Geometric interpretation of subalgebras as non-linear coordinate change? (context of Noether normalization)
$begingroup$
Consider a $k$-algebra morphism $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ defined by $x_imapsto g_i$. In a nice setting, e.g what $k$ is a field and $kto frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ is finitely generated, then we may choose $g$ to be injective, so that its image is a free $k$-subalgebra. Noether normalization says that we may moreover choose $g$ such that the inclusion of its image is a finite module. When $k$ is an infinite field, we may take $g$ to be linear in $x_i$.
Geometrically, I am thinking of $g$ as the morphism $$Xto mathbb A^d_k,;;(t_1,dots ,t_n)mapsto(g_1(t_1,dots ,t_n),dots,g_d(t_1,dots,t_n)).$$ I would like to understand the formal explanation for some pictures.
When $g$ is linear, it seems (?!) that fixing an inner product gives even a $k$-algebra isomorphism $$ operatorname{Im} gcong frac{k[x_1,dots ,x_n]}{(operatorname{Im}g)^perp} .$$
Is there anything similar when $g$ is non-linear? That is, when we can visualize, can we obtain an $n$-variable presentation of a subalgebra via "non-linear orthogonal projection"?
Motivation. In the context of Noether normalization, we usually consider a geometric object specified by a presentation of a $k$-algebra, and then pick new variables $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ (even algebraically independent) which make the "remainder" a finite module. When $g$ is linear we may interpret the generators geometrically as described above, and "linear duality" lets us actually view the finite surjective projection onto an affine space of smaller dimension as a finite surjection projection onto a closed subspace of the original affine space $mathbb A_k^n$. In particular, this seems to be the mechanism which transports geometric intuition from drawing pictures, as in this answer and this answer. However, a linear change of variable is not always available, and even when it is, I'd still like to know what can be done in the non-linear case.
Concrete example. Consider the Noether normalization of the hyperbola given by $(x,y)mapsto x-y$. This corresponds to the subalgebra generated by $overline x-overline y$ in $frac{k[x,y]}{(xy-1)}$ and can be thought of as orthogonally projection the hyperbola onto $mathbf Z(x+y)subsetmathbb A^2_k$. Is there a differential-geometric way of thinking about the subalgebra generated by $overline y-overline x^3infrac{k[x,y]}{(xy-1)}$?
algebraic-geometry commutative-algebra affine-schemes
asked Jan 8 at 14:46
ArrowArrow
5,08121445
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add a comment |
$begingroup$
Consider a $k$-algebra morphism $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ defined by $x_imapsto g_i$. In a nice setting, e.g what $k$ is a field and $kto frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ is finitely generated, then we may choose $g$ to be injective, so that its image is a free $k$-subalgebra. Noether normalization says that we may moreover choose $g$ such that the inclusion of its image is a finite module. When $k$ is an infinite field, we may take $g$ to be linear in $x_i$.
Geometrically, I am thinking of $g$ as the morphism $$Xto mathbb A^d_k,;;(t_1,dots ,t_n)mapsto(g_1(t_1,dots ,t_n),dots,g_d(t_1,dots,t_n)).$$ I would like to understand the formal explanation for some pictures.
When $g$ is linear, it seems (?!) that fixing an inner product gives even a $k$-algebra isomorphism $$ operatorname{Im} gcong frac{k[x_1,dots ,x_n]}{(operatorname{Im}g)^perp} .$$
Is there anything similar when $g$ is non-linear? That is, when we can visualize, can we obtain an $n$-variable presentation of a subalgebra via "non-linear orthogonal projection"?
Motivation. In the context of Noether normalization, we usually consider a geometric object specified by a presentation of a $k$-algebra, and then pick new variables $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ (even algebraically independent) which make the "remainder" a finite module. When $g$ is linear we may interpret the generators geometrically as described above, and "linear duality" lets us actually view the finite surjective projection onto an affine space of smaller dimension as a finite surjection projection onto a closed subspace of the original affine space $mathbb A_k^n$. In particular, this seems to be the mechanism which transports geometric intuition from drawing pictures, as in this answer and this answer. However, a linear change of variable is not always available, and even when it is, I'd still like to know what can be done in the non-linear case.
Concrete example. Consider the Noether normalization of the hyperbola given by $(x,y)mapsto x-y$. This corresponds to the subalgebra generated by $overline x-overline y$ in $frac{k[x,y]}{(xy-1)}$ and can be thought of as orthogonally projection the hyperbola onto $mathbf Z(x+y)subsetmathbb A^2_k$. Is there a differential-geometric way of thinking about the subalgebra generated by $overline y-overline x^3infrac{k[x,y]}{(xy-1)}$?
algebraic-geometry commutative-algebra affine-schemes
asked Jan 8 at 14:46
ArrowArrow
5,08121445
$endgroup$
$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56
add a comment |
$begingroup$
Consider a $k$-algebra morphism $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ defined by $x_imapsto g_i$. In a nice setting, e.g what $k$ is a field and $kto frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ is finitely generated, then we may choose $g$ to be injective, so that its image is a free $k$-subalgebra. Noether normalization says that we may moreover choose $g$ such that the inclusion of its image is a finite module. When $k$ is an infinite field, we may take $g$ to be linear in $x_i$.
