Direct Product cancellation in Hopfian rings
$begingroup$
A ring $R$ is called to be 'Hopfian' if every ring homomorphism of $R$ onto $R$ is an automorphism of $R$
Question:
Given $R$, $S$, $T$, Hopfian rings and
$$R times S cong T times S$$
implies that there exists an isomorphism
$$R cong T$$
My approach:
Let $X1, X2, Y in C$.
According to product rule in Category theory for every object Y and pair of morphisms
$$f1 : Y to X1, spacespace f2 : Y → X2$$
there exists a unique morphism $$f : Y → X1 × X2$$ where $f = <f_1, f_2>$
Given $f: R times S to T times S$, therefore $f = <f_1, f_2>$ where $f_1: R times S to T$ and $f_2: R times S to S$. Since $S$ is Hopfian, therefore $f_2$ is an isomorphic function. I wonder if isomorphism of $f$ does implies isomorphism of $f_1$ and $f_2$. Also please let me know if this is the correct approach or should I look at it in some other way.
P.S: This is my first question on mathexchange, please help me correct any mistakes which I might have made over here.
abstract-algebra ring-theory direct-product
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
$endgroup$
add a comment |
$begingroup$
A ring $R$ is called to be 'Hopfian' if every ring homomorphism of $R$ onto $R$ is an automorphism of $R$
Question:
Given $R$, $S$, $T$, Hopfian rings and
$$R times S cong T times S$$
implies that there exists an isomorphism
$$R cong T$$
My approach:
Let $X1, X2, Y in C$.
According to product rule in Category theory for every object Y and pair of morphisms
$$f1 : Y to X1, spacespace f2 : Y → X2$$
there exists a unique morphism $$f : Y → X1 × X2$$ where $f = <f_1, f_2>$
Given $f: R times S to T times S$, therefore $f = <f_1, f_2>$ where $f_1: R times S to T$ and $f_2: R times S to S$. Since $S$ is Hopfian, therefore $f_2$ is an isomorphic function. I wonder if isomorphism of $f$ does implies isomorphism of $f_1$ and $f_2$. Also please let me know if this is the correct approach or should I look at it in some other way.
P.S: This is my first question on mathexchange, please help me correct any mistakes which I might have made over here.
abstract-algebra ring-theory direct-product
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
$endgroup$
$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50
add a comment |
$begingroup$
A ring $R$ is called to be 'Hopfian' if every ring homomorphism of $R$ onto $R$ is an automorphism of $R$
Question:
Given $R$, $S$, $T$, Hopfian rings and
$$R times S cong T times S$$
implies that there exists an isomorphism
$$R cong T$$
My approach:
Let $X1, X2, Y in C$.
According to product rule in Category theory for every object Y and pair of morphisms
$$f1 : Y to X1, spacespace f2 : Y → X2$$
there exists a unique morphism $$f : Y → X1 × X2$$ where $f = <f_1, f_2>$
Given $f: R times S to T times S$, therefore $f = <f_1, f_2>$ where $f_1: R times S to T$ and $f_2: R times S to S$. Since $S$ is Hopfian, therefore $f_2$ is an isomorphic function. I wonder if isomorphism of $f$ does implies isomorphism of $f_1$ and $f_2$. Also please let me know if this is the correct approach or should I look at it in some other way.
P.S: This is my first question on mathexchange, please help me correct any mistakes which I might have made over here.
abstract-algebra ring-theory direct-product
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
$endgroup$
A ring $R$ is called to be 'Hopfian' if every ring homomorphism of $R$ onto $R$ is an automorphism of $R$
Question:
Given $R$, $S$, $T$, Hopfian rings and
$$R times S cong T times S$$
implies that there exists an isomorphism
$$R cong T$$
My approach:
Let $X1, X2, Y in C$.
According to product rule in Category theory for every object Y and pair of morphisms
$$f1 : Y to X1, spacespace f2 : Y → X2$$
there exists a unique morphism $$f : Y → X1 × X2$$ where $f = <f_1, f_2>$
Given $f: R times S to T times S$, therefore $f = <f_1, f_2>$ where $f_1: R times S to T$ and $f_2: R times S to S$. Since $S$ is Hopfian, therefore $f_2$ is an isomorphic function. I wonder if isomorphism of $f$ does implies isomorphism of $f_1$ and $f_2$. Also please let me know if this is the correct approach or should I look at it in some other way.
P.S: This is my first question on mathexchange, please help me correct any mistakes which I might have made over here.
abstract-algebra ring-theory direct-product
abstract-algebra ring-theory direct-product
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
asked Dec 10 '18 at 7:49
forcehandlerforcehandler
61
61
$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50
add a comment |
$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50
$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50
add a comment |
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$begingroup$
Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:Rtimes Sto S$ an iso?
$endgroup$
– Berci
Dec 10 '18 at 8:27
$begingroup$
Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R times S to S$ where $f_2(r,s) to s$ is a bijective mapping since S is given to be hopfian object.
$endgroup$
– forcehandler
Dec 10 '18 at 8:50