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yell
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Given an algebraically closed field $\mathbb{F}$ of characteristic $0$ and a closed subgroup $G$ of $GL_n(\mathbb{F})$. Let $\{g_1,\ldots, g_r\}$ be a Gr"obner basis for the correpsonding ideal $\mathcal{I}(G)$. I want to find the irreducible component of $1$ (that means i want to find generators for $\mathcal{I}(G^\circ)$). Is there a better way thenthan computing the primary decomposition, pick the component, which contains $1$, and compute its radical? and compute its radical?

Given an algebraically closed field $\mathbb{F}$ of characteristic $0$ and a closed subgroup $G$ of $GL_n(\mathbb{F})$. I want to find the irreducible component of $1$. Is there a better way then computing the primary decomposition, pick the component, which contains $1$, and compute its radical?

Given an algebraically closed field $\mathbb{F}$ of characteristic $0$ and a closed subgroup $G$ of $GL_n(\mathbb{F})$. Let $\{g_1,\ldots, g_r\}$ be a Gr"obner basis for the correpsonding ideal $\mathcal{I}(G)$. I want to find the irreducible component of $1$ (that means i want to find generators for $\mathcal{I}(G^\circ)$). Is there a better way than computing the primary decomposition, pick the component, which contains $1$, and compute its radical?

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Sándor Kovács
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yell
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Computing the connected component without primary decomposition

Given an algebraically closed field $\mathbb{F}$ of characteristic $0$ and a closed subgroup $G$ of $GL_n(\mathbb{F})$. I want to find the irreducible component of $1$. Is there a better way then computing the primary decomposition, pick the component, which contains $1$, and compute its radical?