an expression of the discriminant
$begingroup$
Let $δ := det((X_i^{j−1} ) 1le i,j le n ) ∈ mathbb{Z}[X_1 , · · · , X_n ]$.
Let $δ = sum_{ν∈mathbb{N}^n} a_nu X^nu$ .
prove that $a_nu ne 0 ⇒ sum_{i=1}^n nu_i = n(n − 1)/2$.
I don't understand the expression in bold. What does $X^nu$ represent?
Thank you for your help.
linear-algebra
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
$endgroup$
add a comment |
$begingroup$
Let $δ := det((X_i^{j−1} ) 1le i,j le n ) ∈ mathbb{Z}[X_1 , · · · , X_n ]$.
Let $δ = sum_{ν∈mathbb{N}^n} a_nu X^nu$ .
prove that $a_nu ne 0 ⇒ sum_{i=1}^n nu_i = n(n − 1)/2$.
I don't understand the expression in bold. What does $X^nu$ represent?
Thank you for your help.
linear-algebra
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
$endgroup$
add a comment |
$begingroup$
Let $δ := det((X_i^{j−1} ) 1le i,j le n ) ∈ mathbb{Z}[X_1 , · · · , X_n ]$.
Let $δ = sum_{ν∈mathbb{N}^n} a_nu X^nu$ .
prove that $a_nu ne 0 ⇒ sum_{i=1}^n nu_i = n(n − 1)/2$.
I don't understand the expression in bold. What does $X^nu$ represent?
Thank you for your help.
linear-algebra
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
$endgroup$
Let $δ := det((X_i^{j−1} ) 1le i,j le n ) ∈ mathbb{Z}[X_1 , · · · , X_n ]$.
Let $δ = sum_{ν∈mathbb{N}^n} a_nu X^nu$ .
prove that $a_nu ne 0 ⇒ sum_{i=1}^n nu_i = n(n − 1)/2$.
I don't understand the expression in bold. What does $X^nu$ represent?
Thank you for your help.
linear-algebra
linear-algebra
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
asked Dec 23 '18 at 17:26
PerelManPerelMan
665313
665313
add a comment |
add a comment |
1 Answer
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$begingroup$
It follows this notation. Specifically, $X^{nu}:=X_1^{nu_1}cdotldotscdot X_n^{nu_n}$.
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
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$begingroup$
It follows this notation. Specifically, $X^{nu}:=X_1^{nu_1}cdotldotscdot X_n^{nu_n}$.
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
It follows this notation. Specifically, $X^{nu}:=X_1^{nu_1}cdotldotscdot X_n^{nu_n}$.
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
It follows this notation. Specifically, $X^{nu}:=X_1^{nu_1}cdotldotscdot X_n^{nu_n}$.
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
$endgroup$
It follows this notation. Specifically, $X^{nu}:=X_1^{nu_1}cdotldotscdot X_n^{nu_n}$.
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
answered Dec 23 '18 at 17:33
metamorphymetamorphy
3,7021621
3,7021621
add a comment |
add a comment |
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