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To adjoin an element $alpha$ to a ring $R$, why does this involve quotienting $R[x]$ by a polynomial $f(x)$ satisfying $f(alpha)=0$?
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I'm reading Artin's section on the Adjunction of Elements, and I don't understand what he means by calling polynomial rings the "universal solution" to the "problem of adjoining a new element." He writes:
If $alpha$ is an element of any ring extension $R'$ of $R$, then there is a unique map $R[x]to R'$ which is the identity on $R$ and which carries $x$ to $alpha$. The image of this map will be the subring $R[alpha]$.
Could someone clarify the necessity of the polynomial ring? What is its relevance?
My vague understanding of this is as follows: suppose we want to adjoin $sqrt2$ to the ring $mathbbQ$. Since $sqrt2$ does not formally exist yet, but we know it satisfies $x^2-2=0$ in $mathbbQ[x]$, we can adjoin it by working with $mathbbQ[x]$ instead.
Edit: $R$ is commutative
abstract-algebra ring-theory
asked Jul 30 at 3:17
FakeAnalyst56
175
add a comment |Â
up vote
1
down vote
favorite
I'm reading Artin's section on the Adjunction of Elements, and I don't understand what he means by calling polynomial rings the "universal solution" to the "problem of adjoining a new element." He writes:
If $alpha$ is an element of any ring extension $R'$ of $R$, then there is a unique map $R[x]to R'$ which is the identity on $R$ and which carries $x$ to $alpha$. The image of this map will be the subring $R[alpha]$.
Could someone clarify the necessity of the polynomial ring? What is its relevance?
My vague understanding of this is as follows: suppose we want to adjoin $sqrt2$ to the ring $mathbbQ$. Since $sqrt2$ does not formally exist yet, but we know it satisfies $x^2-2=0$ in $mathbbQ[x]$, we can adjoin it by working with $mathbbQ[x]$ instead.
Edit: $R$ is commutative
abstract-algebra ring-theory
asked Jul 30 at 3:17
FakeAnalyst56
175
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading Artin's section on the Adjunction of Elements, and I don't understand what he means by calling polynomial rings the "universal solution" to the "problem of adjoining a new element." He writes:
If $alpha$ is an element of any ring extension $R'$ of $R$, then there is a unique map $R[x]to R'$ which is the identity on $R$ and which carries $x$ to $alpha$. The image of this map will be the subring $R[alpha]$.
Could someone clarify the necessity of the polynomial ring? What is its relevance?
My vague understanding of this is as follows: suppose we want to adjoin $sqrt2$ to the ring $mathbbQ$. Since $sqrt2$ does not formally exist yet, but we know it satisfies $x^2-2=0$ in $mathbbQ[x]$, we can adjoin it by working with $mathbbQ[x]$ instead.
Edit: $R$ is commutative
abstract-algebra ring-theory
asked Jul 30 at 3:17
FakeAnalyst56
175
I'm reading Artin's section on the Adjunction of Elements, and I don't understand what he means by calling polynomial rings the "universal solution" to the "problem of adjoining a new element." He writes:
If $alpha$ is an element of any ring extension $R'$ of $R$, then there is a unique map $R[x]to R'$ which is the identity on $R$ and which carries $x$ to $alpha$. The image of this map will be the subring $R[alpha]$.
Could someone clarify the necessity of the polynomial ring? What is its relevance?
My vague understanding of this is as follows: suppose we want to adjoin $sqrt2$ to the ring $mathbbQ$. Since $sqrt2$ does not formally exist yet, but we know it satisfies $x^2-2=0$ in $mathbbQ[x]$, we can adjoin it by working with $mathbbQ[x]$ instead.
Edit: $R$ is commutative
abstract-algebra ring-theory
asked Jul 30 at 3:17
FakeAnalyst56
175
edited Jul 30 at 6:10
asked Jul 30 at 3:17
FakeAnalyst56
175
asked Jul 30 at 3:17
FakeAnalyst56
175
asked Jul 30 at 3:17
FakeAnalyst56
175
175
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The extension $S$ of $R$ by an element $a$ is the smallest right, that contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Therefore, $S=R[a]$.
