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edited Aug 8 2011 at 17:49 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
Excellent Counterexamples.
Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.
The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) ={e}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?
The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link
Next one:
- Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|Aut(H)|>|Aut(G)|$
Arturo Magidin posed this question some time ago at MATH.SE
Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$? Answer was provided by Vipul Naik. Link is given here.
Question was posed by Zev Chonoles at $\textbf{MATH.SE}$
I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example? Answer from this link: Take $G = \mathbb{Z}_4 \times \mathbb{Z}_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$. Then $H \cong K \cong \mathbb{Z}_2$ but $G/H \cong \mathbb{}Z_4$ while $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.
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edited Jun 19 2011 at 16:43 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
Excellent Counterexamples.
Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.
The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) ={e}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?
The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link
Next one:
- Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|Aut(H)|>|Aut(G)|$
Arturo Magidin posed this question some time ago at MATH.SE
Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$? Answer was provided by Vipul Naik. Link is given here.
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edited Jun 19 2011 at 10:32 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
Excellent Counterexamples.
Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.
The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) ={e}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?
The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link
Next one:
Arturo Magidin posed this link:question some time ago at MATH.SE
Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$?
Answer was provided by Vipul Naik. Link is given here.
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edited Jun 19 2011 at 10:26 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
Excellent Counterexamples.
Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.
The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) ={e}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?
The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link
Next one:
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edited Jun 19 2011 at 8:40 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
Excellent Counterexamples.
Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.
The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) ={e}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?
The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link
Next one:
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edited Jun 19 2011 at 8:19 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. ) Here is a counterexample. Let $R=\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}$, $S=\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$. $R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
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answered Jun 18 2011 at 8:25 |
Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. )
Here is a counterexample. Let $R=\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}$, $S=\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$. Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.
In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3) http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf
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