Find the number of elements in the factor group $\mathbb Z_5[x]/A$ where $A=\langle x^2+x+3\rangle$

Question

I am very confused about factor groups of polynomials and I would appreciate some help.

I know that in the coset representation $x^2=-x-3$ and so all cosets are in the form $ax+b+A$. Now let us take the coset $x+1+A$ we know it is not the zero element as $x+1 \not \in A$. Now $$ax+b+A=(a(x+1)+b-a)+A=(x+1)+A +(b-a)+A$$ and so when $b=a$ we get $(x+1)+A$. Hence from $25$ elements, we get $21$.

I am almost sure this is wrong, and I am confused what I am doing wrong. Any help would be appreciated. Thanks!

Answer 1

As long as $\,f(x)$ is a monic polynomial of degree $2$, $\Bbb F_5[x]/\langle f\rangle$ will be a vector space of dimension two over $\Bbb F_5$, and thus have $25$ elements.

Indeed, calling $\langle f\rangle=A$, as you did, we see that $1+A$ and $x+A$ make up a basis of the factor group: if $\alpha+\beta x+A$ is zero in the factor group, then $\alpha+\beta x\in A$, i.e. $\alpha+\beta x=g\cdot f$ for some polynomial $g\in\Bbb F_5[x]$, impossible unless $g=0$, which says that $\alpha=\beta=0$, which shows linear independence of $1+A$ and $x+A$. That every $Z\in\Bbb F_5[x]/A$ is of form $\alpha+\beta x+A$ follows by applying Euclidean division to any polynomial $z(x)$ for which $z+A=Z$.

Answer 2

First, let's check if the ideal is prime (maximal) or not. Brute-force checking gives us that $1$ is a root of the polynomial, so the polynomial is $(x+4)(x+2)$. These are coprime ideals, so the Chinese Remainder Theorem gives that the quotient ring is isomorphic to $\mathbb{Z}_5[x]/(x+4)\oplus\mathbb{Z}_5[x]/(x+2)$. Each of these has exactly 5 elements (we're just setting $x=3,1$), so there are 25 elements total.

Answer 3

I think your mistake is when you wrote: $a(x+1)+b-a + A=(x+1)+A+b-a+A$,

why did $a(x+1)$ become $x+1$?