Irreducible algebraic sets of $\mathbb A^2(k)$ I want to show that the irreducible algebraic sets of $\mathbb A^2(k)$ are exactly the following:
$$\mathbb A^2(k),\emptyset,\text{Singleton and algebraic curves}.$$
Of course all of them are irreducible algebraic sets of $\mathbb A^2(k)$, but how can I show that there are exactly those sets?
 A: The case of a general field $k$
If $k$ is a field the points of the scheme $\mathbb A^2_k=\operatorname {Spec}k[X,Y]$ correspond to the prime ideals $\mathfrak p\subset k[X,Y]$.
Since $k[X,Y]$ has dimension $2$ (Matsumura, page 35), the height of such a prime ideal satisfies $0\leq \operatorname {ht}(\mathfrak p)\leq 2$.    
$\boxed {\operatorname {ht}(\mathfrak p)=0}$
Then $\mathfrak p=(0)$ and the corresponding variety is $\mathbb A^2_k$ itself.
$\boxed {\operatorname {ht}(\mathfrak p)=1}$
Any irreducible  polynomial $0\neq f\in \mathfrak p$ generates a prime ideal $(f)$ which, since $(0)\neq (f)\subseteq \mathfrak p$, must be equal to $\mathfrak p$ for height reason.
Hence the subvariety corresponding to $\mathfrak p$ is the curve $V(f)$.
Example for $k=\mathbb R: f=X^2+Y^2+1$, which gives rise to the curve $V(X^2+Y^2+1)\subset \mathbb A^2_k$ .
Beware that this curve has infinitely many points but none of those correspond to a pair $(r,s)\in \mathbb R^2$.
$\boxed {\operatorname {ht}(\mathfrak p)=2}$
Then, since $\dim k[X,Y]=2$, $\mathfrak p$ is a maximal ideal.
Zariski's version of the Nulstellensatz (Atiyah-Macdonald, Prop. 7.9, page 82) then implies that the corresponding subscheme is a single point, but maybe not rational, with residue field $\kappa(\mathfrak p)=k[X,Y]/\mathfrak p$ of finite dimension over $k$.
Example for $k=\mathbb R: \mathfrak p=(X^2+1,Y)$ and $\kappa(\mathfrak p)=\mathbb C$. 
The classical case where $k$ algebraically closed
The above is naturally still valid but the maximal ideals are now all of the form $(X-a,Y-b)$ with $a,b\in k$ and they correpond to the good old points $(a,b)\in k^2$, just as Descartes described them for us in 1637.
A: Let $X\subseteq \mathbb{A}^2$ be an irreducible algebraic set. If $X$ is finite or the ideal $I(X)$ is zero, then $X$ is $\mathbb{A}^2$, $\emptyset$ or a single point. Here
$$
I(X)=\{ f\in k[X,Y]\mid f(p)=0 \text{ for all } p\in X\}.
$$
Otherwise the ideal $I(X)$ in $k[X,Y]$ is non-zero
so there is at least one non-zero polynomial $f\in I(X)$. Now $I(X )$ is prime,
because $X$ is irreducible, so it must also contain an irreducible factor of $f$. So w.l.o.g. let $f$ be irreducible. Then we have that $I(X ) = ⟨f⟩$. To see this, let $h \in I(X ) \setminus ⟨f⟩$. Now $h$ and $f$ are relatively prime, so we can use another result for relatively prime polynomials 
that $V(f,h)=V(f)\cap V(h)$ is finite, so that $X\subseteq V(f,h)$ is finite, which we have excluded above.
