Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 4 down vote accepted

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 $q^n$.

share|improve this answer
    
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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.