Prove that $p(X,Y)mapsto p(i,-i)$'s kernel is the ideal $(X+Y,X^2+1)$ The Next CEO of Stack OverflowZeros in the polynomial ring $mathbbR [X,Y]$Understanding the kernel of the evaluation mapkernel maximal idealevery ideal of $F[x]$ is prime ideal but not maximalProve that the preimage of a prime ideal is also prime.In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?Zeros in the polynomial ring $mathbbR [X,Y]$Intuitive reasons of ring modulo maximal ideal or prime idealIs the Kernel a Maximal Ideal?Ring Homomorphisms, ideals and kernelProving that an inverse ring homomorphism of an ideal is an ideal?
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Prove that $p(X,Y)mapsto p(i,-i)$'s kernel is the ideal $(X+Y,X^2+1)$
The Next CEO of Stack OverflowZeros in the polynomial ring $mathbbR [X,Y]$Understanding the kernel of the evaluation mapkernel maximal idealevery ideal of $F[x]$ is prime ideal but not maximalProve that the preimage of a prime ideal is also prime.In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?Zeros in the polynomial ring $mathbbR [X,Y]$Intuitive reasons of ring modulo maximal ideal or prime idealIs the Kernel a Maximal Ideal?Ring Homomorphisms, ideals and kernelProving that an inverse ring homomorphism of an ideal is an ideal?
$begingroup$
I want to show that $I=(X+Y,X^2+1)in R=mathbbR[X,Y]$ is a prime ideal. Since $I$ a prime ideal is equivalent to $R/I$ an integral domain, I want to create a homomorphism $phi: R to mathbbC$ given by $phi: p(X,Y)mapsto p(i,-i)$. This is a surjective homomorphism, so by the first isomorphism theorem we have that $R/ker(phi)$ isomorphic to $mathbbC$ and since $mathbbC$ is an integral domain, $R/ker(phi)$ is an integral domain, so $ker(phi)$ is a prime ideal.
So what's left to show is that $ker(phi)=I$.
Since $phi(X+Y)=i+(-i)=0$, and $phi(X^2+1)=i^2+1=-1+1=0$, we have that $Isubsetker(phi)$. I now want to show that no other elements but elements of $I$ are in the kernel, but I don't understand the following proof:
Suppose $p(X,Y)inker(phi)$, in other words, $p(i,-i)=0$. This implies that the polynomial $P(X,-X)$ evaluated at $X=i$ is zero and so $p(X,-X)$ is divisible by $X^2+1$. So $P(X,-X)=(X^2+1)Q(X)$ for some $QinmathbbR[X]$. Furthermore, $p(X,Y)-p(X,-X)$ is divisble by $Y-(-X)=Y+X$ in $mathbbR[X,Y]$. Hence $p(X,Y)=p(X,-X)+(Y+X)T(X,Y)=(X^2+1)Q(X)+(Y+X)T(X,Y)$. So $p(X,Y)in(X+Y,X^2+1)$.
I don't understand why $p(i,-i)=0$ implies $p(X,-X)$ evaluated at $X=i$ equals zero (we just fill in the $i$'s?), nor why it then would be divisible by $X^2+1$.
I also don't understand the part starting from "Furthermore". Like why we use that subtraction, or what that means, and why it would be divisible by $Y-(-X)$. Also what is $T(X,Y)$, an equivalent of $Q$?
polynomials ring-theory ideals
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
$endgroup$
|
show 4 more comments
$begingroup$
I want to show that $I=(X+Y,X^2+1)in R=mathbbR[X,Y]$ is a prime ideal. Since $I$ a prime ideal is equivalent to $R/I$ an integral domain, I want to create a homomorphism $phi: R to mathbbC$ given by $phi: p(X,Y)mapsto p(i,-i)$. This is a surjective homomorphism, so by the first isomorphism theorem we have that $R/ker(phi)$ isomorphic to $mathbbC$ and since $mathbbC$ is an integral domain, $R/ker(phi)$ is an integral domain, so $ker(phi)$ is a prime ideal.
So what's left to show is that $ker(phi)=I$.
Since $phi(X+Y)=i+(-i)=0$, and $phi(X^2+1)=i^2+1=-1+1=0$, we have that $Isubsetker(phi)$. I now want to show that no other elements but elements of $I$ are in the kernel, but I don't understand the following proof:
Suppose $p(X,Y)inker(phi)$, in other words, $p(i,-i)=0$. This implies that the polynomial $P(X,-X)$ evaluated at $X=i$ is zero and so $p(X,-X)$ is divisible by $X^2+1$. So $P(X,-X)=(X^2+1)Q(X)$ for some $QinmathbbR[X]$. Furthermore, $p(X,Y)-p(X,-X)$ is divisble by $Y-(-X)=Y+X$ in $mathbbR[X,Y]$. Hence $p(X,Y)=p(X,-X)+(Y+X)T(X,Y)=(X^2+1)Q(X)+(Y+X)T(X,Y)$. So $p(X,Y)in(X+Y,X^2+1)$.
I don't understand why $p(i,-i)=0$ implies $p(X,-X)$ evaluated at $X=i$ equals zero (we just fill in the $i$'s?), nor why it then would be divisible by $X^2+1$.
