To determine which of the given ideals are maximal ideals of $\mathbb{Z}[X]$, we use the property that for a prime number $p$ and a polynomial $f(X) \in \mathbb{Z}[X]$, the ideal generated by $p$ and $f(X)$, denoted as $\langle p, f(X) \rangle$, is a maximal ideal in $\mathbb{Z}[X]$ if and only if the polynomial $f(X)$ is irreducible over the field $\mathbb{Z}_p$. The ring $\mathbb{Z}[X] / \langle p, f(X) \rangle$ is isomorphic to $(\mathbb{Z}/\langle p \rangle)[X] / \langle \overline{f(X)} \rangle = \mathbb{Z}_p[X] / \langle \overline{f(X)} \rangle$, where $\overline{f(X)}$ is the image of $f(X)$ in $\mathbb{Z}_p[X]$. This quotient ring is a field if and only if $\overline{f(X)}$ is irreducible over $\mathbb{Z}_p$. For a polynomial of degree 2 or 3, it is irreducible over a field if and only if it has no roots in that field.

Maximal Ideals Analysis

Let's examine each given ideal:

• Ideal generated by 2 and (1 + X²): This is the ideal $\langle 2, 1 + X^2 \rangle$. We need to check if the polynomial $1 + X^2$ is irreducible over $\mathbb{Z}_2$. Let's evaluate the polynomial for the elements in $\mathbb{Z}_2$, which are 0 and 1.
  • For $X = 0$: $1 + 0^2 = 1 \pmod{2}$
  • For $X = 1$: $1 + 1^2 = 1 + 1 = 2 \equiv 0 \pmod{2}$
  Since $X=1$ is a root of $1 + X^2$ in $\mathbb{Z}_2$, the polynomial $1 + X^2$ is reducible over $\mathbb{Z}_2$. Specifically, $1 + X^2 = (1+X)^2$ in $\mathbb{Z}_2[X]$. Therefore, the ideal $\langle 2, 1 + X^2 \rangle$ is not a maximal ideal in $\mathbb{Z}[X]$.
• Ideal generated by 2 and (1 + X + X²): This is the ideal $\langle 2, 1 + X + X^2 \rangle$. We need to check if the polynomial $1 + X + X^2$ is irreducible over $\mathbb{Z}_2$. Let's evaluate the polynomial for the elements in $\mathbb{Z}_2$, which are 0 and 1.
  • For $X = 0$: $1 + 0 + 0^2 = 1 \pmod{2}$
  • For $X = 1$: $1 + 1 + 1^2 = 1 + 1 + 1 = 3 \equiv 1 \pmod{2}$
  Since there are no roots in $\mathbb{Z}_2$ and the polynomial has degree 2, $1 + X + X^2$ is irreducible over $\mathbb{Z}_2$. Therefore, the ideal $\langle 2, 1 + X + X^2 \rangle$ is a maximal ideal in $\mathbb{Z}[X]$.
• Ideal generated by 3 and (1 + X²): This is the ideal $\langle 3, 1 + X^2 \rangle$. We need to check if the polynomial $1 + X^2$ is irreducible over $\mathbb{Z}_3$. Let's evaluate the polynomial for the elements in $\mathbb{Z}_3$, which are 0, 1, and 2.
  • For $X = 0$: $1 + 0^2 = 1 \pmod{3}$
  • For $X = 1$: $1 + 1^2 = 1 + 1 = 2 \pmod{3}$
  • For $X = 2$: $1 + 2^2 = 1 + 4 = 5 \equiv 2 \pmod{3}$
  Since there are no roots in $\mathbb{Z}_3$ and the polynomial has degree 2, $1 + X^2$ is irreducible over $\mathbb{Z}_3$. Therefore, the ideal $\langle 3, 1 + X^2 \rangle$ is a maximal ideal in $\mathbb{Z}[X]$.
• Ideal generated by 3 and (1 + X + X²): This is the ideal $\langle 3, 1 + X + X^2 \rangle$. We need to check if the polynomial $1 + X + X^2$ is irreducible over $\mathbb{Z}_3$. Let's evaluate the polynomial for the elements in $\mathbb{Z}_3$, which are 0, 1, and 2.
  • For $X = 0$: $1 + 0 + 0^2 = 1 \pmod{3}$
  • For $X = 1$: $1 + 1 + 1^2 = 1 + 1 + 1 = 3 \equiv 0 \pmod{3}$
  • For $X = 2$: $1 + 2 + 2^2 = 1 + 2 + 4 = 7 \equiv 1 \pmod{3}$
  Since $X=1$ is a root of $1 + X + X^2$ in $\mathbb{Z}_3$, the polynomial $1 + X + X^2$ is reducible over $\mathbb{Z}_3$. Specifically, $1 + X + X^2 = (X-1)^2$ in $\mathbb{Z}_3[X]$. Therefore, the ideal $\langle 3, 1 + X + X^2 \rangle$ is not a maximal ideal in $\mathbb{Z}[X]$.

Summary of Maximal Ideals

Based on the irreducibility checks:

The maximal ideals from the given options are $\langle 2, 1 + X + X^2 \rangle$ and $\langle 3, 1 + X^2 \rangle$.