It is wellknown that for a polynomial ring $\mathbb{Z}_p[x]$, $\mathbb{Z}_p[x]/\langle p(x) \rangle$ for prime $p$ is a field if and only if $p(x)$ is irreducible over the given polynomial ring, in turn making $\langle p(x) \rangle$ a maximal ideal. What exactly are the elements of this quotient ring? I have developed an intuition of thinking about quotient groups in terms of cosets, but that doesn't seem to apply as well here, at least on the intuitive level.

  • $\begingroup$ Of course, let me fix that. $\endgroup$ – Andrew Thompson Apr 14 '14 at 10:12

I expect you are familiar with modular arithmetic. One of the intuitive ways that you can think of arithmetic modulo $5$, say, is that it is just normal arithmetic except that $5$ is "the same as" $0$. For example, $$3\times4=12=2\times5+2=2\times0+2=2\ .$$

You can think of this in much the same way: ${\Bbb Z}_n[x]$ is the set of all polynomials with coefficients modulo $n$: for example, in ${\Bbb Z}_5[x]$ we have $$\eqalign{(1+2x+3x^2)(4+3x+2x^2) &=4+11x+20x^2+13x^3+6x^4\cr &=4+x+3x^3+x^4\ ,\cr}$$ because $5$ "is the same as" $0$. In ${\Bbb Z}_n[x]/\langle p(x)\rangle$ you go one step further and say that $p(x)$ also "is the same as" the zero polynomial. For example, continuing the above example, in ${\Bbb Z}_5[x]/\langle x^3+x+1\rangle$ we have $$\eqalign{(1+2x+3x^2)(4+3x+2x^2) &=4+x+3x^3+x^4\cr &=(x^3+x+1)(x+3)+(4x^2+2x+1)\cr &=4x^2+2x+1\ .\cr}$$ Hope this helps!

  • $\begingroup$ Oh, okay, so using this logic, one could consider the quotient ring, which is now a field, to be the "a field of cosets", and we simply set $p(x)$ to act as zero? That does coincide with my intuition of quotient groups and rings. How would one go about, say, finding the number of elements in this field? Is it $125$, due to the fact that we can always factorize out some multiple of $p(x)$ for degrees higher than three (given non-zero coefficients), giving $5^3$ elements? $\endgroup$ – Andrew Thompson Apr 14 '14 at 10:29
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    $\begingroup$ I'm not sure "factorise" is quite the right word, but apart from that you are exactly correct. We are identifying each coset by its unique representative with degree less than $3$, just as in arithmetic modulo $5$ we identify each coset by its unique representative from $0$ to $4$. $\endgroup$ – David Apr 14 '14 at 10:33

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