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Suppose $R$ is a field and $f$ is a polynomial of degree $d$ in $R[x]$. How do you show that each coset in $R[x]/\langle f\rangle$ may be represented by a unique polynomial of degree less than $d$? Secondly, if $R$ is finite with $n$ elements, how do you show that $R[x]/I$ has exactly $n^d$ cosets?

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Do you know about division with remainder in polynomial rings? –  Johannes Kloos Sep 20 '12 at 13:20
    
No, I don't. I'm quite lost in my class... –  Kiyoshi Sep 20 '12 at 13:22
    
Since I suppose this is homework, I'll post an incomplete answer, and you can try to fill in the gaps. –  Johannes Kloos Sep 20 '12 at 13:23
    
Oh wow, thank you Chris. –  Kiyoshi Sep 20 '12 at 13:27
    
If you assume the first part, the second part follows just by counting the number of possible polynomials. –  Ben Millwood Sep 20 '12 at 13:27
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2 Answers 2

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We will use the following fact (it should have been proved in the above-mentioned class): Let $K$ be any field. Then $K[x]$ is an Euclidean ring, i.e., for every two polynomials $f,g \in K[x]$ such that $g \neq 0$, there exist (unique!) polynomials $q, r$ such that $f = g \cdot q + r$ and $\deg r < \deg g$, with $\deg 0 = -\infty$.

Now, let $\overline{g}$ be any coset of $R[x]/\langle f \rangle$, and $g$ its representative. By division with remainder, we have $g = f \cdot q + r$ with $\deg r < \deg f$ and ... (fill in the gaps here).

For the second part, count the polynomials with degree $< d$.

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Hint $\rm\ R[x]/(f)\:$ has a complete system of reps being the least degree elements in the cosets, i.e. the remainders mod $\rm\:f,\:$ which exist (and are unique) by the Polynomial Division Algorithm..

Therefore the cardinality of the quotient ring equals the number of such reps, i.e. the number of polynomials $\rm\in R[x]\:$ with degree smaller than that of $\rm\:f.$

Remark $\ $ This is a generalization of the analogous argument for $\rm\:\Bbb Z/m.\:$ The argument generalizes to any ring with a Division (with Remainder) Algorithm, i.e. any Euclidean domain.

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