Question: $GF(2^2)$ consists of four elements $0,1,\omega,\omega +1$, where $1+\omega+\omega^2=0$. Verify that $f(y)=\omega+\omega y+\omega y^2+y^3$ is an irreducible polynomial over $GF(2^2)$.
My background is weak on this and I cannot find any material that outlines a procedure for $GF(p^n)$. I'm looking for a procedure that I can do by hand.
Attempt: If $f(y)$ is reducible then we can write $f(y)=g(y)h(y)$, where $\deg(g(y)) < 3$ and $\deg(h(y)) < 3$. In fact, we need $\deg(f(y))=\deg(g(y))+\deg(h(y))$; i.e. one of $g(y)$ or $h(y)$ must be degree 1. Hence, a degree 1 polynomial must divide $f(y)$ for it to be reducible.
My plan was to brute force check that every single degree 1 polynomial of $GF(2^2)$ does not divide $f(y)$, and hence $f(y)$ is not reducible. But I feel uneasy about the elements of $GF(2^2)$ and the "rules" that I need to play by. In other words, what are all of the degree 1 polynomials of $GF(2^2)$? Also, is my procedure valid?
I'm sorry if this is a trivial question. I'd appreciate any help.
Your steps so far are valid. Now, because the polynomial lies in a finite field, instead of checking 16 linear polynomials you only need to check that $f(x)\ne0$ for each of the four elements $x\in\mathbb F_4$. $$f(0)=\omega$$ $$f(1)=1+\omega$$ $$f(\omega)=1$$ $$f(1+\omega)=1$$ Therefore $f$ is irreducible over $\mathbb F_4$.