Consider the quotient ring $\mathbb{C}[x]/(x^{2})$. What do the ideals of $\mathbb{C}[x]/(x^{2})$ look like?
Ultimately I want to look at prime ideals of $\mathbb{C}[x]/(x^{2})$ but I thought this would be a good place to start.
Letting $J=(x^2)$ ideal. Elements of $\mathbb{C}[x]/J$ are represented by $h+ J$ for $h \in \mathbb{C}[x]$ where $h$ has degree at most $1$ without loss of generality. I'm at a loss as to where to go from here. Any tips would be helpful thank you.
On
If $A$ is any ring and $I$ an ideal, then the ideals of $A/I$ correspond to ideals of $A$ which contain $I$.
Now let's consider our case, $A=\Bbb C[x]$ and $I=(x^2)$. Ideals of $\Bbb{C}[x]/(x^2)$ correspond to ideals of $\Bbb C[x]$ which contain $x^2$ (in particular, primes correspond to primes). But $\Bbb C[x]$ is a PID which means any ideal is of the form $(f)$ for some (wlog monic) polynomial $f$. But saying $(f)$ contains $x^2$ is the same as saying we can write $x^2=f(x)g(x)$ for some $g$.
From here, you should proceed by cases. If $f$ is constant, then the ideal is the whole ring. If $f$ has degree $1$, then $f(x)=x+a$ and $g(x)=x+b$, and you can use the identity above to show that $(f)=(x)$. Then do this one more time for when $f$ has degree $2$.
Check which of the arising ideals are prime, and you'll have found all the prime ideals.
One approach is to note that the only non-units of this quotient ring are the multiples of $x$.
Note: Why is $1-x$ a unit?