The ring $K[t^2,t^3]$ is not a PID
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I have to show that the ring $R=K[t^2,t^3]$ is not a PID, where $K$ is a field.
Consider the ideal $I=(t^2,t^3)$. If $R$ is a PID then there exist $f(t^2,t^3)in R$ such that $(t^2,t^3)= (f(t^2,t^3))$. Since, $t^2in Iimplies t^2=f(t^2,t^3)g(t^2,t^3) implies $ either $f$ is constant polynomial or $f$ is of degree 2 polynomial. If it is degree 2 polynomial then it can't give $t^3$ but what if it is constant?
abstract-algebra ring-theory
edited Dec 13 '18 at 21:43
user26857
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
$endgroup$
add a comment |
$begingroup$
I have to show that the ring $R=K[t^2,t^3]$ is not a PID, where $K$ is a field.
Consider the ideal $I=(t^2,t^3)$. If $R$ is a PID then there exist $f(t^2,t^3)in R$ such that $(t^2,t^3)= (f(t^2,t^3))$. Since, $t^2in Iimplies t^2=f(t^2,t^3)g(t^2,t^3) implies $ either $f$ is constant polynomial or $f$ is of degree 2 polynomial. If it is degree 2 polynomial then it can't give $t^3$ but what if it is constant?
abstract-algebra ring-theory
edited Dec 13 '18 at 21:43
user26857
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
$endgroup$
$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
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Sorry, but I did not get it.
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– XYZABC
Sep 1 '17 at 0:18
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An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43
add a comment |
$begingroup$
I have to show that the ring $R=K[t^2,t^3]$ is not a PID, where $K$ is a field.
Consider the ideal $I=(t^2,t^3)$. If $R$ is a PID then there exist $f(t^2,t^3)in R$ such that $(t^2,t^3)= (f(t^2,t^3))$. Since, $t^2in Iimplies t^2=f(t^2,t^3)g(t^2,t^3) implies $ either $f$ is constant polynomial or $f$ is of degree 2 polynomial. If it is degree 2 polynomial then it can't give $t^3$ but what if it is constant?
abstract-algebra ring-theory
edited Dec 13 '18 at 21:43
user26857
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
$endgroup$
I have to show that the ring $R=K[t^2,t^3]$ is not a PID, where $K$ is a field.
Consider the ideal $I=(t^2,t^3)$. If $R$ is a PID then there exist $f(t^2,t^3)in R$ such that $(t^2,t^3)= (f(t^2,t^3))$. Since, $t^2in Iimplies t^2=f(t^2,t^3)g(t^2,t^3) implies $ either $f$ is constant polynomial or $f$ is of degree 2 polynomial. If it is degree 2 polynomial then it can't give $t^3$ but what if it is constant?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 13 '18 at 21:43
user26857
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
edited Dec 13 '18 at 21:43
user26857
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
edited Dec 13 '18 at 21:43
user26857
39.4k124183
edited Dec 13 '18 at 21:43
user26857
39.4k124183
edited Dec 13 '18 at 21:43
user26857
39.4k124183
39.4k124183
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
asked Sep 1 '17 at 0:01
XYZABCXYZABC
334110
334110
$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
$begingroup$
Sorry, but I did not get it.
$endgroup$
– XYZABC
Sep 1 '17 at 0:18
$begingroup$
An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43
add a comment |
$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
$begingroup$
Sorry, but I did not get it.
$endgroup$
– XYZABC
Sep 1 '17 at 0:18
$begingroup$
An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43
$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
$begingroup$
Sorry, but I did not get it.
$endgroup$
– XYZABC
Sep 1 '17 at 0:18
$begingroup$
Sorry, but I did not get it.
$endgroup$
– XYZABC
Sep 1 '17 at 0:18
$begingroup$
An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43
add a comment |
1 Answer
1
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oldest
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$begingroup$
Here is another take.
Every PID is a UFD.
$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.
Therefore, $R$ is not a UFD and so cannot be a PID.
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Here is another take.
Every PID is a UFD.
$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.
Therefore, $R$ is not a UFD and so cannot be a PID.
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Here is another take.
Every PID is a UFD.
$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.
Therefore, $R$ is not a UFD and so cannot be a PID.
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Here is another take.
Every PID is a UFD.
$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.
Therefore, $R$ is not a UFD and so cannot be a PID.
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
$endgroup$
Here is another take.
Every PID is a UFD.
$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.
Therefore, $R$ is not a UFD and so cannot be a PID.
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
edited Dec 14 '18 at 11:29
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
answered Dec 14 '18 at 1:01
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
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$begingroup$
A nonzero constant is a unit.
$endgroup$
– quasi
Sep 1 '17 at 0:10
$begingroup$
Sorry, but I did not get it.
$endgroup$
– XYZABC
Sep 1 '17 at 0:18
$begingroup$
An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring.
$endgroup$
– quasi
Sep 1 '17 at 0:26
$begingroup$
Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant.
$endgroup$
– quasi
Sep 1 '17 at 0:28
$begingroup$
Okay, I got it. Thanks :)
$endgroup$
– XYZABC
Sep 1 '17 at 0:43