Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic
Artin Algebra:
Definition of maximal ideals:
Maximal ideals of $F[x]$:
In connection with these, I think $x-2$ is monic irreducible in $mathbb R[x]$ (and other fields if it's absolutely irreducible), so we should have the quotient ring $mathbb R[x]/(x-2)$ to be a field. If we were to prove this from definition of a field, then we satisfy all properties besides multiplicative inverse and now must show that a non-zero element of $mathbb R[x]/(x-2)$ has a multiplicative inverse. Here is what I have tried:
An element of $mathbb R[x]/(x-2)$ has the form $[a+(x-2)], a in mathbb R[x]$, and the multiplicative identity of $mathbb R[x]/(x-2)$ is $[1+(x-2)]$ because $$[a+(x-2)][1+(x-2)] = [(a)(1)+(x-2)] = [a+(x-2)]$$
If $a$ is constant, then $[a+(x-2)]$'s inverse is $[frac 1 a+(x-2)]$.
And now I am stuck.
- For non-constant polynomials like $a=2x^2+1$, what's the multiplicative inverse of $$[2x^2+1+(x-2)]$$?
I think the adding relations become relevant like we introduce the relation $x-2=0$ or something. I sort of forgot, but I think we just replace $x=2$ so $overline{2x^2+1}=[2x^2+1+(x-2)]=[9+(x-2)]=overline{9}$. I think the inverse of $overline{2x^2+1}$ is $overline{frac19}$ then.
- So then the claim in Proposition 11.8.4a is that $F[x]/(p)$ is a field if and only if $p$ is monic irreducible in $F[x]$?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 10:35
Jack Bauer
1,2911631
add a comment |
Artin Algebra:
Definition of maximal ideals:
Maximal ideals of $F[x]$:
In connection with these, I think $x-2$ is monic irreducible in $mathbb R[x]$ (and other fields if it's absolutely irreducible), so we should have the quotient ring $mathbb R[x]/(x-2)$ to be a field. If we were to prove this from definition of a field, then we satisfy all properties besides multiplicative inverse and now must show that a non-zero element of $mathbb R[x]/(x-2)$ has a multiplicative inverse. Here is what I have tried:
An element of $mathbb R[x]/(x-2)$ has the form $[a+(x-2)], a in mathbb R[x]$, and the multiplicative identity of $mathbb R[x]/(x-2)$ is $[1+(x-2)]$ because $$[a+(x-2)][1+(x-2)] = [(a)(1)+(x-2)] = [a+(x-2)]$$
If $a$ is constant, then $[a+(x-2)]$'s inverse is $[frac 1 a+(x-2)]$.
And now I am stuck.
- For non-constant polynomials like $a=2x^2+1$, what's the multiplicative inverse of $$[2x^2+1+(x-2)]$$?
I think the adding relations become relevant like we introduce the relation $x-2=0$ or something. I sort of forgot, but I think we just replace $x=2$ so $overline{2x^2+1}=[2x^2+1+(x-2)]=[9+(x-2)]=overline{9}$. I think the inverse of $overline{2x^2+1}$ is $overline{frac19}$ then.
- So then the claim in Proposition 11.8.4a is that $F[x]/(p)$ is a field if and only if $p$ is monic irreducible in $F[x]$?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 10:35
Jack Bauer
1,2911631
1
No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
Artin Algebra:
Definition of maximal ideals:
Maximal ideals of $F[x]$:
In connection with these, I think $x-2$ is monic irreducible in $mathbb R[x]$ (and other fields if it's absolutely irreducible), so we should have the quotient ring $mathbb R[x]/(x-2)$ to be a field. If we were to prove this from definition of a field, then we satisfy all properties besides multiplicative inverse and now must show that a non-zero element of $mathbb R[x]/(x-2)$ has a multiplicative inverse. Here is what I have tried:
An element of $mathbb R[x]/(x-2)$ has the form $[a+(x-2)], a in mathbb R[x]$, and the multiplicative identity of $mathbb R[x]/(x-2)$ is $[1+(x-2)]$ because $$[a+(x-2)][1+(x-2)] = [(a)(1)+(x-2)] = [a+(x-2)]$$
If $a$ is constant, then $[a+(x-2)]$'s inverse is $[frac 1 a+(x-2)]$.
And now I am stuck.
- For non-constant polynomials like $a=2x^2+1$, what's the multiplicative inverse of $$[2x^2+1+(x-2)]$$?
I think the adding relations become relevant like we introduce the relation $x-2=0$ or something. I sort of forgot, but I think we just replace $x=2$ so $overline{2x^2+1}=[2x^2+1+(x-2)]=[9+(x-2)]=overline{9}$. I think the inverse of $overline{2x^2+1}$ is $overline{frac19}$ then.
