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Computing the kernel of a morphism whose domain and codomain are polynomial algebras
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Let $k$ denote a field.
If $varphi$ is a $k$-linear map between $k$-modules freely generated by finite sets, then the kernel of $varphi$ can easily be found by reinterpreting $varphi$ as a matrix and performing row-reduction.
Suppose instead that $varphi$ is a $k$-algebra homomorphism between $k$-algebras freely generated by finite sets, i.e. between polynomial algebras in finitely-many indeterminates. Is there a way of explicitly computing the kernel of such a map?
Let's look at a really simple example. Suppose we want to show that $$fracmathbbR[x,y](y-x^2) cong mathbbR[x],$$
for example. A reasonable place to start is to define $varphi$ to be the unique morphism of $mathbbR$-algebras $$mathbbR[x,y] rightarrow mathbbR[x]$$ satisfying $varphi(x)=x$ and $varphi(y) = x^2$. This is clearly surjective, and presumably the kernel of this map is $(y-x^2),$ thereby giving the desired result via the first isomorphism theorem.
Part of my question is: if $varphi$ were more complicated, how could we work out what it's kernel was? Is there a systematic approach, like there is in linear algebra?
Part of my question is also about rigor. In the particular example of $varphi$ given above, since $varphi(x) = x$, thus $varphi(x^2) = x^2$, so $varphi(y - x^2) = 0$. Thus $(y-x^2) subseteq mathrmker(varphi)$. I'm a bit unsure how to go the other way. Supposing $varphi(P) = 0$, how can we show that $P in (y - x^2)$? And is there a method that works generally, even for much more complicate examples?
Related.
linear-algebra abstract-algebra polynomials commutative-algebra field-theory
asked Aug 6 at 11:30
goblin
35.5k1153181
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3
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Let $k$ denote a field.
If $varphi$ is a $k$-linear map between $k$-modules freely generated by finite sets, then the kernel of $varphi$ can easily be found by reinterpreting $varphi$ as a matrix and performing row-reduction.
Suppose instead that $varphi$ is a $k$-algebra homomorphism between $k$-algebras freely generated by finite sets, i.e. between polynomial algebras in finitely-many indeterminates. Is there a way of explicitly computing the kernel of such a map?
Let's look at a really simple example. Suppose we want to show that $$fracmathbbR[x,y](y-x^2) cong mathbbR[x],$$
for example. A reasonable place to start is to define $varphi$ to be the unique morphism of $mathbbR$-algebras $$mathbbR[x,y] rightarrow mathbbR[x]$$ satisfying $varphi(x)=x$ and $varphi(y) = x^2$. This is clearly surjective, and presumably the kernel of this map is $(y-x^2),$ thereby giving the desired result via the first isomorphism theorem.
Part of my question is: if $varphi$ were more complicated, how could we work out what it's kernel was? Is there a systematic approach, like there is in linear algebra?
Part of my question is also about rigor. In the particular example of $varphi$ given above, since $varphi(x) = x$, thus $varphi(x^2) = x^2$, so $varphi(y - x^2) = 0$. Thus $(y-x^2) subseteq mathrmker(varphi)$. I'm a bit unsure how to go the other way. Supposing $varphi(P) = 0$, how can we show that $P in (y - x^2)$? And is there a method that works generally, even for much more complicate examples?
Related.
linear-algebra abstract-algebra polynomials commutative-algebra field-theory
asked Aug 6 at 11:30
goblin
35.5k1153181
1
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $k$ denote a field.
If $varphi$ is a $k$-linear map between $k$-modules freely generated by finite sets, then the kernel of $varphi$ can easily be found by reinterpreting $varphi$ as a matrix and performing row-reduction.
Suppose instead that $varphi$ is a $k$-algebra homomorphism between $k$-algebras freely generated by finite sets, i.e. between polynomial algebras in finitely-many indeterminates. Is there a way of explicitly computing the kernel of such a map?
Let's look at a really simple example. Suppose we want to show that $$fracmathbbR[x,y](y-x^2) cong mathbbR[x],$$
for example. A reasonable place to start is to define $varphi$ to be the unique morphism of $mathbbR$-algebras $$mathbbR[x,y] rightarrow mathbbR[x]$$ satisfying $varphi(x)=x$ and $varphi(y) = x^2$. This is clearly surjective, and presumably the kernel of this map is $(y-x^2),$ thereby giving the desired result via the first isomorphism theorem.
Part of my question is: if $varphi$ were more complicated, how could we work out what it's kernel was? Is there a systematic approach, like there is in linear algebra?
Part of my question is also about rigor. In the particular example of $varphi$ given above, since $varphi(x) = x$, thus $varphi(x^2) = x^2$, so $varphi(y - x^2) = 0$. Thus $(y-x^2) subseteq mathrmker(varphi)$. I'm a bit unsure how to go the other way. Supposing $varphi(P) = 0$, how can we show that $P in (y - x^2)$? And is there a method that works generally, even for much more complicate examples?
Related.
linear-algebra abstract-algebra polynomials commutative-algebra field-theory
asked Aug 6 at 11:30
goblin
35.5k1153181
Let $k$ denote a field.
If $varphi$ is a $k$-linear map between $k$-modules freely generated by finite sets, then the kernel of $varphi$ can easily be found by reinterpreting $varphi$ as a matrix and performing row-reduction.
Suppose instead that $varphi$ is a $k$-algebra homomorphism between $k$-algebras freely generated by finite sets, i.e. between polynomial algebras in finitely-many indeterminates. Is there a way of explicitly computing the kernel of such a map?
Let's look at a really simple example. Suppose we want to show that $$fracmathbbR[x,y](y-x^2) cong mathbbR[x],$$
for example. A reasonable place to start is to define $varphi$ to be the unique morphism of $mathbbR$-algebras $$mathbbR[x,y] rightarrow mathbbR[x]$$ satisfying $varphi(x)=x$ and $varphi(y) = x^2$. This is clearly surjective, and presumably the kernel of this map is $(y-x^2),$ thereby giving the desired result via the first isomorphism theorem.
Part of my question is: if $varphi$ were more complicated, how could we work out what it's kernel was? Is there a systematic approach, like there is in linear algebra?
Part of my question is also about rigor. In the particular example of $varphi$ given above, since $varphi(x) = x$, thus $varphi(x^2) = x^2$, so $varphi(y - x^2) = 0$. Thus $(y-x^2) subseteq mathrmker(varphi)$. I'm a bit unsure how to go the other way. Supposing $varphi(P) = 0$, how can we show that $P in (y - x^2)$? And is there a method that works generally, even for much more complicate examples?
Related.
linear-algebra abstract-algebra polynomials commutative-algebra field-theory
asked Aug 6 at 11:30
goblin
35.5k1153181
edited Aug 7 at 7:18
asked Aug 6 at 11:30
goblin
35.5k1153181
asked Aug 6 at 11:30
goblin
35.5k1153181
asked Aug 6 at 11:30
goblin
35.5k1153181
35.5k1153181
1
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44
add a comment |Â
1
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44
1
1
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44
add a comment |Â
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1
In your example, the image is an integral domain, so the kernel is a prime ideal. The height of $(y-x^2)$ is already $1$, so any prime containing it would be maximal. I can't give any general advice for calculating kernels; when forced to do so I generally resort to ad hoc methods.
– John Brevik
Aug 6 at 12:44