find the maximal ideal of the ring ?.












2














find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra







share|cite




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    59 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    57 mins ago






  • 1




    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    – Bernard
    53 mins ago






  • 2




    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    – IAmNoOne
    41 mins ago


















2














find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra







share|cite




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    59 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    57 mins ago






  • 1




    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    – Bernard
    53 mins ago






  • 2




    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    – IAmNoOne
    41 mins ago
















2












2








2


1





find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra







share|cite















find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra




abstract-algebra






share|cite















share|cite













share|cite




share|cite








edited 35 mins ago









Key Flex

7,56241232




7,56241232










asked 1 hour ago









jasmine

1,582416




1,582416








  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    59 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    57 mins ago






  • 1




    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    – Bernard
    53 mins ago






  • 2




    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    – IAmNoOne
    41 mins ago
















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    59 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    57 mins ago






  • 1




    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    – Bernard
    53 mins ago






  • 2




    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    – IAmNoOne
    41 mins ago










3




3




No: $(x^2)$ is not maximal.
– Bernard
59 mins ago




No: $(x^2)$ is not maximal.
– Bernard
59 mins ago












@Bernard im not getting can u elaborate this ?
– jasmine
57 mins ago




@Bernard im not getting can u elaborate this ?
– jasmine
57 mins ago




1




1




The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
– Bernard
53 mins ago




The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
– Bernard
53 mins ago




2




2




@jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
– IAmNoOne
41 mins ago






@jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
– IAmNoOne
41 mins ago












2 Answers
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Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






share|cite|improve this answer





























    2














    There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      3














      Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






      share|cite|improve this answer


























        3














        Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






        share|cite|improve this answer
























          3












          3








          3






          Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






          share|cite|improve this answer












          Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 39 mins ago









          user544921

          607




          607























              2














              There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






              share|cite|improve this answer




























                2














                There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






                share|cite|improve this answer


























                  2












                  2








                  2






                  There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






                  share|cite|improve this answer














                  There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 11 mins ago

























                  answered 37 mins ago









                  Key Flex

                  7,56241232




                  7,56241232






























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