Checking Ideals of Rings
Consider $A = {f(x) in mathbb{R}(x):f'(0)=0}$. Is $A$ an ideal in $mathbb{R}(x)$?
My answer: No it is not an ideal. Consider $g(x)in mathbb{R}(x)$ s.t. $g(x) = x$. And $f(x)in A$ s.t. $f(x)= 1$. $g(x)f(x) = x$ and the $frac{d}{dx} x = 1 neq 0$. So it is not the case that for any $r in mathbb{R}(x)$ and any $x in A$, $xr in A$. So A is not an ideal.
Did I interpret the definition of an ideal correctly?
abstract-algebra proof-verification ring-theory ideals
asked Nov 25 at 2:31
zodross
1546
add a comment |
Consider $A = {f(x) in mathbb{R}(x):f'(0)=0}$. Is $A$ an ideal in $mathbb{R}(x)$?
My answer: No it is not an ideal. Consider $g(x)in mathbb{R}(x)$ s.t. $g(x) = x$. And $f(x)in A$ s.t. $f(x)= 1$. $g(x)f(x) = x$ and the $frac{d}{dx} x = 1 neq 0$. So it is not the case that for any $r in mathbb{R}(x)$ and any $x in A$, $xr in A$. So A is not an ideal.
Did I interpret the definition of an ideal correctly?
abstract-algebra proof-verification ring-theory ideals
asked Nov 25 at 2:31
zodross
1546
add a comment |
Consider $A = {f(x) in mathbb{R}(x):f'(0)=0}$. Is $A$ an ideal in $mathbb{R}(x)$?
My answer: No it is not an ideal. Consider $g(x)in mathbb{R}(x)$ s.t. $g(x) = x$. And $f(x)in A$ s.t. $f(x)= 1$. $g(x)f(x) = x$ and the $frac{d}{dx} x = 1 neq 0$. So it is not the case that for any $r in mathbb{R}(x)$ and any $x in A$, $xr in A$. So A is not an ideal.
Did I interpret the definition of an ideal correctly?
abstract-algebra proof-verification ring-theory ideals
asked Nov 25 at 2:31
zodross
1546
Consider $A = {f(x) in mathbb{R}(x):f'(0)=0}$. Is $A$ an ideal in $mathbb{R}(x)$?
My answer: No it is not an ideal. Consider $g(x)in mathbb{R}(x)$ s.t. $g(x) = x$. And $f(x)in A$ s.t. $f(x)= 1$. $g(x)f(x) = x$ and the $frac{d}{dx} x = 1 neq 0$. So it is not the case that for any $r in mathbb{R}(x)$ and any $x in A$, $xr in A$. So A is not an ideal.
Did I interpret the definition of an ideal correctly?
abstract-algebra proof-verification ring-theory ideals
abstract-algebra proof-verification ring-theory ideals
asked Nov 25 at 2:31
zodross
1546
asked Nov 25 at 2:31
zodross
1546
asked Nov 25 at 2:31
zodross
1546
asked Nov 25 at 2:31
zodross
1546
asked Nov 25 at 2:31
zodross
1546
1546
add a comment |
add a comment |
1 Answer
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I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.
answered Nov 25 at 2:37
mathnoob
1,794422
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
add a comment |
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1 Answer
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1 Answer
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I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.
answered Nov 25 at 2:37
mathnoob
1,794422
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
add a comment |
I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.
answered Nov 25 at 2:37
mathnoob
1,794422
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
add a comment |
I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.
answered Nov 25 at 2:37
mathnoob
1,794422
I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.
answered Nov 25 at 2:37
mathnoob
1,794422
answered Nov 25 at 2:37
mathnoob
1,794422
answered Nov 25 at 2:37
mathnoob
1,794422
answered Nov 25 at 2:37
mathnoob
1,794422
1,794422
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
add a comment |
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
Thank you very much sir or madam.
– zodross
Nov 25 at 2:39
add a comment |
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