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Old October 17th, 2009, 07:40 AM
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Default Multiplication by scalar in K[X]/(f)

Hi:
Let K[X] be the ring of polynomials over the field K and let
f \epsilon K[X]. Let K[X]/(f) be the quotient
ring by (f). In K[X]/(f) how can I define multiplication by
a scalar in K in order to make K[X]/(f) into a K-vector space?
Thanks for reading.
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  #2  
Old October 17th, 2009, 11:59 AM
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Originally Posted by ENRIQUESTEFANINI View Post
Hi:
Let K[X] be the ring of polynomials over the field K and let
f \epsilon K[X]. Let K[X]/(f) be the quotient
ring by (f). In K[X]/(f) how can I define multiplication by
a scalar in K in order to make K[X]/(f) into a K-vector space?
Thanks for reading.
If g(x) is in K[x] and k is in K, simply define k*(g(x) + (f)):= kg(x) + (f). Is easy, though slightly annoying, to check this is well defined.

Tonio
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Old October 17th, 2009, 06:50 PM
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Thank you for your reply.
I see. And I also see that, with this multiplication, K[X]/(f)
is a K-algebra. My question realy is: is this the only way to define a multiplication by scalar
that makes K[X]/(f) into a K-algebra? Or is it that there are more than one way of building
an K-algebra out of K[X]/(f)?

Enrique.
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  #4  
Old October 17th, 2009, 09:47 PM
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Originally Posted by ENRIQUESTEFANINI View Post
Thank you for your reply.
I see. And I also see that, with this multiplication, K[X]/(f)
is a K-algebra. My question realy is: is this the only way to define a multiplication by scalar
that makes K[X]/(f) into a K-algebra? Or is it that there are more than one way of building
an K-algebra out of K[X]/(f)?

Enrique.

Not sure though definitely the above is the most "natural" way since it comes from the definition of scalar multiplication that makes K[x] a K-algebra.

Tonio
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Old October 18th, 2009, 03:44 AM
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Originally Posted by tonio View Post
Not sure though definitely the above is the most "natural" way since it comes from the definition of scalar multiplication that makes K[x] a K-algebra.

Tonio
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