List all elements of $\mathbb{Z}_5[x]/\langle x^2+3x+1\rangle$

Question

List all elements of $\mathbb{Z}_5[x]/\langle x^2+3x+1\rangle$

Is there a simple way to solve these kind of questions (only using pen and paper)?

If I understand it correctly, I can multiply each element of $\mathbb{Z}_5[x]$ with each other and replace all $x^2$ with $2x+4$, then $x^3$ with $4x^2+x+1$, etc.

But there are way too many elements in $\mathbb{Z}_5[x]$. Is there an easier way?

Answer 1

First of all, note that all elements of $\mathbb{Z}_5[x]$ of the form $px+q$ with $p,q\in\mathbb{Z}_5$ are different when considered modulo $x^2+3x+1$ (their difference is only a multiple of $x^2+3x+1$ when they are equal).

Now let $ax^2+bx+c\in \mathbb{Z}_5[x]$ be a quadratic polynomial. Then: $$ ax^2+bx+c-a(x^2+3x+1)=(b-3a)x+(c-a) $$ So we see that: $$ax^2+bx+c\equiv (b-3a)x+(c-a)\mod x^2+3x+1$$ Hence all quadratic polynomials are equivalent to some polynomial of the form $px+q$.

We can inductively reduce the degree of any polynomial of degree $\geq 2$ modulo $x^2+3x+1$ in this way: $$ (a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0)-a_nx^{n-2}(x^2+3x+1)=(a_{n-1}-3a_n)x^{n-1}+\cdots $$ So: $$ (a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0)\equiv (a_{n-1}-3a_n)x^{n-1}+\cdots\mod x^2+3x+1 $$ Therefore we find that any element of $\mathbb{Z}_5[x]$ is (modulo $x^2+3x+1$) equivalent to a "linear" polynomial. The elements of $\mathbb{Z}_5[x]/\langle x^2+3x+1\rangle$ are therefore precisely the equivalence classes of the polynomials of the form $px+q$ with $p,q\in\mathbb{Z}_5$. Since $\mathbb{Z}_5$ has $5$ elements, there are $5\cdot 5=25$ such polynomials.

We did not really use a special property of $x^2+3x+1$ or $\mathbb{Z}_5$ here. In general, when we have a finite finite $F$ with $n$ elements, and a polynomial $P$ of degree $d$ in $F[x]$, then the ring (field iff $P$ is irreducible) $F[x]/\langle P\rangle$ will have $n^d$ elements (you can try to construct the argument in a general case for yourself, you will need that each nonzero element of $F$ has a multiplicative inverse).

Answer 2

Hint: $\Bbb{Z}_5$ is a finite field. $\mathbb{Z}_5[x]/\langle x^2+3x+1\rangle$ is a vector space of finite dimension over that field. If you are being asked this question, you should know how to calculate the number of elements in the field and the dimension of the vector space. That should cut down "way too many" to A to Z with one letter spare.

Answer 3

$\mathbb{Z}_5[x]/\langle x^2+3x+1\rangle$ is the set of classes of polynomials mod $x^2+3x+1$ with coefficients in $\mathbb{Z}_5$.

By the Euclidean division of polynomials, each element has a canonical representative of the form $ax+b$ with $a,b\in \mathbb{Z}_5$.

Therefore, the ring has $5\times 5=25$ elements.