Concerning the evaluation map
$begingroup$
Let $r=r(x,y) in mathbb{C}[x,y]$ and define $e_{alpha,beta} : mathbb{C}[x,y] to mathbb{C}$ by $e_{alpha,beta}(r(x,y)):=r(alpha,beta)$.
If I am not wrong, $e_{alpha,beta}$, the evaluation map, is a ring homomorphism (it is known to be a ring homomorphism for one variable; I guess that it is still a ring homomorphism for more than one variable?).
The kernel of $e_{alpha,beta}$ is the maximal ideal $(x-alpha,y-beta)$.
Now let $p=p(x,y),q=q(x,y) in mathbb{C}[x,y]$, with $n=deg(f) geq 2$ and $m=deg(g) geq 2$.
Can we find a commutative ring $R$ and a ring homomorphism $e: mathbb{C}[x,y] to R$
such that $(p,q)$ is the kernel of $e$?
(The special case where $R$ is an integral domain and $e$ is an epimorphism, in other words, $(p,q)$ is a prime ideal, was dealt with here).
Thank you very much!
polynomials ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
$endgroup$
add a comment |
$begingroup$
Let $r=r(x,y) in mathbb{C}[x,y]$ and define $e_{alpha,beta} : mathbb{C}[x,y] to mathbb{C}$ by $e_{alpha,beta}(r(x,y)):=r(alpha,beta)$.
If I am not wrong, $e_{alpha,beta}$, the evaluation map, is a ring homomorphism (it is known to be a ring homomorphism for one variable; I guess that it is still a ring homomorphism for more than one variable?).
The kernel of $e_{alpha,beta}$ is the maximal ideal $(x-alpha,y-beta)$.
Now let $p=p(x,y),q=q(x,y) in mathbb{C}[x,y]$, with $n=deg(f) geq 2$ and $m=deg(g) geq 2$.
Can we find a commutative ring $R$ and a ring homomorphism $e: mathbb{C}[x,y] to R$
such that $(p,q)$ is the kernel of $e$?
(The special case where $R$ is an integral domain and $e$ is an epimorphism, in other words, $(p,q)$ is a prime ideal, was dealt with here).
Thank you very much!
polynomials ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
$endgroup$
$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17
add a comment |
$begingroup$
Let $r=r(x,y) in mathbb{C}[x,y]$ and define $e_{alpha,beta} : mathbb{C}[x,y] to mathbb{C}$ by $e_{alpha,beta}(r(x,y)):=r(alpha,beta)$.
If I am not wrong, $e_{alpha,beta}$, the evaluation map, is a ring homomorphism (it is known to be a ring homomorphism for one variable; I guess that it is still a ring homomorphism for more than one variable?).
The kernel of $e_{alpha,beta}$ is the maximal ideal $(x-alpha,y-beta)$.
Now let $p=p(x,y),q=q(x,y) in mathbb{C}[x,y]$, with $n=deg(f) geq 2$ and $m=deg(g) geq 2$.
Can we find a commutative ring $R$ and a ring homomorphism $e: mathbb{C}[x,y] to R$
such that $(p,q)$ is the kernel of $e$?
(The special case where $R$ is an integral domain and $e$ is an epimorphism, in other words, $(p,q)$ is a prime ideal, was dealt with here).
Thank you very much!
polynomials ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
$endgroup$
Let $r=r(x,y) in mathbb{C}[x,y]$ and define $e_{alpha,beta} : mathbb{C}[x,y] to mathbb{C}$ by $e_{alpha,beta}(r(x,y)):=r(alpha,beta)$.
If I am not wrong, $e_{alpha,beta}$, the evaluation map, is a ring homomorphism (it is known to be a ring homomorphism for one variable; I guess that it is still a ring homomorphism for more than one variable?).
The kernel of $e_{alpha,beta}$ is the maximal ideal $(x-alpha,y-beta)$.
Now let $p=p(x,y),q=q(x,y) in mathbb{C}[x,y]$, with $n=deg(f) geq 2$ and $m=deg(g) geq 2$.
Can we find a commutative ring $R$ and a ring homomorphism $e: mathbb{C}[x,y] to R$
such that $(p,q)$ is the kernel of $e$?
(The special case where $R$ is an integral domain and $e$ is an epimorphism, in other words, $(p,q)$ is a prime ideal, was dealt with here).
Thank you very much!
polynomials ring-theory commutative-algebra maximal-and-prime-ideals
polynomials ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
edited Dec 26 '18 at 4:19
user237522
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
asked Dec 26 '18 at 4:09
user237522user237522
2,1671617
2,1671617
$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17
add a comment |
$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17
$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $Ato A/I$ is a ring homomorphism whose kernel is $I$.
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $Ato A/I$ is a ring homomorphism whose kernel is $I$.
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
add a comment |
$begingroup$
Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $Ato A/I$ is a ring homomorphism whose kernel is $I$.
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
add a comment |
$begingroup$
Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $Ato A/I$ is a ring homomorphism whose kernel is $I$.
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
$endgroup$
Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $Ato A/I$ is a ring homomorphism whose kernel is $I$.
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
answered Dec 26 '18 at 4:18
Eric WofseyEric Wofsey
186k14214341
186k14214341
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
add a comment |
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
$begingroup$
Thank you. Your answer is ok if we require that $R$ is just an arbitrary commutative ring.
$endgroup$
– user237522
Dec 26 '18 at 4:29
add a comment |
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$begingroup$
Surely you are familiar with quotient rings?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 4:11
$begingroup$
Thanks. Please, could you elaborate a little?
$endgroup$
– user237522
Dec 26 '18 at 4:17