module question

[Maximilian Rogers]

How can I show that for a commutative ring R, R[x]/(f) is a rank n R-
module, where n=deg(f)?
I can see why it is true intuitively, since R[x]/(f) is polynomials of
degree <n, but I am not sure how to prove this rigorously.

[Chip Eastham]

Did you mean to assume that f is monic
(leading coefficient 1)? And that n > 0?

regards, chip

[Maximilian Rogers]

sure,thanks

[Larry Hammick]

"Maximilian Rogers"

]sure,thanks

I would take a look at the free subset {1, x, xx, ..., x^{n-1}} of R[x], and
its image in R[x]/(f) (where f is monic).
LH

[Chip Eastham]

To finish the idea, Larry's suggested set, the
image in R[x]/(f) where f is monic and degree n,
is clearly a spanning set (as an R-module)
since any polynomial p(x) can be expressed as
p(x) = q(x)*f(x) + r(x) where degree r < n.
[Division algorithm can be proved by induction
on degree of p, using f monic.]

To show uniqueness of representations mod f,
suppose that SUM a_i x^i FOR i = 0 to n-1
is in (f), i.e. divisible by f. A standard
degree argument show all a_i are zero, i.e.
degree of f*q = (degree of f)*(degree of q),
making use of f monic.

regards, chip

[Maximilian Rogers]

thanks a lot !!!

i have another quick question: i'm trying to show that the kernel of
the homomorphism R[x,y,z]->R[u,v] f(x,y,z)->f(g(u,v),u,v) for soem
fixed g(u,v) in R[u,v] is principal, but I can't figure out what it is
generated by... Could you please help me with this?

Thanks again!

[Chip Eastham]

Can you think of a low degree (in x) polynomial
which is annihilated by this mapping? That is
my candidate for a generator of a principal ideal.

regards, chip

[Maximilian Rogers]

I suppose it should depend on g(u,v), right? Low degree in x? Should
it not be in x,y,z...sorry I am still confused.

[Chip Eastham]

It involves x,y,z, but we can identify the degree
of the polynomial with respect to x, i.e. the
highest power of x that appears. Consider:

x |-> g(u,v)
y |-> u
z |-> v

Again, what is a polynomial of low degree with
respect to x which is sent to zero by the ring
homomorphism?

regards, chip

[Maximilian Rogers]

Is it x-g(y,z)? If so, how can I show that any polynomial in the
kernel is a multiple of it?
Thanks

[Chip Eastham]

Yes, it is f(x) = x - g(y,z). Try to show "that any
polynomial in the kernel is a multiple of it" by
revisiting some of the ideas (division algorithm)
used in the solution to your original problem in
this thread.

regards, chip

[Maximilian Rogers]

Thanks a lot!

[Chip Eastham]

Here's a bit of a high level perspective on the
mapping discussed above. Let S = R[y,z] and
note that R[x,y,z] is isomorphic to S[x] in a
natural way.

e:R[x,y,z]->R[u,v] is a ring homomorphism defined
by its action on the R-transcendentals x,y,z:

e(x) = g(u,v) /* polynomial in R[u,v] */
e(y) = u
e(z) = v

This is an example of an evaluation homomorphism.
We can simplify the example by identifying y and u,
resp. z and v, so that as we've defined S above,
e:S[x]->S is defined by evaluating polynomials in
x by setting x = g (where now we consider g in S).

The evaluation homomorphism has kernel equal to
the principal ideal (x-g) as one proves directly
from writing any f(x) in the kernel as:

f(x) = (x-g)*q(x) + r

where because the degree of x-g is one (wrt x),
the "remainder" r is a constant (in order to
have degree less than x-g).

regards, chip

[Maximilian Rogers]

Thanks again for the great insight.