On page 587 in Dummit and Foote, we are given that $f_1(x)=x^4+x^3+1$ and $f_2(x)=x^4+x+1$ are two irreducible quartics over $\mathbb{F}_2$.
The authors then claim that a "simple calculation" verifies that $\alpha (x)=x^3+x^2$ is a root of $f_2(x)$ in $\mathbb{F}_{16}=\mathbb{F}_2[x]/(x^4+x^3+1)$, and we may use this fact to verify that $\mathbb{F}_2[x]/(x^4+x+1)\cong\mathbb{F}_2[x]/(x^4+x^3+1)$ with $x\mapsto x^3+x^2$.
However, I am having difficulty verifying this calculation. We have that $f_2(\alpha(x))=(x^3+x^2)^4+(x^3+x^2)+1=x^{12}+x^8+x^3+x^2+1$, yet I don't see why this should equal $0$ in $\mathbb{F}_2[x]/(x^4+x^3+1)$. I know that I will have to use the fact that $x^4+x^3+1=0$ in this field, but I'm not sure how. What am I missing?
The answer is easy. We have $$ x^{12}+x^8+x^3+x^2+1= (x^8 + x^7 + x^6 + x^5 + x^4 + x^2 + 1)(x^4 + x^3 + 1)=0 $$ because $x^4+x^3+1=0$.