There are several versions of the nullstellenzats. We have a particular case from one of those versions: $\hspace{120mm}$ Let $k$ be any algebraically closed field. Let $M$ be a maximal ideal of $k[x,y]$. Then, there exists a point $(a,b)\in\mathbb{A}^{2}(k)$ such that $M=(x-a,y-b)$. Maybe this could be a third idea (this blog is lame): $\hspace{120mm}$ Idea 3. Let ${k}$ be an algebraically closed field. If we have a maximal ideal $M$ of a ring $R$, where the ring depends on ${k}$, then there exist certain points in ${k}$ which correspond in some way to a set of elements in $R$ that generate $M$. $\hspace{120mm}$ Suppose that $f\in{{k}[x,y]}$ is an irreducible polynomial, and define $C_{f}:={k}[x,y]/(f)$. This ring depends on ${k}$. So maybe we can apply our third idea in order to figure out how a maximal ideal of this ring looks like. Since we have a quotient and we know that a maximal ideal in ${k}[x,y]$ is generated by $x-a$ and $y-b$ for some point $(a,b)\in{\overline{k}\times{\overline{k}}}$, we can try the elements $x-a+(f)$ and $y-b+(f)$. From the definitions it follows that an ideal in $k[x,y]$ is maximal if and only if its image in $C_{f}$ is a maximal ideal, and we can verify that $x-a+(f)$ and $y-b+(f)$ also generate the maximal ideal in $R$ if $x-a$ and $y-a$ generate the corresponding maximal ideal in $k[x,y]$. $\hspace{120mm}$ It is interesting to notice that the prime ideal $(f)$ is contained in some maximal ideal $\overline{M}$, and using Nullstellenzats it follows that $\overline{M}=(x-a,y-b)$ for some point $(a,b)\in{\overline{k}\times{\overline{k}}}$. Therefore, $f(x,y)=(x-a)g_{1}(x,y)+(y-b)g_{2}(x,y)$, for some polinomials $g_{1},g_{2}$, and then the point $(a,b)$ is actually a solution of the polynomial $f$. One can prove that there exists a bijection between $Z_{f}(k):=\{(a,b)\in{k\times{k}\;|\;f(a,b)=0}\}$ and $\mathrm{Max}(C_{f}):=\{M\subseteq{C_{f}\;|\;M\mathrm{\;is\;maximal}}\}$. So, this is our fourth idea: $\hspace{120mm}$ Idea 4. If we have a ring $R_{f}$ "depending" on a polynomial $f$ and an algebraically closed field $\overline{k}$, then it's likely to find a bijection between the zeroes of $f$ (over $\overline{k}$) and the maximal ideals of $R_{f}$ .