- So then the claim in Proposition 11.8.4a is that $F[x]/(p)$ is a field if and only if $p$ is monic irreducible in $F[x]$?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
asked Nov 20 at 10:35
Jack Bauer
1,2911631
Artin Algebra:
Definition of maximal ideals:
Maximal ideals of $F[x]$:
In connection with these, I think $x-2$ is monic irreducible in $mathbb R[x]$ (and other fields if it's absolutely irreducible), so we should have the quotient ring $mathbb R[x]/(x-2)$ to be a field. If we were to prove this from definition of a field, then we satisfy all properties besides multiplicative inverse and now must show that a non-zero element of $mathbb R[x]/(x-2)$ has a multiplicative inverse. Here is what I have tried:
An element of $mathbb R[x]/(x-2)$ has the form $[a+(x-2)], a in mathbb R[x]$, and the multiplicative identity of $mathbb R[x]/(x-2)$ is $[1+(x-2)]$ because $$[a+(x-2)][1+(x-2)] = [(a)(1)+(x-2)] = [a+(x-2)]$$
If $a$ is constant, then $[a+(x-2)]$'s inverse is $[frac 1 a+(x-2)]$.
And now I am stuck.
- For non-constant polynomials like $a=2x^2+1$, what's the multiplicative inverse of $$[2x^2+1+(x-2)]$$?
I think the adding relations become relevant like we introduce the relation $x-2=0$ or something. I sort of forgot, but I think we just replace $x=2$ so $overline{2x^2+1}=[2x^2+1+(x-2)]=[9+(x-2)]=overline{9}$. I think the inverse of $overline{2x^2+1}$ is $overline{frac19}$ then.
- So then the claim in Proposition 11.8.4a is that $F[x]/(p)$ is a field if and only if $p$ is monic irreducible in $F[x]$?
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 10:35
Jack Bauer
1,2911631
asked Nov 20 at 10:35
Jack Bauer
1,2911631
asked Nov 20 at 10:35
Jack Bauer
1,2911631
asked Nov 20 at 10:35
Jack Bauer
1,2911631
asked Nov 20 at 10:35
Jack Bauer
1,2911631
1,2911631
1
No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
1
No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
1
1
No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
2 Answers
2
active
oldest
votes
If $p(x)inmathbb{R}[x]$ is such that $p(1)neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $frac1{p(1)}+[x-1]$. That's so because:
$p(x)+[x-1]=p(1)+[x-1]$;
$bigl(p(1)+[x-1]bigr)timesleft(frac1{p(1)}+[x-1]right)=bigl(1+[x-1]bigr)$.
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
add a comment |
You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $cinBbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $deg(r)<1$. This forces $rinBbb R$, in particular $r=f(2)$.
With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)in (x-2)$. The inverse is therefore $frac{1}{9}+(x-2)$.It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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votes
If $p(x)inmathbb{R}[x]$ is such that $p(1)neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $frac1{p(1)}+[x-1]$. That's so because:
$p(x)+[x-1]=p(1)+[x-1]$;
$bigl(p(1)+[x-1]bigr)timesleft(frac1{p(1)}+[x-1]right)=bigl(1+[x-1]bigr)$.
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
add a comment |
If $p(x)inmathbb{R}[x]$ is such that $p(1)neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $frac1{p(1)}+[x-1]$. That's so because:
$p(x)+[x-1]=p(1)+[x-1]$;
$bigl(p(1)+[x-1]bigr)timesleft(frac1{p(1)}+[x-1]right)=bigl(1+[x-1]bigr)$.
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
add a comment |
If $p(x)inmathbb{R}[x]$ is such that $p(1)neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $frac1{p(1)}+[x-1]$. That's so because:
$p(x)+[x-1]=p(1)+[x-1]$;
$bigl(p(1)+[x-1]bigr)timesleft(frac1{p(1)}+[x-1]right)=bigl(1+[x-1]bigr)$.
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
If $p(x)inmathbb{R}[x]$ is such that $p(1)neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $frac1{p(1)}+[x-1]$. That's so because:
$p(x)+[x-1]=p(1)+[x-1]$;
$bigl(p(1)+[x-1]bigr)timesleft(frac1{p(1)}+[x-1]right)=bigl(1+[x-1]bigr)$.
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
edited Nov 20 at 11:23
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
answered Nov 20 at 10:40
José Carlos Santos
149k22117219
149k22117219
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
add a comment |
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
$overline {frac{1}{p(1)}}$ is so compact! Thank you!
– Jack Bauer
Nov 20 at 11:07
add a comment |
You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $cinBbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $deg(r)<1$. This forces $rinBbb R$, in particular $r=f(2)$.
With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)in (x-2)$. The inverse is therefore $frac{1}{9}+(x-2)$.It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $cinBbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $deg(r)<1$. This forces $rinBbb R$, in particular $r=f(2)$.
With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)in (x-2)$. The inverse is therefore $frac{1}{9}+(x-2)$.It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $cinBbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $deg(r)<1$. This forces $rinBbb R$, in particular $r=f(2)$.
With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)in (x-2)$. The inverse is therefore $frac{1}{9}+(x-2)$.It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $cinBbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $deg(r)<1$. This forces $rinBbb R$, in particular $r=f(2)$.
With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)in (x-2)$. The inverse is therefore $frac{1}{9}+(x-2)$.It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
answered Nov 20 at 10:47
cansomeonehelpmeout
6,6433834
6,6433834
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10
add a comment |
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No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either.
– DonAntonio
Nov 20 at 10:41
@DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $mathbb Z$? Thank you!
– Jack Bauer
Nov 20 at 11:10