Show that $\mathbb{F}_p[x]/ ⟨f(x)⟩$ is a field of characteristic $p$

Question

Let $\mathbb{F}_p[x]/⟨f(x)⟩$ with $\mathbb{F}_p= \mathbb{Z}/5\mathbb{Z}$, $\mathbb{F}_p[x]= \mathbb{Z}/5\mathbb{Z}$, $\mathbb{F}_p[x]$ the ring of polynomials with coefficients in $\mathbb{F}_p$, and $f(x)=x^2-ax-b \in \mathbb{F}_p[x]$. I have to show that this set is a field. Thing is, I can't figure out what an object inside this set looks like so it makes it hard to check if it's a field. Any ideas? Thanks a lot!

Answer 1

The ring $\mathbb{F}_p[x]/\langle f(x) \rangle$ is usually defined to be the set of polynomials in $\mathbb{F}_p[x]$ of degree smaller than $\deg f(x)$ with addition in the ring being ordinary polynomial addition (which doesn't cause the degree of the sum to increase to $\deg f(x)$ or more, and multiplication in the ring being defined as polynomial multiplication (in $\mathbb{F}_p[x]$) followed by computation of the residue of this product modulo $f(x)$ so that the result is once again of degree smaller than $\deg f(x)$. Thus, $\mathbb{F}_p[x]/\langle f(x) \rangle$ is closed under polynomial addition and polynomial multiplication modulo $f(x)$.

$\mathbb{F}_p[x]/\langle f(x) \rangle$ is not a field unless $f(x)$ is an irreducible polynomial, that is, $f(x)$ cannot be factored into two polynomials both of which have positive degrees. So, what the OP is trying to prove is not true in general but will be true for certain choices of $a$ and $b$ in $\mathbb F_p$.

Answer 2

An arbitrary element in this can be represented by a degree one polynomial with coefficients in $\mathbb{F}_{5}$, that is because upon taking the quotient, you have $x^2=ax+b$, so given any polynomial, whenever you see $x^{2}$ term simply replace it with $ax+b$. (similarly, when $x^3=x^2x=(ax+b)(x)=ax^2+bx=a(ax+b)+b$.)

Moreover, it's well known that the quotient ring is a field iff the polynomial is irreducible. If $x^2-ax-b$ were reducible, then if you attempt to use the quadratic formula, you have $x= \frac{a\pm \sqrt{a^{2}+4b}}{2}\equiv3(a\pm\sqrt{a^2+4b})$. This has root, say take $a=1,b=1$,so your statement is not true in general.