Maximal and Prime Ideals I was assigned these problems for homework to designate if they were maximal, prime or neither.  I was able to determine that (a) was solely prime by showing $\mathbb{Z}[x] /(x-1)$ is isomorphic to $\mathbb{Z}$ which is an integral domain.  I am having troubles in answering the other 3 though and could use some help. Thanks.
(a) $(x − 1)$ in $\mathbb{Z}[x]$
(b) $(x − 1)$ in $\mathbb{R}[x]$
(c) $(x^2 + 1)$ in $\mathbb{R}[x]$
(d) $(x^2 + 1)$ in $\mathbb{C}[x]$
 A: Hints: 
(b) See if you can show that $\mathbb{R}[x]/(x-1) \cong \mathbb{R}$ in a similar manner to (a). From this, you conclude...
(c) Is $(x^{2}+1)$ reducible over $\mathbb{R}[x]$? What do you know about irreducible elements of PIDs?
(d) Is $(x^{2}+1)$ reducible over $\mathbb{C}[x]$?
A: For the others, it's not too bad. For $\Bbb R[x]/(x-1)$ consider the map
$$\phi_1: \begin{cases}\Bbb R[x]\to\Bbb R \\ f(x)\mapsto f(1)\end{cases}$$
The kernel is all polynomials which are a multiple of $x-1$, and the map is clearly onto, hence the image is isomorphic to the desired quotient by the first isomorphism theorem. Since the quotient is a field, the ideal is maximal.
For the next one, Consider the map
$$\phi_2:\begin{cases}\Bbb R[x]\to \Bbb C \\ f(x)\mapsto f(i)\end{cases}$$
Again the kernel is generated by the polynomial in question and the map is onto, hence we have an isomorphism, and again the ideal is maximal.
Finally, we define a third map
$$\phi_3:\begin{cases}\Bbb C[x]\to\Bbb C \\ f(x)\mapsto f(i)\end{cases}$$
Clearly our ideal is contained in the kernel, but because $x^2+1=(x+i)(x-i)$ we see there are zero divisors in our image, hence the ideal is not prime.
