$mathbb Z [X] / (X)$ isomorphic to $mathbb Z[X] / (X+1)$ isomorphic to $mathbb Z [X] / (X+2015)$
$begingroup$
I have to prove that $mathbb Z [X] / (X) cong mathbb Z[X] / (X+1) cong mathbb Z [X] / (X+2015)$.
I think that one answer could be that $mathbb Z[X]/(X) cong mathbb Z(0)$, $mathbb Z[X]/(X+1)cong Z(-1)$ and $mathbb Z [X] /(X+2015)cong mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015in mathbb Z$, then all of them are isomorphic to $mathbb Z $, hence isomorphic between them.
But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
abstract-algebra ring-theory
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
$endgroup$
add a comment |
$begingroup$
I have to prove that $mathbb Z [X] / (X) cong mathbb Z[X] / (X+1) cong mathbb Z [X] / (X+2015)$.
I think that one answer could be that $mathbb Z[X]/(X) cong mathbb Z(0)$, $mathbb Z[X]/(X+1)cong Z(-1)$ and $mathbb Z [X] /(X+2015)cong mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015in mathbb Z$, then all of them are isomorphic to $mathbb Z $, hence isomorphic between them.
But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
abstract-algebra ring-theory
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
$endgroup$
add a comment |
$begingroup$
I have to prove that $mathbb Z [X] / (X) cong mathbb Z[X] / (X+1) cong mathbb Z [X] / (X+2015)$.
I think that one answer could be that $mathbb Z[X]/(X) cong mathbb Z(0)$, $mathbb Z[X]/(X+1)cong Z(-1)$ and $mathbb Z [X] /(X+2015)cong mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015in mathbb Z$, then all of them are isomorphic to $mathbb Z $, hence isomorphic between them.
But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
abstract-algebra ring-theory
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
$endgroup$
I have to prove that $mathbb Z [X] / (X) cong mathbb Z[X] / (X+1) cong mathbb Z [X] / (X+2015)$.
I think that one answer could be that $mathbb Z[X]/(X) cong mathbb Z(0)$, $mathbb Z[X]/(X+1)cong Z(-1)$ and $mathbb Z [X] /(X+2015)cong mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015in mathbb Z$, then all of them are isomorphic to $mathbb Z $, hence isomorphic between them.
But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
asked Mar 30 '15 at 10:37
mkspkmkspk
588617
588617
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Some ideas:
Define
$$phi:Bbb Z[x]toBbb Z[x]/langle x+1rangle;,;;phi(p(x)):=p(x+1)+langle x+1rangle$$
Show that this is a homomorphism, and now: what is its kernel? You may want to apply the first isomorphism theorem.
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some ideas:
Define
$$phi:Bbb Z[x]toBbb Z[x]/langle x+1rangle;,;;phi(p(x)):=p(x+1)+langle x+1rangle$$
Show that this is a homomorphism, and now: what is its kernel? You may want to apply the first isomorphism theorem.
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
$endgroup$
add a comment |
$begingroup$
Some ideas:
Define
$$phi:Bbb Z[x]toBbb Z[x]/langle x+1rangle;,;;phi(p(x)):=p(x+1)+langle x+1rangle$$
Show that this is a homomorphism, and now: what is its kernel? You may want to apply the first isomorphism theorem.
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
$endgroup$
add a comment |
$begingroup$
Some ideas:
Define
$$phi:Bbb Z[x]toBbb Z[x]/langle x+1rangle;,;;phi(p(x)):=p(x+1)+langle x+1rangle$$
Show that this is a homomorphism, and now: what is its kernel? You may want to apply the first isomorphism theorem.
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
$endgroup$
Some ideas:
Define
$$phi:Bbb Z[x]toBbb Z[x]/langle x+1rangle;,;;phi(p(x)):=p(x+1)+langle x+1rangle$$
Show that this is a homomorphism, and now: what is its kernel? You may want to apply the first isomorphism theorem.
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
edited Mar 30 '15 at 11:12
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
answered Mar 30 '15 at 10:41
TimbucTimbuc
31k22145
31k22145
add a comment |
add a comment |
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