For the "evaluation" map from $operatornameev_a:R[x]to R[a]$ it is routine to check that sends $a_nx^n+a_n-1x^n-1...+a_0$ to $a_na^n+a_n-1a^n-1...+a_0$ is a ring homomorphism.
Since $S$ is the smallest extension of $R$ containing $a$, if $R'$ is another extension of $R$ containing $a$, then $Ssubset R'$. Therefore, you have the same map $R[x]to Shookrightarrow R'$ obtained by composing $operatornameev_a$ with the inclusion $S=R[a]hookrightarrow R'$.
In some cases, when $a$ satisfies some algebraic relation with coefficients in $R$, it will happen that $operatornameev_a$ is not injective, and sends some non-zero polynomials to the zero element of $R$. The set of all elements sent to zero by $operatornameev_a$ is an ideal of $R[x]$. If that ideal happens to be principal, then it is genrated by one polynomial $f(x)in R[x]$.
In your example of $a=sqrt2$ and $R=mathbbQ$, it results $f(x)=x^2-2$.
Look at the example $x=pi$ with the same $R$ for a case in which there is no such $f$.
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
add a comment |Â
up vote
2
down vote
As you have access to Artin's book let me say just the right amount to understand what is contained in that book.
The only operations on a ring are addition and multiplication: we are looking at enlarged ring from $R$ by adjoining an extra element from a larger ring $R'$. So the enlargement should minimally contain all powers of the introduced new element, multiplied by old elements and then the sums of these.
This is how the polynomial ring arises (express in symbols what my previous sentence states. That these alone together keep it a ring is also clear).
In case the new element introduced satisfies some condition every polynomial will not lead to a new element.
That is if $f(alpha) = 0$, then $g(alpha) f(alpha)$ will be zero etc.
This shows that the polynomial ring has to be quotiented by that ideal.
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The extension $S$ of $R$ by an element $a$ is the smallest right, that contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Therefore, $S=R[a]$.
For the "evaluation" map from $operatornameev_a:R[x]to R[a]$ it is routine to check that sends $a_nx^n+a_n-1x^n-1...+a_0$ to $a_na^n+a_n-1a^n-1...+a_0$ is a ring homomorphism.
Since $S$ is the smallest extension of $R$ containing $a$, if $R'$ is another extension of $R$ containing $a$, then $Ssubset R'$. Therefore, you have the same map $R[x]to Shookrightarrow R'$ obtained by composing $operatornameev_a$ with the inclusion $S=R[a]hookrightarrow R'$.
In some cases, when $a$ satisfies some algebraic relation with coefficients in $R$, it will happen that $operatornameev_a$ is not injective, and sends some non-zero polynomials to the zero element of $R$. The set of all elements sent to zero by $operatornameev_a$ is an ideal of $R[x]$. If that ideal happens to be principal, then it is genrated by one polynomial $f(x)in R[x]$.
In your example of $a=sqrt2$ and $R=mathbbQ$, it results $f(x)=x^2-2$.
Look at the example $x=pi$ with the same $R$ for a case in which there is no such $f$.
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
add a comment |Â
up vote
1
down vote
accepted
The extension $S$ of $R$ by an element $a$ is the smallest right, that contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Therefore, $S=R[a]$.
For the "evaluation" map from $operatornameev_a:R[x]to R[a]$ it is routine to check that sends $a_nx^n+a_n-1x^n-1...+a_0$ to $a_na^n+a_n-1a^n-1...+a_0$ is a ring homomorphism.
Since $S$ is the smallest extension of $R$ containing $a$, if $R'$ is another extension of $R$ containing $a$, then $Ssubset R'$. Therefore, you have the same map $R[x]to Shookrightarrow R'$ obtained by composing $operatornameev_a$ with the inclusion $S=R[a]hookrightarrow R'$.
