi( (x^2 + 1) + (x+1)) = Phi( (2) + (1+x)) = (2) $ hence $ mathbb{Z} [X] / (X+1) cong mathbb{Z} / (2) $ if i am not mistaken
$endgroup$
– Psylex
Jan 24 at 9:21
add a comment |
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