Inverses of elements for (Part One) an ideal generated by a reducible polynomial and (Part Two) an...
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Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)
In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic
Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.
Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?
In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".
In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)
Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.
For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.
To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".
What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?
( I think I will know the answer if I know the answer to 2 in Part One.)
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 11:42
Jack Bauer
1,2531631
|
show 5 more comments
up vote
0
down vote
favorite
Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)
In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic
Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.
Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?
In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".
In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)
Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.
For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.
To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".
What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?
( I think I will know the answer if I know the answer to 2 in Part One.)
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 11:42
Jack Bauer
1,2531631
1
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
1
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
1
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
1
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
1
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25
|
show 5 more comments
up vote
0
down vote
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up vote
0
down vote
favorite
Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)
In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic
Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.
Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?
In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".
In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)
Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.
For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.
To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".
What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?
( I think I will know the answer if I know the answer to 2 in Part One.)
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 11:42
Jack Bauer
1,2531631
Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)
In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic
Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.
Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?
In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".
In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)
Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.
For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.
To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.
What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".
What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?
( I think I will know the answer if I know the answer to 2 in Part One.)
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 11:42
Jack Bauer
1,2531631
asked Nov 20 at 11:42
Jack Bauer
1,2531631
edited Nov 20 at 13:47
asked Nov 20 at 11:42
Jack Bauer
1,2531631
asked Nov 20 at 11:42
Jack Bauer
1,2531631
asked Nov 20 at 11:42
Jack Bauer
1,2531631
1,2531631
1
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
1
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
1
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
1
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
1
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25
|
show 5 more comments
1
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
1
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
1
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
1
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
1
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25
1
1
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
1
1
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
1
1
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
1
1
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
1
1
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25
|
show 5 more comments
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1
Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46
1
$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58
1
$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54
1
You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21
1
For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25