In some cases, when $a$ satisfies some algebraic relation with coefficients in $R$, it will happen that $operatornameev_a$ is not injective, and sends some non-zero polynomials to the zero element of $R$. The set of all elements sent to zero by $operatornameev_a$ is an ideal of $R[x]$. If that ideal happens to be principal, then it is genrated by one polynomial $f(x)in R[x]$.
In your example of $a=sqrt2$ and $R=mathbbQ$, it results $f(x)=x^2-2$.
Look at the example $x=pi$ with the same $R$ for a case in which there is no such $f$.
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The extension $S$ of $R$ by an element $a$ is the smallest right, that contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Therefore, $S=R[a]$.
For the "evaluation" map from $operatornameev_a:R[x]to R[a]$ it is routine to check that sends $a_nx^n+a_n-1x^n-1...+a_0$ to $a_na^n+a_n-1a^n-1...+a_0$ is a ring homomorphism.
Since $S$ is the smallest extension of $R$ containing $a$, if $R'$ is another extension of $R$ containing $a$, then $Ssubset R'$. Therefore, you have the same map $R[x]to Shookrightarrow R'$ obtained by composing $operatornameev_a$ with the inclusion $S=R[a]hookrightarrow R'$.
In some cases, when $a$ satisfies some algebraic relation with coefficients in $R$, it will happen that $operatornameev_a$ is not injective, and sends some non-zero polynomials to the zero element of $R$. The set of all elements sent to zero by $operatornameev_a$ is an ideal of $R[x]$. If that ideal happens to be principal, then it is genrated by one polynomial $f(x)in R[x]$.
In your example of $a=sqrt2$ and $R=mathbbQ$, it results $f(x)=x^2-2$.
Look at the example $x=pi$ with the same $R$ for a case in which there is no such $f$.
The extension $S$ of $R$ by an element $a$ is the smallest right, that contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Therefore, $S=R[a]$.
For the "evaluation" map from $operatornameev_a:R[x]to R[a]$ it is routine to check that sends $a_nx^n+a_n-1x^n-1...+a_0$ to $a_na^n+a_n-1a^n-1...+a_0$ is a ring homomorphism.
Since $S$ is the smallest extension of $R$ containing $a$, if $R'$ is another extension of $R$ containing $a$, then $Ssubset R'$. Therefore, you have the same map $R[x]to Shookrightarrow R'$ obtained by composing $operatornameev_a$ with the inclusion $S=R[a]hookrightarrow R'$.
In some cases, when $a$ satisfies some algebraic relation with coefficients in $R$, it will happen that $operatornameev_a$ is not injective, and sends some non-zero polynomials to the zero element of $R$. The set of all elements sent to zero by $operatornameev_a$ is an ideal of $R[x]$. If that ideal happens to be principal, then it is genrated by one polynomial $f(x)in R[x]$.
In your example of $a=sqrt2$ and $R=mathbbQ$, it results $f(x)=x^2-2$.
Look at the example $x=pi$ with the same $R$ for a case in which there is no such $f$.
edited Jul 30 at 3:50
answered Jul 30 at 3:44
user580373
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
add a comment |Â
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Regarding your example about $pi$, I find this very interesting: is the takeaway that given a ring $R$, we can only adjoin elements that are roots of polynomials in $R[x]$?
– FakeAnalyst56
Jul 30 at 4:03
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
Careful! $ev_a$ is a ring homomorphism only if $a$ commutes with all the elements of $R$. True in the OP's context commutativity was probably included in the assumptions but it wasn't listed!
– Jyrki Lahtonen
Jul 30 at 5:20
1
1
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
@FakeAnalyst56 You can adjoin any sort of element. When you add to a ring like $mathbbQ$, and element like $pi$ that doesn't satisfy any polynomial relation with coefficients in $mathbbQ$, then you get a ring $mathbbQ[pi]$ that is isomorphic to $mathbbQ[x]$.
