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Show that following quotient ring is field.
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I need to show that following ring is field.$$R[x]/(x^2+1)$$
I know $(x^2+1)$ is an ideal the unity of the quotient ring is $1+(x^2+1)$.
Hence I need to show that for any $f(x)in R[x]$ I can find $g(x)in R[x]$ such that $$f(x)+(x^2+1)g(x)+(x^2+1) =1+(x^2+1)$$
Expanding on left gives us $$f(x)g(x)+(x^2+1)$$
Hence I need to show that there is a $g(x)$ such that $f(x)g(x)-1in (x^2+1)$.
abstract-algebra polynomials ring-theory ideals
edited 5 mins ago
José Carlos Santos
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
add a comment |Â
up vote
1
down vote
favorite
I need to show that following ring is field.$$R[x]/(x^2+1)$$
I know $(x^2+1)$ is an ideal the unity of the quotient ring is $1+(x^2+1)$.
Hence I need to show that for any $f(x)in R[x]$ I can find $g(x)in R[x]$ such that $$f(x)+(x^2+1)g(x)+(x^2+1) =1+(x^2+1)$$
Expanding on left gives us $$f(x)g(x)+(x^2+1)$$
Hence I need to show that there is a $g(x)$ such that $f(x)g(x)-1in (x^2+1)$.
abstract-algebra polynomials ring-theory ideals
edited 5 mins ago
José Carlos Santos
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
4
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
3
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
1
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to show that following ring is field.$$R[x]/(x^2+1)$$
I know $(x^2+1)$ is an ideal the unity of the quotient ring is $1+(x^2+1)$.
Hence I need to show that for any $f(x)in R[x]$ I can find $g(x)in R[x]$ such that $$f(x)+(x^2+1)g(x)+(x^2+1) =1+(x^2+1)$$
Expanding on left gives us $$f(x)g(x)+(x^2+1)$$
Hence I need to show that there is a $g(x)$ such that $f(x)g(x)-1in (x^2+1)$.
abstract-algebra polynomials ring-theory ideals
edited 5 mins ago
José Carlos Santos
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
I need to show that following ring is field.$$R[x]/(x^2+1)$$
I know $(x^2+1)$ is an ideal the unity of the quotient ring is $1+(x^2+1)$.
Hence I need to show that for any $f(x)in R[x]$ I can find $g(x)in R[x]$ such that $$f(x)+(x^2+1)g(x)+(x^2+1) =1+(x^2+1)$$
Expanding on left gives us $$f(x)g(x)+(x^2+1)$$
Hence I need to show that there is a $g(x)$ such that $f(x)g(x)-1in (x^2+1)$.
abstract-algebra polynomials ring-theory ideals
abstract-algebra polynomials ring-theory ideals
edited 5 mins ago
José Carlos Santos
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
edited 5 mins ago
José Carlos Santos
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
edited 5 mins ago
José Carlos Santos
133k17107196
edited 5 mins ago
José Carlos Santos
133k17107196
edited 5 mins ago
José Carlos Santos
133k17107196
133k17107196
asked 41 mins ago
Piyush Divyanakar
3,331222
asked 41 mins ago
Piyush Divyanakar
3,331222
asked 41 mins ago
Piyush Divyanakar
3,331222
3,331222
4
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
3
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
1
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago
add a comment |Â
4
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
3
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
1
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago
4
4
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
3
3
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
1
1
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Since $x^2+1nmid f(x)$ and $1+x^2$ is irreducible (in $mathbbR[x]$), $x^2+1$ and $f(x)$ are coprime in $mathbbR[x]$ and therefore thre are $g(x),h(x)inmathbbR[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)in1+(x^2+1)$.
answered 31 mins ago
José Carlos Santos
133k17107196
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
add a comment |Â
up vote
1
down vote
Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?
answered 28 mins ago
Teddan the Terran
1,178210
add a comment |Â
up vote
1
down vote
A different view would be to consider the ring substitution homomorphism $phi:Bbb R[x]rightarrow Bbb C$ with $phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since $Bbb C = Bbb R[i] = Bbb R(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, $Bbb R[x]/ker phi$ is isomorphic to $Bbb C$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over $Bbb R$. Qed.
answered 13 mins ago
Wuestenfux
1,936139
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $x^2+1nmid f(x)$ and $1+x^2$ is irreducible (in $mathbbR[x]$), $x^2+1$ and $f(x)$ are coprime in $mathbbR[x]$ and therefore thre are $g(x),h(x)inmathbbR[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)in1+(x^2+1)$.
