$R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ is not PID












2












$begingroup$


Suppose $R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ subring of the principal domain $mathbb{Q}[x]$.



I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 + a_2x^2 + a_3x^3+...+a_nx^n$ and hence if $x^2=ab$, one of them have to have degree 0, hence unit.



I have already proven that $(x^2)$ is not a prime ideal in $R$. Since $a=x^3$, $b=x^5$, $abin (x^2)$ and $a notin (x^2)$, $b notin (x^2)$.



Now I need to deduce that $R$ is not PID. It should be easy but I can not see it.






ring-theory commutative-algebra principal-ideal-domains







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
    $endgroup$
    – Seewoo Lee
    Dec 13 '18 at 15:28






  • 2




    $begingroup$
    @rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 16:11






  • 1




    $begingroup$
    @TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:22








  • 1




    $begingroup$
    @TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:26








  • 1




    $begingroup$
    @rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 20:49
















2












$begingroup$


Suppose $R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ subring of the principal domain $mathbb{Q}[x]$.



I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 + a_2x^2 + a_3x^3+...+a_nx^n$ and hence if $x^2=ab$, one of them have to have degree 0, hence unit.



I have already proven that $(x^2)$ is not a prime ideal in $R$. Since $a=x^3$, $b=x^5$, $abin (x^2)$ and $a notin (x^2)$, $b notin (x^2)$.



Now I need to deduce that $R$ is not PID. It should be easy but I can not see it.






ring-theory commutative-algebra principal-ideal-domains







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
    $endgroup$
    – Seewoo Lee
    Dec 13 '18 at 15:28






  • 2




    $begingroup$
    @rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 16:11






  • 1




    $begingroup$
    @TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:22








  • 1




    $begingroup$
    @TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:26








  • 1




    $begingroup$
    @rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 20:49














2












2








2


1



$begingroup$


Suppose $R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ subring of the principal domain $mathbb{Q}[x]$.



I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 + a_2x^2 + a_3x^3+...+a_nx^n$ and hence if $x^2=ab$, one of them have to have degree 0, hence unit.



I have already proven that $(x^2)$ is not a prime ideal in $R$. Since $a=x^3$, $b=x^5$, $abin (x^2)$ and $a notin (x^2)$, $b notin (x^2)$.



Now I need to deduce that $R$ is not PID. It should be easy but I can not see it.






ring-theory commutative-algebra principal-ideal-domains







share|cite|improve this question









$endgroup$




Suppose $R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ subring of the principal domain $mathbb{Q}[x]$.



I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 + a_2x^2 + a_3x^3+...+a_nx^n$ and hence if $x^2=ab$, one of them have to have degree 0, hence unit.



I have already proven that $(x^2)$ is not a prime ideal in $R$. Since $a=x^3$, $b=x^5$, $abin (x^2)$ and $a notin (x^2)$, $b notin (x^2)$.



Now I need to deduce that $R$ is not PID. It should be easy but I can not see it.






ring-theory commutative-algebra principal-ideal-domains




ring-theory commutative-algebra principal-ideal-domains






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 15:22









idriskameniidriskameni

745321




745321








  • 1




    $begingroup$
    I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
    $endgroup$
    – Seewoo Lee
    Dec 13 '18 at 15:28






  • 2




    $begingroup$
    @rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 16:11






  • 1




    $begingroup$
    @TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:22








  • 1




    $begingroup$
    @TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:26








  • 1




    $begingroup$
    @rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 20:49














  • 1




    $begingroup$
    I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
    $endgroup$
    – Seewoo Lee
    Dec 13 '18 at 15:28






  • 2




    $begingroup$
    @rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 16:11






  • 1




    $begingroup$
    @TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:22








  • 1




    $begingroup$
    @TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:26








  • 1




    $begingroup$
    @rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
    $endgroup$
    – Trevor Gunn
    Dec 13 '18 at 20:49








1




1




$begingroup$
I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
$endgroup$
– Seewoo Lee
Dec 13 '18 at 15:28




$begingroup$
I think $x^{5} = x^{2}cdot x^{3}$ and $x^{3}in R$.
$endgroup$
– Seewoo Lee
Dec 13 '18 at 15:28




2




2




$begingroup$
@rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
$endgroup$
– Trevor Gunn
Dec 13 '18 at 16:11




$begingroup$
@rschwieb I'm pretty sure $R = mathbb{Q}[x^2,x^3],(cong mathbb{Q}[s,t]/(s^3 - t^2))$ and $R/(x^2) cong R[y]/(y^2)$ via $x^3 mapsto y$.
$endgroup$
– Trevor Gunn
Dec 13 '18 at 16:11




