Mixing
Describe $Bbb{R}[x]/(x^2 + 1)$
$begingroup$
First by previous knowledge, I do know that $Bbb{R}[x]/(x^2 + 1) cong Bbb{C}$, so this might seem trivial. But I am not here to ask about that and I don't want to use this fact
I am reading Dummit and Foote on Chapter 13 and this example (pg 515) describes the field as "an extension of degree $2$ of $Bbb{R}$ in which $x^2 + 1$ has a root". I am not sure what this sentence means.
First there was a theorem that stated the basis elements of $K = Bbb{R}[x]/(p(x))$ as described by $theta equiv xpmod{p(x)}$ and preceding paragraphs describe multiplication in the quotient ring.
Okay here is the actual question, they described the elements as $a + btheta$ with $a,bin Bbb{R}[x]$. How do we arrive at that form? There is no talk about the isomorphism to $Bbb{C}$ until much later.
I do not not see how setting $x^2 + 1 equiv 0$ turns $a_0 + a_1x + a_2x^2 + dots a_nx^nto a_0 + a_1x$
.
polynomials ring-theory field-theory roots extension-field
edited Jan 9 at 11:04
Thomas Shelby
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
$endgroup$
add a comment |
$begingroup$
First by previous knowledge, I do know that $Bbb{R}[x]/(x^2 + 1) cong Bbb{C}$, so this might seem trivial. But I am not here to ask about that and I don't want to use this fact
I am reading Dummit and Foote on Chapter 13 and this example (pg 515) describes the field as "an extension of degree $2$ of $Bbb{R}$ in which $x^2 + 1$ has a root". I am not sure what this sentence means.
First there was a theorem that stated the basis elements of $K = Bbb{R}[x]/(p(x))$ as described by $theta equiv xpmod{p(x)}$ and preceding paragraphs describe multiplication in the quotient ring.
Okay here is the actual question, they described the elements as $a + btheta$ with $a,bin Bbb{R}[x]$. How do we arrive at that form? There is no talk about the isomorphism to $Bbb{C}$ until much later.
I do not not see how setting $x^2 + 1 equiv 0$ turns $a_0 + a_1x + a_2x^2 + dots a_nx^nto a_0 + a_1x$
.
polynomials ring-theory field-theory roots extension-field
edited Jan 9 at 11:04
Thomas Shelby
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
$endgroup$
$begingroup$
$x^2 +1 equiv 0$ means $x^2 equiv -1$.
$endgroup$
– Stockfish
Jan 9 at 10:04
$begingroup$
Divide the polynomial by $x^2+1$ and consider its remainder.
$endgroup$
– Seewoo Lee
Jan 9 at 10:08
$begingroup$
Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
$endgroup$
– drhab
Jan 9 at 10:08
$begingroup$
@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
$endgroup$
– Hawk
Jan 9 at 10:20
add a comment |
$begingroup$
First by previous knowledge, I do know that $Bbb{R}[x]/(x^2 + 1) cong Bbb{C}$, so this might seem trivial. But I am not here to ask about that and I don't want to use this fact
I am reading Dummit and Foote on Chapter 13 and this example (pg 515) describes the field as "an extension of degree $2$ of $Bbb{R}$ in which $x^2 + 1$ has a root". I am not sure what this sentence means.
First there was a theorem that stated the basis elements of $K = Bbb{R}[x]/(p(x))$ as described by $theta equiv xpmod{p(x)}$ and preceding paragraphs describe multiplication in the quotient ring.
Okay here is the actual question, they described the elements as $a + btheta$ with $a,bin Bbb{R}[x]$. How do we arrive at that form? There is no talk about the isomorphism to $Bbb{C}$ until much later.
I do not not see how setting $x^2 + 1 equiv 0$ turns $a_0 + a_1x + a_2x^2 + dots a_nx^nto a_0 + a_1x$
.
polynomials ring-theory field-theory roots extension-field
edited Jan 9 at 11:04
Thomas Shelby
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
$endgroup$
First by previous knowledge, I do know that $Bbb{R}[x]/(x^2 + 1) cong Bbb{C}$, so this might seem trivial. But I am not here to ask about that and I don't want to use this fact
I am reading Dummit and Foote on Chapter 13 and this example (pg 515) describes the field as "an extension of degree $2$ of $Bbb{R}$ in which $x^2 + 1$ has a root". I am not sure what this sentence means.
