 Solution

The polynomial $f(x)=x^5+2x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb{Q}$ and is the minimal polynomial of every one of its roots. Part (a) givesButIn the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an integral domain, so $(5)$ is not a prime ideal.

For ring homomorphisms preserving the identity, an image $\beta$ of $\alpha$ in the Gaussian integers would satisfy $f(\beta )=0$. The minimal polynomial of $\beta$ over $\mathbb{Q}$ would then be $f$, of degree five. This is impossible because $\beta \in \mathbb{Q}(i)$ has degree at most two. Thus there is no unital homomorphism $\mathbb{Z}[\alpha ]\to \mathbb{Z}[i]$. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.