Maximal and Prime Ideals

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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 $\mathbb{C}[x]$

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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 $\mathbb{C}[x]$?

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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.