Show that there is no isomorphism between $K[X]\big/(X^2)$ and $K$ 
Let $K$ be a field and consider $K[X]\big/(X^2)$ and $K$. I want to show that there is no isomorphism between $K[X]\big/(X^2)$ and $K$ as rings.

My attempt was the following:
Proof Let us assume that there exists $$\phi:K[X]\big/(X^2)\stackrel{\sim}{\rightarrow}K$$as rings.
Now using the prime ideal correspondence for quotients we know that $$\left\{\text{prime ideals of}~~ K[X]\big/\left(X^2\right)\right\}\stackrel{\sim}{=}\left\{\text{prime ideals of}~~K[X]~~\text{containing}~~\left(X^2\right)\right\}=\left\{(X)\right\}$$
Now pick $(X)\subset K[X]\big/(X^2)$. Then since $\phi$ is an isomorphism $\phi\big((X)\big)$ is a prime ideal of $K$. Since $K$ is a field this means $\phi\big((X)\big)=(0)\Leftrightarrow(X)=\phi^{-1}\big((0)\big)\Rightarrow (X)=(0)$ which gives a contradiction. Hence there is no isomorphism.
Does this work?
If not could someone show me how to do it using prime ideals?
 A: $K[x]/(x^2)$ is NEVER a field since it is not an integral domain.
($x*x=0$ in the quotient ring, but $x\neq 0$)
A: You cannot use the set (or space) of prime ideals alone. For every commutative ring $R$ and every ideal $I$ and $n \geq 1$, the projection $R/I^n \to R/I$ induces a bijection $\mathrm{Spec}(R/I) \to \mathrm{Spec}(R/I^n)$. It is even a homeomorphism. In particular, $\mathrm{Spec}(K[X]/X^2)$ and $\mathrm{Spec}(K)$ are homeomorphic. That being said, we need to use the specific algebraic structure to differentiate the rings. For example, you can use that $K$ is a field and $K[X]/X^2$ is not (clearly, $[X]$ here is nilpotent and $\neq0$  and hence not invertible). And clearly, this is much easier than using the prime ideals.
A: As supposed, $\mathbb{K}$ is a field. Suppose $\mathbb{K} \cong \frac{\mathbb{K}[x]}{(x^2)}$, then the latter must be a field. This implies that $(x^2)$ is a maximal ideal. However, we know $(x)$ is some proper ideal strictly larger than $(x^2)$. Thus $(x^2)$ cannot be maximal. Thus contradiction arises and there cannot be such an isomorphism.
