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
    1 hour ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    1 hour 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
    1 hour 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
    55 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
    1 hour ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    1 hour 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
    1 hour 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
    55 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 49 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
    1 hour ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    1 hour 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
    1 hour 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
    55 mins ago
















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    1 hour ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    1 hour 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
    1 hour 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
    55 mins ago










3




3




No: $(x^2)$ is not maximal.
– Bernard
1 hour ago




No: $(x^2)$ is not maximal.
– Bernard
1 hour ago












@Bernard im not getting can u elaborate this ?
– jasmine
1 hour ago




@Bernard im not getting can u elaborate this ?
– jasmine
1 hour 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
1 hour 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
1 hour 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
55 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
55 mins ago












3 Answers
3






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





























    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































      0














      Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



      Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains either $x^2$ or $x$) thus getting a nonzero constant (i.e. a unit).



      Let me clarify the last part by an example: so, why wouldn't $(x^2,x-6)$ be maximal? Well, multiply $(x-6)$ by $x$. Get $x^2-6x$. Subtract it from $x^2$. Get $6x$. From there we get $x$, then $6$, then $1$.






      share|cite





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        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 53 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 25 mins ago

























                    answered 51 mins ago









                    Key Flex

                    7,56241232




                    7,56241232























                        0














                        Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                        Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains either $x^2$ or $x$) thus getting a nonzero constant (i.e. a unit).



                        Let me clarify the last part by an example: so, why wouldn't $(x^2,x-6)$ be maximal? Well, multiply $(x-6)$ by $x$. Get $x^2-6x$. Subtract it from $x^2$. Get $6x$. From there we get $x$, then $6$, then $1$.






                        share|cite


























                          0














                          Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                          Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains either $x^2$ or $x$) thus getting a nonzero constant (i.e. a unit).



                          Let me clarify the last part by an example: so, why wouldn't $(x^2,x-6)$ be maximal? Well, multiply $(x-6)$ by $x$. Get $x^2-6x$. Subtract it from $x^2$. Get $6x$. From there we get $x$, then $6$, then $1$.






                          share|cite
























                            0












                            0








                            0






                            Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                            Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains either $x^2$ or $x$) thus getting a nonzero constant (i.e. a unit).



                            Let me clarify the last part by an example: so, why wouldn't $(x^2,x-6)$ be maximal? Well, multiply $(x-6)$ by $x$. Get $x^2-6x$. Subtract it from $x^2$. Get $6x$. From there we get $x$, then $6$, then $1$.






                            share|cite












                            Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                            Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains either $x^2$ or $x$) thus getting a nonzero constant (i.e. a unit).



                            Let me clarify the last part by an example: so, why wouldn't $(x^2,x-6)$ be maximal? Well, multiply $(x-6)$ by $x$. Get $x^2-6x$. Subtract it from $x^2$. Get $6x$. From there we get $x$, then $6$, then $1$.







                            share|cite












                            share|cite



                            share|cite










                            answered 1 min ago









                            Chris Custer

                            10.9k3824




                            10.9k3824






























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