– user580373
Jul 30 at 11:22
add a comment |Â
up vote
2
down vote
As you have access to Artin's book let me say just the right amount to understand what is contained in that book.
The only operations on a ring are addition and multiplication: we are looking at enlarged ring from $R$ by adjoining an extra element from a larger ring $R'$. So the enlargement should minimally contain all powers of the introduced new element, multiplied by old elements and then the sums of these.
This is how the polynomial ring arises (express in symbols what my previous sentence states. That these alone together keep it a ring is also clear).
In case the new element introduced satisfies some condition every polynomial will not lead to a new element.
That is if $f(alpha) = 0$, then $g(alpha) f(alpha)$ will be zero etc.
This shows that the polynomial ring has to be quotiented by that ideal.
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
add a comment |Â
up vote
2
down vote
As you have access to Artin's book let me say just the right amount to understand what is contained in that book.
The only operations on a ring are addition and multiplication: we are looking at enlarged ring from $R$ by adjoining an extra element from a larger ring $R'$. So the enlargement should minimally contain all powers of the introduced new element, multiplied by old elements and then the sums of these.
This is how the polynomial ring arises (express in symbols what my previous sentence states. That these alone together keep it a ring is also clear).
In case the new element introduced satisfies some condition every polynomial will not lead to a new element.
That is if $f(alpha) = 0$, then $g(alpha) f(alpha)$ will be zero etc.
This shows that the polynomial ring has to be quotiented by that ideal.
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As you have access to Artin's book let me say just the right amount to understand what is contained in that book.
The only operations on a ring are addition and multiplication: we are looking at enlarged ring from $R$ by adjoining an extra element from a larger ring $R'$. So the enlargement should minimally contain all powers of the introduced new element, multiplied by old elements and then the sums of these.
This is how the polynomial ring arises (express in symbols what my previous sentence states. That these alone together keep it a ring is also clear).
In case the new element introduced satisfies some condition every polynomial will not lead to a new element.
That is if $f(alpha) = 0$, then $g(alpha) f(alpha)$ will be zero etc.
This shows that the polynomial ring has to be quotiented by that ideal.
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
As you have access to Artin's book let me say just the right amount to understand what is contained in that book.
The only operations on a ring are addition and multiplication: we are looking at enlarged ring from $R$ by adjoining an extra element from a larger ring $R'$. So the enlargement should minimally contain all powers of the introduced new element, multiplied by old elements and then the sums of these.
This is how the polynomial ring arises (express in symbols what my previous sentence states. That these alone together keep it a ring is also clear).
In case the new element introduced satisfies some condition every polynomial will not lead to a new element.
That is if $f(alpha) = 0$, then $g(alpha) f(alpha)$ will be zero etc.
This shows that the polynomial ring has to be quotiented by that ideal.
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
answered Jul 30 at 3:52
P Vanchinathan
13.9k12035
13.9k12035
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
add a comment |Â
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
Would you mind clarifying your paragraph "In case the new element [...] will be zero etc." I'm not sure why your implication shows that every polynomial will not lead to a new element, or how your final sentence follows. What precisely are you trying to show in that paragraph?
– FakeAnalyst56
Jul 30 at 22:15
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
If you adjoin square root of 2 then the polynomials $x^4+1$ and $3x^2-1$ will both lead to same element namely 5.
– P Vanchinathan
Jul 31 at 1:49
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
Two polynomials $f(x)$ and $g(x)$ will evaluate to same value at $x=sqrt2$ if the polynomial $f(x)-g(x)$ is of the form $(x^2-2) h(x)$ for some other polynomial $h(x)$. That is if $f(x)$ and $g(x)$ belong to the same coset for the ideal consisting of $h(x)(x^2-2)$ with $h(x)$ an arbitrary polynomial.
– P Vanchinathan
Jul 31 at 2:31
add a comment |Â
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