Consider the polynomial $f(x) = x$ in $\mathbb{Z}/6\mathbb{Z}[x]$, which is reducible as it factors as $(3x+4)(4x+3)$.
Determine all the factorizations of $f(x)$ in $\mathbb{Z}/6\mathbb{Z}[x]$. [Use the Chinese Remainder Theorem.]
I'm having trouble with this. We know the factorization of $f(x)$ modulo $(2)$ must be $1 \cdot x$ and the factorization of $f(x)$ modulo $(3)$ is also $1 \cdot x$ since $f(x)$ is irreducible in these fields. By the Chinese Remainder Theorem we know $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. We can make a homomorphism
\begin{align} \phi: \mathbb{Z}/6\mathbb{Z}[x] &\to \mathbb{Z}/2\mathbb{Z}[x] \times \mathbb{Z}/3\mathbb{Z}[x] \\ p(x) &\mapsto (p(x)+(2), \ p(x)+(3)). \end{align}
I don't know what to do from here, any suggestions?
$$ \def\Z{\mathbb{Z}} $$
The image of $x$ under the homomorphism you gave is the pair $(x, x)$. Since $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are fields, the polynomial $x$ is irreducible over them (degree), and so this pair can be factored in a total of four ways (where we are counting $2x\cdot2$ and $x\cdot1$ as different factorizations in $\Z/3\Z$ i.e. we are not ignoring units): $$ (1, 1) \cdot (x, x) \hspace{1pc} (1, 2) \cdot (x, 2x) \hspace{1pc} (1, x) \cdot(x, 1) \hspace{1pc} (x, 2) \cdot (1, 2x) $$ Now we have to lift these factorizations to $\mathbb{Z}/6\mathbb{Z}[x]$, i.e. invert the CRT isomorphism, which is the fun part. Note that under the map $$ \mathbb{Z}/6\mathbb{Z} \to \Z/2\Z \times \Z/3\Z $$ the image of $3$ is $(1, 0)$ and the image of $4$ is $(0, 1)$. Therefore, the inverse of this map must be $(a, b) \mapsto 3a + 4b \mod 6$. So our factorizations lift to: $$ 1\cdot x \hspace{1pc} 5(5x) \hspace{1pc} (3 + 4x)(3x + 4) \hspace{1pc} (3x+2)(3+2x). $$ (Again, two of these factorizations only differ by the unit $5 \in \mathbb{Z}/6\mathbb{Z}$, so should probably be counted as the same.)