If a field $F$ is infinite, show that the ring homomorphism $eta : F[x]to C(F)$ is one-to-one.
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Here is the question I am trying to attempt:
I feel like there is a typo... it says let $p(x)=a_{n}x^{n}$. Shouldn't it be $p(x) = a_{n}^{n}cdots + a_{1}x+a_{0}$? I feel like I must use the roots of $p(x)$ in my answer, but I am not sure how and where. Any hints on what I should do?
Edit: The duplicate answer did not help me because I am trying to prove injectivity...it has already been assumed that $eta$ is a ring homomorphism.
abstract-algebra ring-theory ring-homomorphism
asked Nov 21 at 2:40
numericalorange
1,619311
add a comment |
up vote
4
down vote
favorite
Here is the question I am trying to attempt:
I feel like there is a typo... it says let $p(x)=a_{n}x^{n}$. Shouldn't it be $p(x) = a_{n}^{n}cdots + a_{1}x+a_{0}$? I feel like I must use the roots of $p(x)$ in my answer, but I am not sure how and where. Any hints on what I should do?
Edit: The duplicate answer did not help me because I am trying to prove injectivity...it has already been assumed that $eta$ is a ring homomorphism.
abstract-algebra ring-theory ring-homomorphism
asked Nov 21 at 2:40
numericalorange
1,619311
1
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
1
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Here is the question I am trying to attempt:
I feel like there is a typo... it says let $p(x)=a_{n}x^{n}$. Shouldn't it be $p(x) = a_{n}^{n}cdots + a_{1}x+a_{0}$? I feel like I must use the roots of $p(x)$ in my answer, but I am not sure how and where. Any hints on what I should do?
Edit: The duplicate answer did not help me because I am trying to prove injectivity...it has already been assumed that $eta$ is a ring homomorphism.
abstract-algebra ring-theory ring-homomorphism
asked Nov 21 at 2:40
numericalorange
1,619311
Here is the question I am trying to attempt:
I feel like there is a typo... it says let $p(x)=a_{n}x^{n}$. Shouldn't it be $p(x) = a_{n}^{n}cdots + a_{1}x+a_{0}$? I feel like I must use the roots of $p(x)$ in my answer, but I am not sure how and where. Any hints on what I should do?
Edit: The duplicate answer did not help me because I am trying to prove injectivity...it has already been assumed that $eta$ is a ring homomorphism.
abstract-algebra ring-theory ring-homomorphism
abstract-algebra ring-theory ring-homomorphism
asked Nov 21 at 2:40
numericalorange
1,619311
asked Nov 21 at 2:40
numericalorange
1,619311
edited Nov 21 at 6:54
asked Nov 21 at 2:40
numericalorange
1,619311
asked Nov 21 at 2:40
numericalorange
1,619311
asked Nov 21 at 2:40
numericalorange
1,619311
1,619311
1
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
1
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56
add a comment |
1
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
1
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56
1
1
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
1
1
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.
answered Nov 21 at 2:54
John Nash
6218
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
add a comment |
up vote
1
down vote
Your thoughts on the typo are right. It should be $p(x)=a_nx^n+cdots+a_0$.
Now it asks you to show that $eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.
Well, suppose $eta(p)=0$ for a polynomial $p$. Then $p(alpha)=0$ for all $alphain F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $deg p$ roots). Thus we have $p=0$, so $eta$ is injective.
answered Nov 21 at 16:46
jgon
9,82611638
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.
answered Nov 21 at 2:54
John Nash
6218
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
add a comment |
up vote
1
down vote
Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.
answered Nov 21 at 2:54
John Nash
6218
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.
answered Nov 21 at 2:54
John Nash
6218
Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.
answered Nov 21 at 2:54
John Nash
6218
answered Nov 21 at 2:54
John Nash
6218
answered Nov 21 at 2:54
John Nash
6218
answered Nov 21 at 2:54
John Nash
6218
6218
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
add a comment |
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
Since $F$ is infinite, does that mean the kernel of $eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(eta)$, but then why is $ker(eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(eta)=F$? I guess because not all polynomials evaluate to zero.
– numericalorange
Nov 21 at 6:52
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $ker eta = 0$ then $eta$ is injective," which is almost what it says in the image in the question.
– jgon
Nov 21 at 16:43
add a comment |
up vote
1
down vote
Your thoughts on the typo are right. It should be $p(x)=a_nx^n+cdots+a_0$.
Now it asks you to show that $eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.
Well, suppose $eta(p)=0$ for a polynomial $p$. Then $p(alpha)=0$ for all $alphain F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $deg p$ roots). Thus we have $p=0$, so $eta$ is injective.
answered Nov 21 at 16:46
jgon
9,82611638
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
add a comment |
up vote
1
down vote
Your thoughts on the typo are right. It should be $p(x)=a_nx^n+cdots+a_0$.
Now it asks you to show that $eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.
Well, suppose $eta(p)=0$ for a polynomial $p$. Then $p(alpha)=0$ for all $alphain F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $deg p$ roots). Thus we have $p=0$, so $eta$ is injective.
answered Nov 21 at 16:46
jgon
9,82611638
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
add a comment |
up vote
1
down vote
up vote
1
down vote
Your thoughts on the typo are right. It should be $p(x)=a_nx^n+cdots+a_0$.
Now it asks you to show that $eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.
Well, suppose $eta(p)=0$ for a polynomial $p$. Then $p(alpha)=0$ for all $alphain F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $deg p$ roots). Thus we have $p=0$, so $eta$ is injective.
answered Nov 21 at 16:46
jgon
9,82611638
Your thoughts on the typo are right. It should be $p(x)=a_nx^n+cdots+a_0$.
Now it asks you to show that $eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.
Well, suppose $eta(p)=0$ for a polynomial $p$. Then $p(alpha)=0$ for all $alphain F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $deg p$ roots). Thus we have $p=0$, so $eta$ is injective.
answered Nov 21 at 16:46
jgon
9,82611638
answered Nov 21 at 16:46
jgon
9,82611638
answered Nov 21 at 16:46
jgon
9,82611638
answered Nov 21 at 16:46
jgon
9,82611638
9,82611638
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
add a comment |
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
I see, and so the kernal of $eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator?
– numericalorange
Nov 21 at 18:00
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
@numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0.
– jgon
Nov 21 at 18:12
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $alpha$, then $(x-alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $eta$ when the field is finite.
– jgon
Nov 21 at 18:14
add a comment |
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1
If ker(η)=0 then η is injective.
– John Nash
Nov 21 at 2:49
1
Possible duplicate of How to prove that the evaluation map is a ring homomorphism?
– Monstrous Moonshiner
Nov 21 at 2:50
Yes, it means $p(x)=a_nx^n+ldots a_1x+a_0.$
– spaceisdarkgreen
Nov 21 at 2:56