First there was a theorem that stated the basis elements of $K = Bbb{R}[x]/(p(x))$ as described by $theta equiv xpmod{p(x)}$ and preceding paragraphs describe multiplication in the quotient ring.
Okay here is the actual question, they described the elements as $a + btheta$ with $a,bin Bbb{R}[x]$. How do we arrive at that form? There is no talk about the isomorphism to $Bbb{C}$ until much later.
I do not not see how setting $x^2 + 1 equiv 0$ turns $a_0 + a_1x + a_2x^2 + dots a_nx^nto a_0 + a_1x$
.
polynomials ring-theory field-theory roots extension-field
polynomials ring-theory field-theory roots extension-field
edited Jan 9 at 11:04
Thomas Shelby
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
edited Jan 9 at 11:04
Thomas Shelby
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
edited Jan 9 at 11:04
Thomas Shelby
2,765421
edited Jan 9 at 11:04
Thomas Shelby
2,765421
edited Jan 9 at 11:04
Thomas Shelby
2,765421
2,765421
asked Jan 9 at 10:04
HawkHawk
5,4971139103
asked Jan 9 at 10:04
HawkHawk
5,4971139103
asked Jan 9 at 10:04
HawkHawk
5,4971139103
5,4971139103
$begingroup$
$x^2 +1 equiv 0$ means $x^2 equiv -1$.
$endgroup$
– Stockfish
Jan 9 at 10:04
$begingroup$
Divide the polynomial by $x^2+1$ and consider its remainder.
$endgroup$
– Seewoo Lee
Jan 9 at 10:08
$begingroup$
Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
$endgroup$
– drhab
Jan 9 at 10:08
$begingroup$
@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
$endgroup$
– Hawk
Jan 9 at 10:20
add a comment |
$begingroup$
$x^2 +1 equiv 0$ means $x^2 equiv -1$.
$endgroup$
– Stockfish
Jan 9 at 10:04
$begingroup$
Divide the polynomial by $x^2+1$ and consider its remainder.
$endgroup$
– Seewoo Lee
Jan 9 at 10:08
$begingroup$
Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
$endgroup$
– drhab
Jan 9 at 10:08
$begingroup$
@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
$endgroup$
– Hawk
Jan 9 at 10:20
$begingroup$
$x^2 +1 equiv 0$ means $x^2 equiv -1$.
$endgroup$
– Stockfish
Jan 9 at 10:04
$begingroup$
$x^2 +1 equiv 0$ means $x^2 equiv -1$.
$endgroup$
– Stockfish
Jan 9 at 10:04
$begingroup$
Divide the polynomial by $x^2+1$ and consider its remainder.
$endgroup$
– Seewoo Lee
Jan 9 at 10:08
$begingroup$
Divide the polynomial by $x^2+1$ and consider its remainder.
$endgroup$
– Seewoo Lee
Jan 9 at 10:08
$begingroup$
Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
$endgroup$
– drhab
Jan 9 at 10:08
$begingroup$
Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
$endgroup$
– drhab
Jan 9 at 10:08
$begingroup$
@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
$endgroup$
– Hawk
Jan 9 at 10:20
$begingroup$
@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
$endgroup$
– Hawk
Jan 9 at 10:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In $mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.
And if $p(x)inmathbb{R}[x]$ there are polynomials $q(x),r(x)inmathbb{R}[x]$ such that $p(x)=q(x)times(x^2+1)+r(x)$ and $deg r(x)leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$bigl[p(x)bigr]=bigl[r(x)bigr]=[a+bx]=a+b[x].$$
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
$begingroup$
Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
$endgroup$
– Thomas Shelby
Jan 9 at 11:14
add a comment |
$begingroup$
Well, if $x^2+1equiv 0$ then $x^2equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3equiv -x$, $x^4equiv -x^2equiv 1$, $x^5equiv -x^3equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.
(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
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$begingroup$
In $mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.
And if $p(x)inmathbb{R}[x]$ there are polynomials $q(x),r(x)inmathbb{R}[x]$ such that $p(x)=q(x)times(x^2+1)+r(x)$ and $deg r(x)leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$bigl[p(x)bigr]=bigl[r(x)bigr]=[a+bx]=a+b[x].$$
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
$begingroup$
Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
$endgroup$
– Thomas Shelby
Jan 9 at 11:14
add a comment |
$begingroup$
In $mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.
