Mixing
Finitely generated $k$-algebras beginner examples.
$begingroup$
I just found out about finitely generated $k$-algebras (where $k$ is a field). So it is an algebra $A$ for which we have a finite set of elements $(a_1,...,a_n)$ such that every element in $A$ can be expressed as $p(a_1,...,a_n)$ where $p$ is a polynomial $p in k[x_1,...,x_n]$. This is pretty abstract for the moment so I am trying to illustrate with some examples. So I understand $k[x_1,...,x_n]$ itself is an example where the generators are $x_1,...,x_n$. It seems that in this particular case, the generators are even algebraically independent. What would be a nice example where the set of generators are not algebraically independent?
commutative-algebra finitely-generated algebras
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
$endgroup$
add a comment |
$begingroup$
I just found out about finitely generated $k$-algebras (where $k$ is a field). So it is an algebra $A$ for which we have a finite set of elements $(a_1,...,a_n)$ such that every element in $A$ can be expressed as $p(a_1,...,a_n)$ where $p$ is a polynomial $p in k[x_1,...,x_n]$. This is pretty abstract for the moment so I am trying to illustrate with some examples. So I understand $k[x_1,...,x_n]$ itself is an example where the generators are $x_1,...,x_n$. It seems that in this particular case, the generators are even algebraically independent. What would be a nice example where the set of generators are not algebraically independent?
commutative-algebra finitely-generated algebras
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
$endgroup$
1
$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
1
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20
add a comment |
$begingroup$
I just found out about finitely generated $k$-algebras (where $k$ is a field). So it is an algebra $A$ for which we have a finite set of elements $(a_1,...,a_n)$ such that every element in $A$ can be expressed as $p(a_1,...,a_n)$ where $p$ is a polynomial $p in k[x_1,...,x_n]$. This is pretty abstract for the moment so I am trying to illustrate with some examples. So I understand $k[x_1,...,x_n]$ itself is an example where the generators are $x_1,...,x_n$. It seems that in this particular case, the generators are even algebraically independent. What would be a nice example where the set of generators are not algebraically independent?
commutative-algebra finitely-generated algebras
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
$endgroup$
I just found out about finitely generated $k$-algebras (where $k$ is a field). So it is an algebra $A$ for which we have a finite set of elements $(a_1,...,a_n)$ such that every element in $A$ can be expressed as $p(a_1,...,a_n)$ where $p$ is a polynomial $p in k[x_1,...,x_n]$. This is pretty abstract for the moment so I am trying to illustrate with some examples. So I understand $k[x_1,...,x_n]$ itself is an example where the generators are $x_1,...,x_n$. It seems that in this particular case, the generators are even algebraically independent. What would be a nice example where the set of generators are not algebraically independent?
commutative-algebra finitely-generated algebras
commutative-algebra finitely-generated algebras
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
asked Jan 27 at 0:13
roi_saumonroi_saumon
62838
62838
1
$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
1
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20
add a comment |
1
$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
1
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20
1
1
$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
1
1
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20
add a comment |
1 Answer
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Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
$endgroup$
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
add a comment |
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1 Answer
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$begingroup$
Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
$endgroup$
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
add a comment |
$begingroup$
Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
$endgroup$
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
add a comment |
$begingroup$
Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
$endgroup$
Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
answered Jan 27 at 0:16
Mr. ChipMr. Chip
3,2851129
3,2851129
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
add a comment |
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
1
1
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial.
$endgroup$
– Matt Samuel
Jan 27 at 1:19
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
@Mr.chip Yes, thank you. The generators are $bar{x}$ and $bar{y}$ but there is an algebraic relation $bar{y}=bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$?
$endgroup$
– roi_saumon
Jan 27 at 12:23
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
$begingroup$
Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules...
$endgroup$
– roi_saumon
Jan 27 at 12:29
add a comment |
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$begingroup$
Any quotient of your example; a simple example would be $k[x,y]/(xy)$.
$endgroup$
– Servaes
Jan 27 at 0:16
1
$begingroup$
Maybe this is to concrete for you, but how about $mathbb{Q}[sqrt{2}, pi]$? The generators $sqrt{2}, pi$ are not algebraically independent, because $$ (sqrt{2})^2 + 0 cdot pi - 2 =0.$$
$endgroup$
– Severin Schraven
Jan 27 at 0:20