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 $R_{f}$ .

Nullstellenzats

The other day I was trying to explain the Nullstellenzats theorem to my mother. It was somewhat hard because she had never taken a course in mathematics, and it was even worse because I'm still having problems understanding and using this theorem LOL, but anyway I managed to draw a simple picture which illustrates it: $\hspace{100mm}$

I explained to her the next nulltellensatz version: Let $k$ be an algebraically closed field, and let $I$ be an ideal of the ring of polynomials $k[x_{1},\dots,x_{n}]$. If a polynomial $f\in{k[x_{1},\dots,x_{n}]}$ is such that $f(\alpha)=0$ for all $\alpha\in{Z(I)}$, then $f\in{\sqrt{I}}$, where $Z(I)=\{\beta\in{k}\;|\;g(\beta)=0\;\forall\;g\in{I}\}$. It's simple: We draw two bags. The first bag on the left hand side represents $I$, and each point inside it represents an element $f\in{I}$. The second bag represents $Z(I)$, and each point inside it represents an element $\beta\in{Z(I)}$. An arrow that goes from a point belonging to the first bag (i.e., an element $f\in{I}$) to a point belonging to the second bag (i.e., $\beta\in{Z(I)}$) indicates that the first point "vanishes" on the second point (i.e., $f(\beta)=0$). From the hypothesis, we find that there is arrow from each point belonging to the first bag to a fixed point belonging to the second bag, and this happens for each fixed point in the second bag. We also have a point outside the first bag (i.e., the polynomial $f$) which vanishes on every point belonging to the second bag (so, there's an arrow that goes from the point $f$ to each point inside the second bag). So, Nullstellezats basically says that somehow $f$ "belongs to" the first bag, namely, choosing an appropriate positive integer $n$ such that $f^{n}\in{I}$. And finally, the white background of the picture reminds us that we can do all this only if $k$ is algebraically closed :D

Easy problem (it was hard for me though lol): $\hspace{120mm}$ Let $M_{i}\subseteq{\mathbb{Z}}$, $i=1,2,\dots$, be a sequence of subsets such that: $\;\;\;$ (i) $M_{1}\supseteq{M_{2}}\supseteq{\dots}$ $\hspace{70mm}$ (ii) If $a,b\in{\mathbb{Z}}$ are such that $ab\in{M_{i}}$, then either $a\in{M_{i}}$ or $b\in{M_{i}}$. $\hspace{90mm}$ Prove that, if $ab\in{\cap_{i=1}^{\infty}}M_{i}$, then either $a\in{\cap_{i=1}^{\infty}}M_{i}$ or $b\in{\cap_{i=1}^{\infty}}M_{i}$. $\hspace{100mm}$ $\hspace{100mm}$ Solution: From the hypothesis, we obtain that $ab\in{M_{i}}$ $\forall$ ${i=1,2,\dots}$. If we have $a\in{M_{i}}$ $\forall$ $i$ or $b\in{M_{i}}$ $\forall$ $i$, then we're done. If not, let $r$ the least positive integer such that $a\in{M_{r}}$ but $a\notin{M_{r+1}}$ (it follows that $a\notin{M_{i}}$ $\forall$ $i\geq{r+1}$, since $M_{r+1}\supseteq{M_{r+2}}\supseteq{\dots}$). Similarly, let $s$ the least positive integer such that $b\in{M_{s}}$ but $b\notin{M_{s+1}}$ (it follows again that $b\notin{M_{i}}$ $\forall$ $i\geq{s+1}$). Since $ab\in{M_{r+1}}$ and $a\notin{M_{r+1}}$, we find that $b\in{M_{r+1}}$, and then by the definition of $s$ it follows that $r+1\leq{s}$. Analogously, we obtain that $s+1\leq{r}$. But this implies that $r+1\leq{s}<{s+1}\leq{r}$, which is a contradiction. Now we're done. $\;\;\;\;\;$ $\hspace{100mm}$ (I think...)

ideas algunas

---------------------------------------------------------------------------------- Idea 1. Let $k$ be any field. Suppose that $x,y$ are elements such that $k(x),k(y)$ are isomorphic as $k$-algebras to the field of rational functions. If you adjoin the elements $x,y$ to the field $k$ and occurs that $k(x,y)/k(x)$ is a finite extension (i.e., $y$ is algebraic over $k(x)$), then there must be some property concerning to a field extension of $k(y)$ in order to prove that $x$ is algebraic over $k(y)$, and therefore, that $k(x,y)/k(y)$ is also finite. ---------------------------------------------------------------------------------- Idea 2. If you have a finite extension $k(x)/k$, then take the minimal polynomial of $x$ over $k$. Reciprocally, if you want an extension $k(x)/k$ to be finite, try to prove that $x$ is algebraic over $k$ (and make sure it actually happens over $k$). ---------------------------------------------------------------------------------- Example: Let $k$ be any field, and let $L/k(x)$ and $K/k(y)$ be two finite extensions. Assume that both $k(x)$ and $k(y)$ are isomorphic to the field of rational functions in one variable. Assume that $K\subseteq{L}$. Then $L/K$ is a finite extension.____________________________ Solution: We can try to apply our idea 1, since $L/k(x)$ is finite and $k(x)\subseteq{k(x,y)}\subseteq{L}$ both implies that $k(x,y)/k(x)$ is finite. We would like to show that $k(x,y)/k(y)$ is finite (no matter what is the problem asking us to prove), so following our idea 1, we must find some property that shows what we're looking for. We find it using our idea 2. Taking $f(Y)$ the minimal polynomial of $y$ over $k(x)$, we notice that since $y$ is algebraic over $k(x)$, then it follows that $x$ is algebraic over $k(y)$ (and therefore $k(x,y)/k(y)$ is finite) if and only if $f(Y)\notin{k[Y]}$ (otherwise we couldn't get a polynomial in terms of a variable $X$ with coefficients over $k(y)$ where $y$ vanishes). Since $f(Y)\notin{k[Y]}$ because $y$ is not algebraic over $k$, we're done.