1




1




$begingroup$
@TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
$endgroup$
– rschwieb
Dec 13 '18 at 16:22






$begingroup$
@TrevorGunn Yeah, I always forget about these crazy terms after $x^2$. You're probably right.
$endgroup$
– rschwieb
Dec 13 '18 at 16:22






1




1




$begingroup$
@TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
$endgroup$
– rschwieb
Dec 13 '18 at 16:26






$begingroup$
@TrevorGunn Well... I'm not so sure about what the quotient looks like now, but it's still true that $R=mathbb Q+I$ where $I$ is the subset of $mathbb Q[x]$ with lowest exponent at least $2$, right? Is that really what $mathbb Q[x^2,x^3]$ looks like? I always envisioned it as being more complicated.
$endgroup$
– rschwieb
Dec 13 '18 at 16:26






1




1




$begingroup$
@rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
$endgroup$
– Trevor Gunn
Dec 13 '18 at 20:49




$begingroup$
@rschwieb That part is still true and there $I = (x^2, x^3)$. This ring shows up as a common problem in an algebraic geometry course (show that ${(t^2,t^3) : t in k}$ is a variety). It is clear that $x^3 - y^2$ vanishes on this set and given $(x,y)$ with $x^3 = y^2$ one sets $t = y/x$. From the representation $k[x^2,x^3] = k[s,t]/(s^3 - t^2)$, one computes $k[x^2,x^3]/(x^2) = k[y]/(y^2)$ and $k[x^2,x^3] = k[y]/(y^3)$. It should also be possible to get this by looking at derivatives of the parametric representation $(t^2,t^3)$.
$endgroup$
– Trevor Gunn
Dec 13 '18 at 20:49










2 Answers
2






active

oldest

votes


















2












$begingroup$

Consider the ideal generated by $x^2$ and $x^3$ it is not principal.



Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)leq 2$, if $deg(p)=1$, $p'(0)neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
    $endgroup$
    – idriskameni
    Dec 13 '18 at 15:41



















2












$begingroup$

Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 cdot x^2 in (x^2)$. What you want is $(x^3)^2 in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:04













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Consider the ideal generated by $x^2$ and $x^3$ it is not principal.



Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)leq 2$, if $deg(p)=1$, $p'(0)neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
    $endgroup$
    – idriskameni
    Dec 13 '18 at 15:41
















2












$begingroup$

Consider the ideal generated by $x^2$ and $x^3$ it is not principal.



Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)leq 2$, if $deg(p)=1$, $p'(0)neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
    $endgroup$
    – idriskameni
    Dec 13 '18 at 15:41














2












2








2





$begingroup$

Consider the ideal generated by $x^2$ and $x^3$ it is not principal.



Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)leq 2$, if $deg(p)=1$, $p'(0)neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)neq 0$.






share|cite|improve this answer











$endgroup$



Consider the ideal generated by $x^2$ and $x^3$ it is not principal.



Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)leq 2$, if $deg(p)=1$, $p'(0)neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)neq 0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 15:55

























answered Dec 13 '18 at 15:29









Tsemo AristideTsemo Aristide

58.8k11445




58.8k11445












  • $begingroup$
    Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
    $endgroup$
    – idriskameni
    Dec 13 '18 at 15:41


















  • $begingroup$
    Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
    $endgroup$
    – idriskameni
    Dec 13 '18 at 15:41
















$begingroup$
Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
$endgroup$
– idriskameni
Dec 13 '18 at 15:41




$begingroup$
Is there any test to see that it is not principal or you just have to compute it and see that it is not possible to generate it with one element?
$endgroup$
– idriskameni
Dec 13 '18 at 15:41











2












$begingroup$

Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 cdot x^2 in (x^2)$. What you want is $(x^3)^2 in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:04


















2












$begingroup$

Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 cdot x^2 in (x^2)$. What you want is $(x^3)^2 in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:04
















2












2








2





$begingroup$

Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 cdot x^2 in (x^2)$. What you want is $(x^3)^2 in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.






share|cite|improve this answer









$endgroup$



Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 cdot x^2 in (x^2)$. What you want is $(x^3)^2 in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 16:02









Trevor GunnTrevor Gunn

14.8k32047




14.8k32047








  • 1




    $begingroup$
    +1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:04
















  • 1




    $begingroup$
    +1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
    $endgroup$
    – rschwieb
    Dec 13 '18 at 16:04










1




1




$begingroup$
+1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
$endgroup$
– rschwieb
Dec 13 '18 at 16:04






$begingroup$
+1 I was trying to come up with a pithy way of explaining $x^3$ is irreducible (or prime) to say just this, that $x^6$ has distinct factorizations.
$endgroup$
– rschwieb
Dec 13 '18 at 16:04




















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