And if $p(x)inmathbb{R}[x]$ there are polynomials $q(x),r(x)inmathbb{R}[x]$ such that $p(x)=q(x)times(x^2+1)+r(x)$ and $deg r(x)leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$bigl[p(x)bigr]=bigl[r(x)bigr]=[a+bx]=a+b[x].$$
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
$begingroup$
Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
$endgroup$
– Thomas Shelby
Jan 9 at 11:14
add a comment |
$begingroup$
In $mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.
And if $p(x)inmathbb{R}[x]$ there are polynomials $q(x),r(x)inmathbb{R}[x]$ such that $p(x)=q(x)times(x^2+1)+r(x)$ and $deg r(x)leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$bigl[p(x)bigr]=bigl[r(x)bigr]=[a+bx]=a+b[x].$$
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
In $mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.
And if $p(x)inmathbb{R}[x]$ there are polynomials $q(x),r(x)inmathbb{R}[x]$ such that $p(x)=q(x)times(x^2+1)+r(x)$ and $deg r(x)leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$bigl[p(x)bigr]=bigl[r(x)bigr]=[a+bx]=a+b[x].$$
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 10:10
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
$begingroup$
Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
$endgroup$
– Thomas Shelby
Jan 9 at 11:14
add a comment |
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
$begingroup$
Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
$endgroup$
– Thomas Shelby
Jan 9 at 11:14
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
Setting $b = x + (x^2 + 1) = x + I$. It is a root because $b^2 + 1 = (x + I)^2 + 1 = (x^2 + I) + 1 = (x^2 + 1) + I = I equiv 0.$(I think we wrote the same thing). And I think the last statement is a reinstatement of corollary 5 in the book in some way.
$endgroup$
– Hawk
Jan 9 at 10:37
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
In this case $Bbb R[x]/ (1+x^2) simeq Bbb R[i]$ since $1+x^2$ is a least degree monic polynomial or irreducible polynomial over $Bbb R$ having $i$ as a root. Since degree of the minimal polynomial is $2,$ $Bbb R[i] = Bbb R + Bbb R.i = {a + bi : a,b in Bbb R } = Bbb C$.
$endgroup$
– Dbchatto67
Jan 9 at 10:48
$begingroup$
The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
$endgroup$
– Dbchatto67
Jan 9 at 10:58
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The general fact is that suppose you are in a commutative ring $R$ and suppose that $alpha notin R$. If $alpha$ satisfies a irreducible polynomial $f(x)$ over $R$ of degree $n$ having $alpha$ as a root then $R[x]/(f(x)) simeq R[alpha] = {r_0 + r_1 alpha + r_2 {alpha}^2 + cdots + r_{n-1} {alpha}^{n-1} : r_i in R text {for} i=0,1,2, cdots , n-1 } ( simeq R^n text {as a group w.r.t.} +)$.
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– Dbchatto67
Jan 9 at 10:58
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Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
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– Thomas Shelby
Jan 9 at 11:14
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Since $Bbb R$ is a field, $Bbb{R}[x] $ is a Euclidean Domain. This fact justifies the second paragraph.
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– Thomas Shelby
Jan 9 at 11:14
add a comment |
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Well, if $x^2+1equiv 0$ then $x^2equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3equiv -x$, $x^4equiv -x^2equiv 1$, $x^5equiv -x^3equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.
(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
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add a comment |
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Well, if $x^2+1equiv 0$ then $x^2equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3equiv -x$, $x^4equiv -x^2equiv 1$, $x^5equiv -x^3equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.
(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
$begingroup$
Well, if $x^2+1equiv 0$ then $x^2equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3equiv -x$, $x^4equiv -x^2equiv 1$, $x^5equiv -x^3equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.
(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
$endgroup$
Well, if $x^2+1equiv 0$ then $x^2equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3equiv -x$, $x^4equiv -x^2equiv 1$, $x^5equiv -x^3equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.
(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
answered Jan 15 at 4:31
Eric WofseyEric Wofsey
184k13212338
184k13212338
add a comment |
add a comment |
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$x^2 +1 equiv 0$ means $x^2 equiv -1$.
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– Stockfish
Jan 9 at 10:04
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Divide the polynomial by $x^2+1$ and consider its remainder.
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– Seewoo Lee
Jan 9 at 10:08
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Note that $x^{2k}=(-1)^k$ and $x^{2k+1}=(-1)^kx$. Btw, if LHS is $a_0+a_1x+cdots a_nx^n$ then $a_0+a_1x$ on RHS is not correct in general.
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– drhab
Jan 9 at 10:08
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@SeewooLee Stockfish, those are good points. They helped me see the collapsation.
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– Hawk
Jan 9 at 10:20