Count of elements in $\Bbb{Z}_7[x]/(3x^2+2x)$ Hi I have some problem how to get count of elements in $\Bbb{Z}_7[x]/(3x^2+2x)$. I think there belong to only polynomials which are indivisible with $3x^2+2x$ ($\gcd=1$). I think it is so as far I know that for example in every $\Bbb{Z}_m$, $m$ prime, is the count of belonging elements eqauls to $\phi(m)$. But I really dont know how to get these polynomials in some efficient way.
 A: The result follows from the following Theorem:
Theorem. Let $F$ be a finite field of order $q$ and let $p(x)$ be a polynomial in $F[x]$ of degree $n \geq 1$. Then $F[x]/(p(x))$ has order $q^n$.
A: Sima: Working with polynomials with coefficients in ${\mathbb Z}_7$ (but the same is true for any field), any polynomial $p$, when divided by a polynomial $q$ of degree larger than $0$, produces a remainder $r$ that is a polynomial of degree strictly less than $q$. If $q(x)=3x^2+2x$, the remainder $r$ is then a polynomial of degree 1 or less, i.e., it has the form $ax+b$ where $a,b$ are elements of ${\mathbb Z}_7$.
Two elements of ${\mathbb Z}_7[x]$ are identified in the quotient by $3x^2+2x$ iff they have the same remainder, so the elements of ${\mathbb Z}_7[x]/3x^2+2x$ are in correspondence with the remainders that, by the paragraph above, are precisely the linear polynomials $ax+b$. There are 7 possibilities for $a$ and 7 for $b$, for a total of $7^2=49$ elements.  
A: HINT $\ $ Use the division algorithm in $\rm\ \mathbb Z_7[x]\ $ to show that every polynomial $\rm\ f(x)\ \in\ \mathbb Z_7[x]\ $ is congruent $\rm\ mod\ \ 3\ x^2 + 2\ x\ $ to a unique polynomial of degree $\:\le 1\:$, viz. $\rm\ f(x)\ \ mod\ \ 3\ x^2 + 2\ x\:,$ analogous to the fact that every elt of $\rm\ \mathbb{Z}/m\ $ has a unique representative in $\rm\:\{0,1,\cdots,\:m-1\}\:.$
The analogous result holds true over any ring if the leading coefficient of the polynomial is a unit, i.e $\rm\ |R[x]/(f(x))|\ =\ |R|^n\ $ for any $\rm\ f(x) \in R[x]\ $ having degree $\rm\:n\:$ and unit leading coefficient. The hypothesis on the leading coefficient guarantees that one can divide by $\rm\:f(x)\:$ with unique remainder. Indeed, the standard high-school long division algorithm clearly works, and if there were two unequal remainders of degree $\rm < n$ then their difference would be divisible by $\rm\:f\:,$ which is impossible since multiples of $\rm\:f\:$ have degree $\ge n$ (else the leading coefficient of $\rm\:f\:$ is a zero-divisor, not a unit).
