A question about the ideals of a ring of power series.












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Let $F$ a field and we consider the ring of power series $F[[x]]$. After proving that $F[[x]]$ is a P.I.D. and after proving that the ideals of $F[[x]]$ are of the form $(x^k)$ for $kge0$, why $F[x]supset (x)supset(x^2)$? Why $F[[x]]ne (x)$ and $(x)ne (x^2)$.



Thanks!






abstract-algebra proof-verification proof-explanation







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  • 2




    $begingroup$
    In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
    $endgroup$
    – rschwieb
    Dec 10 '18 at 15:05


















3












$begingroup$


Let $F$ a field and we consider the ring of power series $F[[x]]$. After proving that $F[[x]]$ is a P.I.D. and after proving that the ideals of $F[[x]]$ are of the form $(x^k)$ for $kge0$, why $F[x]supset (x)supset(x^2)$? Why $F[[x]]ne (x)$ and $(x)ne (x^2)$.



Thanks!






abstract-algebra proof-verification proof-explanation







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
    $endgroup$
    – rschwieb
    Dec 10 '18 at 15:05
















3












3








3





$begingroup$


Let $F$ a field and we consider the ring of power series $F[[x]]$. After proving that $F[[x]]$ is a P.I.D. and after proving that the ideals of $F[[x]]$ are of the form $(x^k)$ for $kge0$, why $F[x]supset (x)supset(x^2)$? Why $F[[x]]ne (x)$ and $(x)ne (x^2)$.



Thanks!






abstract-algebra proof-verification proof-explanation







share|cite|improve this question









$endgroup$




Let $F$ a field and we consider the ring of power series $F[[x]]$. After proving that $F[[x]]$ is a P.I.D. and after proving that the ideals of $F[[x]]$ are of the form $(x^k)$ for $kge0$, why $F[x]supset (x)supset(x^2)$? Why $F[[x]]ne (x)$ and $(x)ne (x^2)$.



Thanks!






abstract-algebra proof-verification proof-explanation




abstract-algebra proof-verification proof-explanation






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 10 '18 at 14:56









Jack J.Jack J.

4532419




4532419








  • 2




    $begingroup$
    In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
    $endgroup$
    – rschwieb
    Dec 10 '18 at 15:05
















  • 2




    $begingroup$
    In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
    $endgroup$
    – rschwieb
    Dec 10 '18 at 15:05










2




2




$begingroup$
In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
$endgroup$
– rschwieb
Dec 10 '18 at 15:05






$begingroup$
In general, for any integral domain, and any nonzero nonunit $x$, $(x^{n+1})subsetneq (x^n)$. For if not, $x^n=x^{n+1}r$ would imply $xr=1$ after cancellation, contradicting that $x$ isn't a unit. But I think Matt's raw description of what each of the ideals $(x^n)$ is like makes it patently clear already for this case.
$endgroup$
– rschwieb
Dec 10 '18 at 15:05












1 Answer
1






active

oldest

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$begingroup$

$(x)$ is the ideal consisting of series with no constant term. Certainly there exist series with nonzero constant term, hence the inclusion is proper. Similarly, $(x^2)$ is the ideal of series with no constant or linear term. Since there exist series with constant terms, or with no constant term but a linear term, the inclusion is proper.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
    $endgroup$
    – Jack J.
    Dec 10 '18 at 15:23








  • 1




    $begingroup$
    @Jack Sure , that works.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 15:45











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4












$begingroup$

$(x)$ is the ideal consisting of series with no constant term. Certainly there exist series with nonzero constant term, hence the inclusion is proper. Similarly, $(x^2)$ is the ideal of series with no constant or linear term. Since there exist series with constant terms, or with no constant term but a linear term, the inclusion is proper.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
    $endgroup$
    – Jack J.
    Dec 10 '18 at 15:23








  • 1




    $begingroup$
    @Jack Sure , that works.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 15:45
















4












$begingroup$

$(x)$ is the ideal consisting of series with no constant term. Certainly there exist series with nonzero constant term, hence the inclusion is proper. Similarly, $(x^2)$ is the ideal of series with no constant or linear term. Since there exist series with constant terms, or with no constant term but a linear term, the inclusion is proper.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
    $endgroup$
    – Jack J.
    Dec 10 '18 at 15:23








  • 1




    $begingroup$
    @Jack Sure , that works.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 15:45














4












4








4





$begingroup$

$(x)$ is the ideal consisting of series with no constant term. Certainly there exist series with nonzero constant term, hence the inclusion is proper. Similarly, $(x^2)$ is the ideal of series with no constant or linear term. Since there exist series with constant terms, or with no constant term but a linear term, the inclusion is proper.






share|cite|improve this answer









$endgroup$



$(x)$ is the ideal consisting of series with no constant term. Certainly there exist series with nonzero constant term, hence the inclusion is proper. Similarly, $(x^2)$ is the ideal of series with no constant or linear term. Since there exist series with constant terms, or with no constant term but a linear term, the inclusion is proper.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 15:03









Matt SamuelMatt Samuel

38.1k63767




38.1k63767












  • $begingroup$
    Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
    $endgroup$
    – Jack J.
    Dec 10 '18 at 15:23








  • 1




    $begingroup$
    @Jack Sure , that works.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 15:45


















  • $begingroup$
    Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
    $endgroup$
    – Jack J.
    Dec 10 '18 at 15:23








  • 1




    $begingroup$
    @Jack Sure , that works.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 15:45
















$begingroup$
Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
$endgroup$
– Jack J.
Dec 10 '18 at 15:23






$begingroup$
Thanks for your answer. So, if it were $F[[x]]=(x)$ we would be asserting that in $F[[x]]$ every formal power series has costant term null, but this is absurd, because just take $1=1+0x+0x^2+cdots$. Moreover, if it were $(x)=(x^2)$ we would asserting that in $(x)$ every formal series has linear term null, but this is absurd, in fact just take $x=0+1x+0x^2cdots$. It's correct?
$endgroup$
– Jack J.
Dec 10 '18 at 15:23






1




1




$begingroup$
@Jack Sure , that works.
$endgroup$
– Matt Samuel
Dec 10 '18 at 15:45




$begingroup$
@Jack Sure , that works.
$endgroup$
– Matt Samuel
Dec 10 '18 at 15:45


















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