Prove that $\mathbb{F}_5[x]/\left\langle x^3+3x^2+4x+1\right\rangle$ is a commutative ring and a finite field. 
I want to prove if $R = \mathbb{F}_{5}[x]/(x^3 + 3x^2 + 4x + 1)\mathbb{F}_{5}[x]$ a commutative ring with unity. Later I want to show that $R$ is a finite field.

I started with researching the criteria for a Ring (with unity):

*

*$(R, +)$ is an abelian group.

*$(R, *)$ is a semi-group.

*The distributive laws hold in $R$.

Now I am not sure which would be the best way to show that. Should I take elements like $a,b, c \in R$, and prove with these elements the properties $1-3$. Or what would be the best approach?
Thank you
 A: What should be the case before answering a problem like this is that you would have previously seen a proof that the quotient ring $R=S/I$, where $I$ is an ideal of $S$, has a ring structure.  like this.
And at that point it would be obvious $R$ is commutative if $S$ is.
Then all that would be left is asking if this particular quotient is a field or not, which has many duplicates on the site, like this one.
It kind of looks like you are just jumping on a problem without having read any basic exposition.  Sometimes it is ok to get started struggling with a subject, but it looks premature in your case.
Perhaps it would be worth your while to survey what is in a basic text, finding one of you don’t actually have one.
A: *

*$\mathbb F_5$ is a field.


*$\mathbb F_5[x]$ is a ring.


*$(x^3 - 3x^2 + 4x + 1)$ is an ideal of the ring $\mathbb F_5[x]$.


*Therefore $\mathbb F_5[x]/(x^3 - 3x^2 + 4x + 1)$ is a ring.
By the general fact that if $I$ is an ideal of a ring $R$ then $R/I$ is a ring.
You can prove this by showing that each of the ring axioms hold for elements in the quotient $r + I$, $s + I$.

A ring quotiented by a maximal ideal is a field.
For a field $K$, the maximals ideals of $K[X]$ are the ideals $(f)$ for irreducible polynomials $f$. (For an explanation of why see here https://en.wikipedia.org/wiki/Principal_ideal_domain )
So to prove that this quotient gives a field it suffices to show that the polynomial is irreducible.

You can show that $f(x)$ is irreducible by proving that the translate $f(x-1) = x^3 + x - 1$ is. This can be seen to have no factors since it has no roots mod 5.
