Is $frac{R[x]}{(f(x))}$ finite for $R$ finite and $f$ not monic polynomial?
$begingroup$
If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?
I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.
Any Help will be appreciated
abstract-algebra polynomials ring-theory finite-rings
edited Jan 19 at 16:44
user26857
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
$endgroup$
add a comment |
$begingroup$
If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?
I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.
Any Help will be appreciated
abstract-algebra polynomials ring-theory finite-rings
edited Jan 19 at 16:44
user26857
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
$endgroup$
add a comment |
$begingroup$
If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?
I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.
Any Help will be appreciated
abstract-algebra polynomials ring-theory finite-rings
edited Jan 19 at 16:44
user26857
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
$endgroup$
If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?
I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.
Any Help will be appreciated
abstract-algebra polynomials ring-theory finite-rings
abstract-algebra polynomials ring-theory finite-rings
edited Jan 19 at 16:44
user26857
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
edited Jan 19 at 16:44
user26857
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
edited Jan 19 at 16:44
user26857
39.3k124183
edited Jan 19 at 16:44
user26857
39.3k124183
edited Jan 19 at 16:44
user26857
39.3k124183
39.3k124183
asked Jan 18 at 11:09
SRJSRJ
1,7831620
asked Jan 18 at 11:09
SRJSRJ
1,7831620
asked Jan 18 at 11:09
SRJSRJ
1,7831620
1,7831620
add a comment |
add a comment |
1 Answer
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In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.
answered Jan 18 at 11:13
ArthurArthur
115k7116198
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.
answered Jan 18 at 11:13
ArthurArthur
115k7116198
$endgroup$
add a comment |
$begingroup$
In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.
answered Jan 18 at 11:13
ArthurArthur
115k7116198
$endgroup$
add a comment |
$begingroup$
In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.
answered Jan 18 at 11:13
ArthurArthur
115k7116198
$endgroup$
In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.
answered Jan 18 at 11:13
ArthurArthur
115k7116198
answered Jan 18 at 11:13
ArthurArthur
115k7116198
answered Jan 18 at 11:13
ArthurArthur
115k7116198
answered Jan 18 at 11:13
ArthurArthur
115k7116198
115k7116198
add a comment |
add a comment |
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