Note: $\epsilon$ satisfies $\epsilon^2=0$.
In previous exercises I proved $K[\epsilon]\cong K[X]/(X^2)$ and $K[\epsilon]$ contains precisely three ideals.
Moreover I know the ideals in $K[\epsilon]$ are in the set $J'=\{(0),(\epsilon),K[\epsilon]\}$. Originally I proved their existence using the bijection of the ideals of $K[\epsilon]\cong K[X]/(X^2)$ and the ideals of $K[X]$ that contain $(X^2)$, namely the set $J=\{(1),(X),(X^2)\}$.
Knowing this, there is a bijection $\phi$ between the sets and which gives the relation
$K[X]/j\cong K[\epsilon]/\phi(j)$ for $j \in J$
Now I'm stuck on how to prove $K[\epsilon]^{\times}\cong K^{\times}\times K^+$ . I suspect I should use the Chinese remainder theorem: for $I,J$ coprime ideals of a unitary commutative ring, $I\cap J=I\cdot J$ and $R/(I\cdot J) \cong (R/I)\times(R/J)$.
It seems that there must be some pair $j_1,j_2 \in J'$ with $K[\epsilon]/(j_1\cap j_2) \cong K[\epsilon]/j_1 \times K[\epsilon]/j_2$ giving the required result but I can't see it. Can anyone help?
Since you already know the ideals of $K[\epsilon]$, look at how they are included in one another:
$$(0) \subset (\epsilon) \subset K[\epsilon]$$
We readily see that there is a single maximal ideal - $(\epsilon)$. In fact, this means that $K[\epsilon]$ is a local ring.
Now, recall that invertible elements are precisely the ones that are not contained in any maximal ideal. Thus, invertible elements of $K[\epsilon]$ are of the form $a+b\epsilon$ with $a \neq 0$.
Now, $a+b\epsilon=a\left(1+\frac{b}{a}\epsilon\right)$. So, all invertible elements of $K[\epsilon]$ can be decomposed as $a(1+u\epsilon)$ with $a\neq 0$. Consider how these are multiplied:
$$a(1+u\epsilon) \cdot c(1+w\epsilon)=ac \cdot (1 +(u+v)\epsilon + uv\epsilon^2) = ac (1+(u+v)\epsilon)$$
So, $a$ and $c$ get multiplied, as elements of $K^\times$, while $u$ and $v$ are added, as in $K^+$. Conclude that $K[\epsilon]^\times \cong K^\times \times K^+$.