I also don't understand the part starting from "Furthermore". Like why we use that subtraction, or what that means, and why it would be divisible by $Y-(-X)$. Also what is $T(X,Y)$, an equivalent of $Q$?
polynomials ring-theory ideals
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
$endgroup$
$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
1
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago
|
show 4 more comments
$begingroup$
I want to show that $I=(X+Y,X^2+1)in R=mathbbR[X,Y]$ is a prime ideal. Since $I$ a prime ideal is equivalent to $R/I$ an integral domain, I want to create a homomorphism $phi: R to mathbbC$ given by $phi: p(X,Y)mapsto p(i,-i)$. This is a surjective homomorphism, so by the first isomorphism theorem we have that $R/ker(phi)$ isomorphic to $mathbbC$ and since $mathbbC$ is an integral domain, $R/ker(phi)$ is an integral domain, so $ker(phi)$ is a prime ideal.
So what's left to show is that $ker(phi)=I$.
Since $phi(X+Y)=i+(-i)=0$, and $phi(X^2+1)=i^2+1=-1+1=0$, we have that $Isubsetker(phi)$. I now want to show that no other elements but elements of $I$ are in the kernel, but I don't understand the following proof:
Suppose $p(X,Y)inker(phi)$, in other words, $p(i,-i)=0$. This implies that the polynomial $P(X,-X)$ evaluated at $X=i$ is zero and so $p(X,-X)$ is divisible by $X^2+1$. So $P(X,-X)=(X^2+1)Q(X)$ for some $QinmathbbR[X]$. Furthermore, $p(X,Y)-p(X,-X)$ is divisble by $Y-(-X)=Y+X$ in $mathbbR[X,Y]$. Hence $p(X,Y)=p(X,-X)+(Y+X)T(X,Y)=(X^2+1)Q(X)+(Y+X)T(X,Y)$. So $p(X,Y)in(X+Y,X^2+1)$.
I don't understand why $p(i,-i)=0$ implies $p(X,-X)$ evaluated at $X=i$ equals zero (we just fill in the $i$'s?), nor why it then would be divisible by $X^2+1$.
I also don't understand the part starting from "Furthermore". Like why we use that subtraction, or what that means, and why it would be divisible by $Y-(-X)$. Also what is $T(X,Y)$, an equivalent of $Q$?
polynomials ring-theory ideals
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
$endgroup$
I want to show that $I=(X+Y,X^2+1)in R=mathbbR[X,Y]$ is a prime ideal. Since $I$ a prime ideal is equivalent to $R/I$ an integral domain, I want to create a homomorphism $phi: R to mathbbC$ given by $phi: p(X,Y)mapsto p(i,-i)$. This is a surjective homomorphism, so by the first isomorphism theorem we have that $R/ker(phi)$ isomorphic to $mathbbC$ and since $mathbbC$ is an integral domain, $R/ker(phi)$ is an integral domain, so $ker(phi)$ is a prime ideal.
So what's left to show is that $ker(phi)=I$.
Since $phi(X+Y)=i+(-i)=0$, and $phi(X^2+1)=i^2+1=-1+1=0$, we have that $Isubsetker(phi)$. I now want to show that no other elements but elements of $I$ are in the kernel, but I don't understand the following proof:
Suppose $p(X,Y)inker(phi)$, in other words, $p(i,-i)=0$. This implies that the polynomial $P(X,-X)$ evaluated at $X=i$ is zero and so $p(X,-X)$ is divisible by $X^2+1$. So $P(X,-X)=(X^2+1)Q(X)$ for some $QinmathbbR[X]$. Furthermore, $p(X,Y)-p(X,-X)$ is divisble by $Y-(-X)=Y+X$ in $mathbbR[X,Y]$. Hence $p(X,Y)=p(X,-X)+(Y+X)T(X,Y)=(X^2+1)Q(X)+(Y+X)T(X,Y)$. So $p(X,Y)in(X+Y,X^2+1)$.
I don't understand why $p(i,-i)=0$ implies $p(X,-X)$ evaluated at $X=i$ equals zero (we just fill in the $i$'s?), nor why it then would be divisible by $X^2+1$.
I also don't understand the part starting from "Furthermore". Like why we use that subtraction, or what that means, and why it would be divisible by $Y-(-X)$. Also what is $T(X,Y)$, an equivalent of $Q$?
polynomials ring-theory ideals
polynomials ring-theory ideals
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
asked Mar 27 at 18:12
The Coding WombatThe Coding Wombat
280110
280110
$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
1
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago
|
show 4 more comments
$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
1
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago
$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
1
1
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago
|
show 4 more comments
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$begingroup$
Note that $P (X,-X) $ is a polynomial in $X $. So we can write $P (X,-X)=F (X)$. Now $i $ is a root of $F (X) $ as $P (i, -i)=0$. Since $x^2+1$ is the minimal polynomial of $i $ over $Bbb R $, $x^2+1$ divides $F (X) $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:31
$begingroup$
Writing out $P (X,Y )-P (X,-X)$ for smaller degree polynomials will be useful to understand the statement about divisibility by $Y+X $.
$endgroup$
– Thomas Shelby
Mar 27 at 18:38
$begingroup$
Why does $P(i,-i)=0$ mean that $i$ is a root of $F(X)$? Which one do you fill in, $i$ or $-i$ into $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
$begingroup$
@ThomasShelby And what do you mean by minimal polynomial "of i over $mathbbR$"? And how does that make $x^2+1$ divide $F(X)$?
$endgroup$
– The Coding Wombat
2 days ago
1
$begingroup$
In your case, you only need to prove that if $i $ is a root of a polynomial $f(X) $ in $Bbb R[X]$, then $X^2+1$ divides $f (X) $. This can be easily proved by using Euclidean algorithm.
$endgroup$
– Thomas Shelby
2 days ago