Geometrically, I am thinking of $g$ as the morphism $$Xto mathbb A^d_k,;;(t_1,dots ,t_n)mapsto(g_1(t_1,dots ,t_n),dots,g_d(t_1,dots,t_n)).$$ I would like to understand the formal explanation for some pictures.
When $g$ is linear, it seems (?!) that fixing an inner product gives even a $k$-algebra isomorphism $$ operatorname{Im} gcong frac{k[x_1,dots ,x_n]}{(operatorname{Im}g)^perp} .$$
Is there anything similar when $g$ is non-linear? That is, when we can visualize, can we obtain an $n$-variable presentation of a subalgebra via "non-linear orthogonal projection"?
Motivation. In the context of Noether normalization, we usually consider a geometric object specified by a presentation of a $k$-algebra, and then pick new variables $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ (even algebraically independent) which make the "remainder" a finite module. When $g$ is linear we may interpret the generators geometrically as described above, and "linear duality" lets us actually view the finite surjective projection onto an affine space of smaller dimension as a finite surjection projection onto a closed subspace of the original affine space $mathbb A_k^n$. In particular, this seems to be the mechanism which transports geometric intuition from drawing pictures, as in this answer and this answer. However, a linear change of variable is not always available, and even when it is, I'd still like to know what can be done in the non-linear case.
Concrete example. Consider the Noether normalization of the hyperbola given by $(x,y)mapsto x-y$. This corresponds to the subalgebra generated by $overline x-overline y$ in $frac{k[x,y]}{(xy-1)}$ and can be thought of as orthogonally projection the hyperbola onto $mathbf Z(x+y)subsetmathbb A^2_k$. Is there a differential-geometric way of thinking about the subalgebra generated by $overline y-overline x^3infrac{k[x,y]}{(xy-1)}$?
algebraic-geometry commutative-algebra affine-schemes
asked Jan 8 at 14:46
ArrowArrow
5,08121445
$endgroup$
Consider a $k$-algebra morphism $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ defined by $x_imapsto g_i$. In a nice setting, e.g what $k$ is a field and $kto frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ is finitely generated, then we may choose $g$ to be injective, so that its image is a free $k$-subalgebra. Noether normalization says that we may moreover choose $g$ such that the inclusion of its image is a finite module. When $k$ is an infinite field, we may take $g$ to be linear in $x_i$.
Geometrically, I am thinking of $g$ as the morphism $$Xto mathbb A^d_k,;;(t_1,dots ,t_n)mapsto(g_1(t_1,dots ,t_n),dots,g_d(t_1,dots,t_n)).$$ I would like to understand the formal explanation for some pictures.
When $g$ is linear, it seems (?!) that fixing an inner product gives even a $k$-algebra isomorphism $$ operatorname{Im} gcong frac{k[x_1,dots ,x_n]}{(operatorname{Im}g)^perp} .$$
Is there anything similar when $g$ is non-linear? That is, when we can visualize, can we obtain an $n$-variable presentation of a subalgebra via "non-linear orthogonal projection"?
Motivation. In the context of Noether normalization, we usually consider a geometric object specified by a presentation of a $k$-algebra, and then pick new variables $k[x_1,dots ,x_d]overset{g}{longrightarrow }frac{k[x_1,dots ,x_n]}{(f_1,dots ,f_m)}$ (even algebraically independent) which make the "remainder" a finite module. When $g$ is linear we may interpret the generators geometrically as described above, and "linear duality" lets us actually view the finite surjective projection onto an affine space of smaller dimension as a finite surjection projection onto a closed subspace of the original affine space $mathbb A_k^n$. In particular, this seems to be the mechanism which transports geometric intuition from drawing pictures, as in this answer and this answer. However, a linear change of variable is not always available, and even when it is, I'd still like to know what can be done in the non-linear case.
Concrete example. Consider the Noether normalization of the hyperbola given by $(x,y)mapsto x-y$. This corresponds to the subalgebra generated by $overline x-overline y$ in $frac{k[x,y]}{(xy-1)}$ and can be thought of as orthogonally projection the hyperbola onto $mathbf Z(x+y)subsetmathbb A^2_k$. Is there a differential-geometric way of thinking about the subalgebra generated by $overline y-overline x^3infrac{k[x,y]}{(xy-1)}$?
algebraic-geometry commutative-algebra affine-schemes
algebraic-geometry commutative-algebra affine-schemes
asked Jan 8 at 14:46
ArrowArrow
5,08121445
asked Jan 8 at 14:46
ArrowArrow
5,08121445
asked Jan 8 at 14:46
ArrowArrow
5,08121445
asked Jan 8 at 14:46
ArrowArrow
5,08121445
asked Jan 8 at 14:46
ArrowArrow
5,08121445
5,08121445
$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56
add a comment |
$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56
$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56
$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56
add a comment |
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$begingroup$
Probably, you meant to say $k[underline{x}] to $ is finite instead of $k to $ is (in your second sentence). I did not understand what you meant by the inner product. However, the direct summand conjecture (now theorem) states that $g$ splits as an $k[x_1,dots,x_d]$-module. See item 6 en.wikipedia.org/wiki/….
$endgroup$
– Youngsu
Jan 9 at 18:56