answered 31 mins ago
José Carlos Santos
133k17107196
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
add a comment |Â
up vote
1
down vote
accepted
Since $x^2+1nmid f(x)$ and $1+x^2$ is irreducible (in $mathbbR[x]$), $x^2+1$ and $f(x)$ are coprime in $mathbbR[x]$ and therefore thre are $g(x),h(x)inmathbbR[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)in1+(x^2+1)$.
answered 31 mins ago
José Carlos Santos
133k17107196
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $x^2+1nmid f(x)$ and $1+x^2$ is irreducible (in $mathbbR[x]$), $x^2+1$ and $f(x)$ are coprime in $mathbbR[x]$ and therefore thre are $g(x),h(x)inmathbbR[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)in1+(x^2+1)$.
answered 31 mins ago
José Carlos Santos
133k17107196
Since $x^2+1nmid f(x)$ and $1+x^2$ is irreducible (in $mathbbR[x]$), $x^2+1$ and $f(x)$ are coprime in $mathbbR[x]$ and therefore thre are $g(x),h(x)inmathbbR[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)in1+(x^2+1)$.
answered 31 mins ago
José Carlos Santos
133k17107196
answered 31 mins ago
José Carlos Santos
133k17107196
answered 31 mins ago
José Carlos Santos
133k17107196
answered 31 mins ago
José Carlos Santos
133k17107196
133k17107196
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
add a comment |Â
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Is this bezout's identity for polynomial rings/
– Piyush Divyanakar
28 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
Yes, that is what I am using here.
– José Carlos Santos
25 mins ago
add a comment |Â
up vote
1
down vote
Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?
answered 28 mins ago
Teddan the Terran
1,178210
add a comment |Â
up vote
1
down vote
Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?
answered 28 mins ago
Teddan the Terran
1,178210
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?
answered 28 mins ago
Teddan the Terran
1,178210
Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?
answered 28 mins ago
Teddan the Terran
1,178210
answered 28 mins ago
Teddan the Terran
1,178210
answered 28 mins ago
Teddan the Terran
1,178210
answered 28 mins ago
Teddan the Terran
1,178210
1,178210
add a comment |Â
add a comment |Â
up vote
1
down vote
A different view would be to consider the ring substitution homomorphism $phi:Bbb R[x]rightarrow Bbb C$ with $phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since $Bbb C = Bbb R[i] = Bbb R(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, $Bbb R[x]/ker phi$ is isomorphic to $Bbb C$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over $Bbb R$. Qed.
answered 13 mins ago
Wuestenfux
1,936139
add a comment |Â
up vote
1
down vote
A different view would be to consider the ring substitution homomorphism $phi:Bbb R[x]rightarrow Bbb C$ with $phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since $Bbb C = Bbb R[i] = Bbb R(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, $Bbb R[x]/ker phi$ is isomorphic to $Bbb C$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over $Bbb R$. Qed.
answered 13 mins ago
Wuestenfux
1,936139
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A different view would be to consider the ring substitution homomorphism $phi:Bbb R[x]rightarrow Bbb C$ with $phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since $Bbb C = Bbb R[i] = Bbb R(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, $Bbb R[x]/ker phi$ is isomorphic to $Bbb C$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over $Bbb R$. Qed.
answered 13 mins ago
Wuestenfux
1,936139
A different view would be to consider the ring substitution homomorphism $phi:Bbb R[x]rightarrow Bbb C$ with $phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since $Bbb C = Bbb R[i] = Bbb R(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, $Bbb R[x]/ker phi$ is isomorphic to $Bbb C$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over $Bbb R$. Qed.
answered 13 mins ago
Wuestenfux
1,936139
answered 13 mins ago
Wuestenfux
1,936139
answered 13 mins ago
Wuestenfux
1,936139
answered 13 mins ago
Wuestenfux
1,936139
1,936139
add a comment |Â
add a comment |Â
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4
Hint: I am pretty certain you have heard about (and worked in) this field before, but usually we don't use $x$, but another letter.
– Arthur
37 mins ago
3
Alternativly show this is a mximal Ideal
– Sheve
33 mins ago
1
Or consider a specific mapping $fcolon mathbbR[x]rightarrowmathbbC$
– Hypertrooper
32 mins ago