Confused about real polynomial ring generated by an element











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Claim: $(x)neq langle x rangle$ in $mathbb{R}[x].$



Notation:



$(x)$ is the principal ideal generated by x.



$langle x rangle$ is the same as in group theory, namely cyclic group generated by x.



My idea: Since the set of all polynomials with constant term $0$ is equal to $(x)$ and an ideal of $mathbb{R}[x]$ (I have proved this) and then showing that $langle x rangle$ is not an ideal, the above claim is true.



However, I am confused on what $langle x rangle$ is in $mathbb{R}[x]$. Is it just the set ${x, 2x, 3x,..., nx: lvert nin mathbb{N}}$? But isn't it generating the polynomial ring? What is it?










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  • 1




    Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
    – Ethan Bolker
    Nov 20 at 1:02






  • 1




    Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
    – xbh
    Nov 20 at 1:02






  • 1




    Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
    – xbh
    Nov 20 at 1:04










  • @xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
    – Tomás Palamás
    Nov 20 at 1:08










  • Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
    – Randall
    Nov 20 at 1:54

















up vote
0
down vote

favorite












Claim: $(x)neq langle x rangle$ in $mathbb{R}[x].$



Notation:



$(x)$ is the principal ideal generated by x.



$langle x rangle$ is the same as in group theory, namely cyclic group generated by x.



My idea: Since the set of all polynomials with constant term $0$ is equal to $(x)$ and an ideal of $mathbb{R}[x]$ (I have proved this) and then showing that $langle x rangle$ is not an ideal, the above claim is true.



However, I am confused on what $langle x rangle$ is in $mathbb{R}[x]$. Is it just the set ${x, 2x, 3x,..., nx: lvert nin mathbb{N}}$? But isn't it generating the polynomial ring? What is it?










share|cite|improve this question




















  • 1




    Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
    – Ethan Bolker
    Nov 20 at 1:02






  • 1




    Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
    – xbh
    Nov 20 at 1:02






  • 1




    Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
    – xbh
    Nov 20 at 1:04










  • @xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
    – Tomás Palamás
    Nov 20 at 1:08










  • Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
    – Randall
    Nov 20 at 1:54















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Claim: $(x)neq langle x rangle$ in $mathbb{R}[x].$



Notation:



$(x)$ is the principal ideal generated by x.



$langle x rangle$ is the same as in group theory, namely cyclic group generated by x.



My idea: Since the set of all polynomials with constant term $0$ is equal to $(x)$ and an ideal of $mathbb{R}[x]$ (I have proved this) and then showing that $langle x rangle$ is not an ideal, the above claim is true.



However, I am confused on what $langle x rangle$ is in $mathbb{R}[x]$. Is it just the set ${x, 2x, 3x,..., nx: lvert nin mathbb{N}}$? But isn't it generating the polynomial ring? What is it?










share|cite|improve this question















Claim: $(x)neq langle x rangle$ in $mathbb{R}[x].$



Notation:



$(x)$ is the principal ideal generated by x.



$langle x rangle$ is the same as in group theory, namely cyclic group generated by x.



My idea: Since the set of all polynomials with constant term $0$ is equal to $(x)$ and an ideal of $mathbb{R}[x]$ (I have proved this) and then showing that $langle x rangle$ is not an ideal, the above claim is true.



However, I am confused on what $langle x rangle$ is in $mathbb{R}[x]$. Is it just the set ${x, 2x, 3x,..., nx: lvert nin mathbb{N}}$? But isn't it generating the polynomial ring? What is it?







abstract-algebra ring-theory






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edited Nov 20 at 1:10

























asked Nov 20 at 0:54









Tomás Palamás

373211




373211








  • 1




    Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
    – Ethan Bolker
    Nov 20 at 1:02






  • 1




    Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
    – xbh
    Nov 20 at 1:02






  • 1




    Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
    – xbh
    Nov 20 at 1:04










  • @xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
    – Tomás Palamás
    Nov 20 at 1:08










  • Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
    – Randall
    Nov 20 at 1:54
















  • 1




    Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
    – Ethan Bolker
    Nov 20 at 1:02






  • 1




    Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
    – xbh
    Nov 20 at 1:02






  • 1




    Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
    – xbh
    Nov 20 at 1:04










  • @xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
    – Tomás Palamás
    Nov 20 at 1:08










  • Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
    – Randall
    Nov 20 at 1:54










1




1




Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
– Ethan Bolker
Nov 20 at 1:02




Not sure this is what you are looking for, but $x^2$ is in the ideal generated by $x$.
– Ethan Bolker
Nov 20 at 1:02




1




1




Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
– xbh
Nov 20 at 1:02




Isn't $langle x rangle = Bbb Zx$ as an additive subgroup?
– xbh
Nov 20 at 1:02




1




1




Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
– xbh
Nov 20 at 1:04




Then easy to see that $sqrt 2 x in (x), notin langle x rangle$.
– xbh
Nov 20 at 1:04












@xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
– Tomás Palamás
Nov 20 at 1:08




@xbh Not sure about your first comment since I'm dealing with real polynomials. Regarding your second comment, that's what I was thinking but I wasn't sure if I am interpreting $langle x rangle$ correctly.
– Tomás Palamás
Nov 20 at 1:08












Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
– Randall
Nov 20 at 1:54






Fun fact: $(a) = langle a rangle$ for all $a in R$ if and only if $R$ is $mathbb{Z}$ or $mathbb{Z}_n$ with the usual operations (up to iso).
– Randall
Nov 20 at 1:54












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As you mentioned, $langle x rangle$ is the set ${x, 2x, 3x,...,nx,... : lvert nin mathbb{N}}$. But $(x) = {rx lvert rin mathbb{R}[x]}$. Then you can see that $x^2 in (x)$,but $x^2notin langle x rangle$ as mentioned in the comment. In general, $langle x rangle subseteq (x) $.






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    As you mentioned, $langle x rangle$ is the set ${x, 2x, 3x,...,nx,... : lvert nin mathbb{N}}$. But $(x) = {rx lvert rin mathbb{R}[x]}$. Then you can see that $x^2 in (x)$,but $x^2notin langle x rangle$ as mentioned in the comment. In general, $langle x rangle subseteq (x) $.






    share|cite|improve this answer

























      up vote
      2
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      As you mentioned, $langle x rangle$ is the set ${x, 2x, 3x,...,nx,... : lvert nin mathbb{N}}$. But $(x) = {rx lvert rin mathbb{R}[x]}$. Then you can see that $x^2 in (x)$,but $x^2notin langle x rangle$ as mentioned in the comment. In general, $langle x rangle subseteq (x) $.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        As you mentioned, $langle x rangle$ is the set ${x, 2x, 3x,...,nx,... : lvert nin mathbb{N}}$. But $(x) = {rx lvert rin mathbb{R}[x]}$. Then you can see that $x^2 in (x)$,but $x^2notin langle x rangle$ as mentioned in the comment. In general, $langle x rangle subseteq (x) $.






        share|cite|improve this answer












        As you mentioned, $langle x rangle$ is the set ${x, 2x, 3x,...,nx,... : lvert nin mathbb{N}}$. But $(x) = {rx lvert rin mathbb{R}[x]}$. Then you can see that $x^2 in (x)$,but $x^2notin langle x rangle$ as mentioned in the comment. In general, $langle x rangle subseteq (x) $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 2:17









        Thomas Shelby

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        1,172116






























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