$R={f(x)in mathbb{Q}[x] | f'(0)=0 }$ 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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038145%2fr-fx-in-mathbbqx-f0-0-is-not-pid%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038145%2fr-fx-in-mathbbqx-f0-0-is-not-pid%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more