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?
abstract-algebra ring-theory
asked Nov 20 at 0:54
Tomás Palamás
351211
add a comment |
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?
abstract-algebra ring-theory
asked Nov 20 at 0:54
Tomás Palamás
351211
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
add a comment |
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?
abstract-algebra ring-theory
asked Nov 20 at 0:54
Tomás Palamás
351211
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
abstract-algebra ring-theory
asked Nov 20 at 0:54
Tomás Palamás
351211
asked Nov 20 at 0:54
Tomás Palamás
351211
edited Nov 20 at 1:10
asked Nov 20 at 0:54
Tomás Palamás
351211
asked Nov 20 at 0:54
Tomás Palamás
351211
asked Nov 20 at 0:54
Tomás Palamás
351211
351211
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
add a comment |
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
add a comment |
1 Answer
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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) $.
answered Nov 20 at 2:17
Thomas Shelby
12510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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) $.
answered Nov 20 at 2:17
Thomas Shelby
12510
add a comment |
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) $.
answered Nov 20 at 2:17
Thomas Shelby
12510
add a comment |
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) $.
answered Nov 20 at 2:17
Thomas Shelby
12510
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) $.
answered Nov 20 at 2:17
Thomas Shelby
12510
answered Nov 20 at 2:17
Thomas Shelby
12510
answered Nov 20 at 2:17
Thomas Shelby
12510
answered Nov 20 at 2:17
Thomas Shelby
12510
12510
add a comment |
add a comment |
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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