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 ring-theory







share|cite|improve this question




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    yesterday










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    yesterday






  • 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
    yesterday






  • 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
    yesterday


















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 ring-theory







share|cite|improve this question




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    yesterday










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    yesterday






  • 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
    yesterday






  • 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
    yesterday
















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 ring-theory







share|cite|improve this question















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 ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









J. W. Tanner

466




466










asked yesterday









jasmine

1,591416




1,591416








  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    yesterday










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    yesterday






  • 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
    yesterday






  • 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
    yesterday
















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    yesterday










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    yesterday






  • 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
    yesterday






  • 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
    yesterday










3




3




No: $(x^2)$ is not maximal.
– Bernard
yesterday




No: $(x^2)$ is not maximal.
– Bernard
yesterday












@Bernard im not getting can u elaborate this ?
– jasmine
yesterday




@Bernard im not getting can u elaborate this ?
– jasmine
yesterday




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
yesterday




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
yesterday




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
yesterday






@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
yesterday












4 Answers
4






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














    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































      3














      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 $x$) thus getting a nonzero constant (i.e. a unit).



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






      share|cite|improve this answer































        0














        Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






        share|cite|improve this answer





















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          4 Answers
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          4 Answers
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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 yesterday









              user544921

              607




              607























                  3














                  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




























                    3














                    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


























                      3












                      3








                      3






                      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 yesterday

























                      answered yesterday









                      Key Flex

                      7,53441232




                      7,53441232























                          3














                          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 $x$) thus getting a nonzero constant (i.e. a unit).



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






                          share|cite|improve this answer




























                            3














                            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 $x$) thus getting a nonzero constant (i.e. a unit).



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






                            share|cite|improve this answer


























                              3












                              3








                              3






                              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 $x$) thus getting a nonzero constant (i.e. a unit).



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






                              share|cite|improve this answer














                              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 $x$) thus getting a nonzero constant (i.e. a unit).



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







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited yesterday

























                              answered yesterday









                              Chris Custer

                              10.9k3824




                              10.9k3824























                                  0














                                  Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                  share|cite|improve this answer


























                                    0














                                    Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                      share|cite|improve this answer












                                      Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered yesterday









                                      Es.Ro

                                      75228




                                      75228






























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