Show homomorphism $f mapsto f^2 $ injective but not surjective.
$begingroup$
Let $R=mathbb F_2[x]$, let $psi:R to R$ given by $f mapsto f^2 $ be homomorphism.
How can I show that $psi$ is injective but not surjective?
Could I show that the image of $psi$ does not equal $R$ for surjectivity? If so, how can I explicitly demonstrate this?
For injectivity could I show $ker(psi) = {0_R}$?
EDIT: Extra question, am I correct in thinking $mathbb F_2[x]$ is isomorphic to a proper subring of itself? Is there a way to show this using $psi$ above and the first isomorphism theorem?
abstract-algebra ring-theory field-theory
edited Dec 26 '18 at 21:32
user26857
39.4k124183
$endgroup$
add a comment |
$begingroup$
Let $R=mathbb F_2[x]$, let $psi:R to R$ given by $f mapsto f^2 $ be homomorphism.
How can I show that $psi$ is injective but not surjective?
Could I show that the image of $psi$ does not equal $R$ for surjectivity? If so, how can I explicitly demonstrate this?
For injectivity could I show $ker(psi) = {0_R}$?
EDIT: Extra question, am I correct in thinking $mathbb F_2[x]$ is isomorphic to a proper subring of itself? Is there a way to show this using $psi$ above and the first isomorphism theorem?
abstract-algebra ring-theory field-theory
edited Dec 26 '18 at 21:32
user26857
39.4k124183
$endgroup$
add a comment |
$begingroup$
Let $R=mathbb F_2[x]$, let $psi:R to R$ given by $f mapsto f^2 $ be homomorphism.
How can I show that $psi$ is injective but not surjective?
Could I show that the image of $psi$ does not equal $R$ for surjectivity? If so, how can I explicitly demonstrate this?
For injectivity could I show $ker(psi) = {0_R}$?
EDIT: Extra question, am I correct in thinking $mathbb F_2[x]$ is isomorphic to a proper subring of itself? Is there a way to show this using $psi$ above and the first isomorphism theorem?
abstract-algebra ring-theory field-theory
edited Dec 26 '18 at 21:32
user26857
39.4k124183
$endgroup$
Let $R=mathbb F_2[x]$, let $psi:R to R$ given by $f mapsto f^2 $ be homomorphism.
How can I show that $psi$ is injective but not surjective?
Could I show that the image of $psi$ does not equal $R$ for surjectivity? If so, how can I explicitly demonstrate this?
For injectivity could I show $ker(psi) = {0_R}$?
EDIT: Extra question, am I correct in thinking $mathbb F_2[x]$ is isomorphic to a proper subring of itself? Is there a way to show this using $psi$ above and the first isomorphism theorem?
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
39.4k124183
asked May 1 '17 at 11:53
user415105
add a comment |
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2 Answers
2
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oldest
votes
$begingroup$
Just use the definition for injectivity.
For surjectivity, I think it is sufficient to show that there is no $a in R$ with $a^2 = x$.
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
$endgroup$
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
add a comment |
$begingroup$
Note:
$deg(p^2) = deg(p) + deg(p) = 2deg(p)$
Injectivity:
The above implies that the kernel is trivial.
Non-Surjectivity:
The above implies that odd degree polynomials do not exist in the image.
edited Dec 26 '18 at 21:32
user26857
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just use the definition for injectivity.
For surjectivity, I think it is sufficient to show that there is no $a in R$ with $a^2 = x$.
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
$endgroup$
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
add a comment |
$begingroup$
Just use the definition for injectivity.
For surjectivity, I think it is sufficient to show that there is no $a in R$ with $a^2 = x$.
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
$endgroup$
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
add a comment |
$begingroup$
Just use the definition for injectivity.
For surjectivity, I think it is sufficient to show that there is no $a in R$ with $a^2 = x$.
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
$endgroup$
Just use the definition for injectivity.
For surjectivity, I think it is sufficient to show that there is no $a in R$ with $a^2 = x$.
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
answered May 1 '17 at 11:57
Pieter21Pieter21
2,8451613
2,8451613
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
add a comment |
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
1
1
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
$begingroup$
For injectivity it is worth noticing that $psi$ is linear, so one just need to check that there is no polynomial $p$ with $p^2 = 0$.
$endgroup$
– user171326
May 1 '17 at 13:00
add a comment |
$begingroup$
Note:
$deg(p^2) = deg(p) + deg(p) = 2deg(p)$
Injectivity:
The above implies that the kernel is trivial.
Non-Surjectivity:
The above implies that odd degree polynomials do not exist in the image.
edited Dec 26 '18 at 21:32
user26857
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Note:
$deg(p^2) = deg(p) + deg(p) = 2deg(p)$
Injectivity:
The above implies that the kernel is trivial.
Non-Surjectivity:
The above implies that odd degree polynomials do not exist in the image.
edited Dec 26 '18 at 21:32
user26857
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Note:
$deg(p^2) = deg(p) + deg(p) = 2deg(p)$
Injectivity:
The above implies that the kernel is trivial.
Non-Surjectivity:
The above implies that odd degree polynomials do not exist in the image.
edited Dec 26 '18 at 21:32
user26857
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
$endgroup$
Note:
$deg(p^2) = deg(p) + deg(p) = 2deg(p)$
Injectivity:
The above implies that the kernel is trivial.
Non-Surjectivity:
The above implies that odd degree polynomials do not exist in the image.
edited Dec 26 '18 at 21:32
user26857
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
edited Dec 26 '18 at 21:32
user26857
39.4k124183
39.4k124183
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
answered Dec 25 '18 at 6:57
MetricMetric
1,25159
1,25159
add a comment |
add a comment |
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