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Let $k$ be a field. Then $k[x]/(x^2)$ has only one prime ideal.

It is clear that $(X)$ is the maximal ideal thus prime where $X$ is the class elements $x$. How do I know there aren't other prime ideals? I tried to define homomorphism $\phi':k[x]/(x^2)\to k$ by evaluation but it seems there is no other way to define the map well defined other than evaluating at 0 due to $\phi'([f])=\phi'([g])$ not equal for different representatives.

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    Accepted solution at linked question explains this very well. – 2017-02-15

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Suppose $\mathfrak{p}$ is a prime ideal. Then $x\cdot x = 0\in \mathfrak{p}$, so $x\in \mathfrak{p}$, and $(x)\subseteq \mathfrak{p}$. Since, as you note, $(x)$ is maximal, $(x) = \mathfrak{p}$.

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Don't forget nontrivial assumption. Anyway suppose the representative ax+b is contained. Then as will (1/b×x)(ax+b). This tells you the class of 1 is there too after subtraction of ax and division by b. That then says everything is. Notice b was nonzero for this. That just leaves you with (x) as you already have.

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    By (1/b×x), you mean $b^{-1}x$, right? Then $b^{-1}x(ax+b) = x$, not $1$. – 2017-02-11
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    This approach can be turned into a solution, but a few more details would need to be filled in (e.g. why can we assume that $ax+b$ with $b\neq 0$ is in our ideal?) – 2017-02-11