# What is the dimension of this vector space

• Oct 22nd 2011, 05:50 AM
gotmejerry
What is the dimension of this vector space
Let p(t) be a polynomial of degree n, in the F[t] polynomial ring, where F is a field. Let ß be a congruence relation for: r(t) ß s(t) <==> p(t)|r(t) - s(t).

What is the dimension of the vector space whose elements are the elements of F[t]/ß. What is the basis of the vector space?
• Oct 22nd 2011, 06:27 AM
Deveno
Re: What is the dimension of this vector space
well, clearly, every multiple of p(t) is congruent to the 0-polynomial. and since we can write, for ANY polynomial f(t) in F[t]:

f(t) = p(t)q(t) + r(t), where the degree of r is less than the degree of p, it's clear that every element of F[t]/β can be written:

r(t) + B, where B is the equivalence class of 0 (and thus p(t)) under β. and since r has degree n-1 or less:

\$\displaystyle \{1+B, t+B, t^2+B,....,t^{n-1}+B\}\$ is a spanning set for F[t]/β. so dim(F[t]/β) ≤ n.

now suppose \$\displaystyle c_0(1+B) + c_1(t+B) +\dots+c_{n-1}(t^{n-1}+B)\$

\$\displaystyle = (c_0 + c_1t +\dots+c_{n-1}t^{n-1}) + B = 0 +B \$

this implies p(t) divides \$\displaystyle c_0 + c_1t +\dots+c_{n-1}t^{n-1}\$,

and since the latter polynomial has degree < degree p(t), it must be the 0-polynomial. that is:

\$\displaystyle c_0 = c_1 = \dots = c_{n-1} = 0\$, so dim(F[t]/β) ≥ n. that leaves just one possibility....
• Nov 11th 2011, 12:53 PM
gotmejerry
Re: What is the dimension of this vector space
Thank you!