# 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.

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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 $q^n$.

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Yes, many thanks, it is what I need..Am I right this is the galois field? And I need to get all these polynomials..I remember it is over any table, but I dont know exactly how to do that.. Now I think I know..I take all polynomials of degree less than degree n above..And I simply multiply that..AM I right? – simekadam Jan 9 '11 at 20:10
Wait..I found in my math book that , this theorem is valid only iff the polynomial p(x) is irreducible…Either I am too confused, or this one is not irreducible???Because of x(3x+2), or not?? //to ashamed:) – simekadam Jan 9 '11 at 20:28
If the polynomial is irreducible then the resulting quotient ring is a field. But the statement about the size of the quotient ring is true in the general case. – Zhen Lin Jan 9 '11 at 20:51
ok ok..My stack has overflowed:)but thanks I hope I got it.. – simekadam Jan 9 '11 at 21:11

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.

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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).

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