This is an Exercise in An Invitation to Algebraic Geometry by Karen Smith. I'm not sure what the hint means thus have no clue how to approach. Any thought please? Thanks very much!
[Math] Show that the Zariski topology on $A^2$ is not the product topology on $A^1 \times A^1$. (Hint: consider the diagonal.)
algebraic-geometry
Related Solutions
Let $Z\subset\mathbb{A}^n=X$ a closed subset; by definition $$ \exists f_1,\dots,f_r\in\mathbb{K}[x_1,\dots,x_n]=R\mid Z=V(f_1,\dots,f_r)=\bigcap_{i=1}^rV(f_r), $$ withous loss of the generality we can assume $r=1$ and we put $f_1=f$ so that $Z=V(f)$; because $R$ is an U.F.D., then \begin{gather*} \exists g_1,\dots,g_s\in R\mid f=g_1\cdot\dots\cdot g_s,\,g_j\,\text{is a prime polynomial}\\ Z=V(g_1\cdot\dots\cdot g_s)=\bigcup_{j=1}^sV(g_j), \end{gather*} without loss of generality we can assume $s=1$, in other words $f$ is a prime polynomial and $Z$ is an irreducible closed subset of $X$.
Let $\{U_i\}_{i\in I}$ an open covering of $Z$, for exact: $$ Z\subseteq\bigcup_{i\in I}U_i, $$ by previous reasoning, we can assume (without loss of the generality) that $U_i$'s are irreducible; by definition: \begin{gather*} \forall i\in I,\exists f_i\in R\mid U_i=D(f_i)=X\setminus V(f_i),\\ \bigcup_{i\in I}U_i=\dots=X\setminus\bigcap_{i\in I}V(f_i)=X\setminus W; \end{gather*} we know that $W$ is a closed subset of $X$, let $I(W)$ be the associated ideal of $W$, by Hilbert's Base theorem, it is finitely generated; by this statement \begin{gather*} \exists I_F\subseteq I\,\text{finite,}\,\{f_i\in R\}_{i\in I_F}\mid I(W)=(f_i\mid i\in I_F)\Rightarrow\\ \Rightarrow Z\subseteq\bigcup_{i\in I}U_i=X\setminus V(I(W))=X\setminus\bigcap_{i\in I_F}V(f_i)=\bigcup_{i\in I_F}U_i \end{gather*} and the claims follows. (Q.E.D.) $\Box$
Remark: I had use only the hypothesys that $X$ is an affine space over a field, independently from its characteristic and other algebraic properties!
A degenerate conic is given by an equation that factors: that is, the conic $V(ax^2+bxy+cy^2+dxz+eyz+fz^2)$ is degenerate iff $$ax^2+bxy+cy^2+dxz+eyz+fz^2 = (px+qy+rz)(sx+ty+uz).$$ Now, just as you identified the six coefficients of the equation of the conic with points in $\Bbb P^5$, you can identify the coefficients on each of the linear equations with points in $\Bbb P^2$. Expanding the product $$(px+qy+rz)(sx+ty+uz) = (ps)x^2+(pt+qs)xy+(qt)y^2+(pu+rs)xz+(qu+rt)yz+(ru)z^2$$ we see that all the entries on the right hand side are sums of products of one coordinate from the first entry and one coordinate from the second entry. As these products are exactly the coordinates on the Segre embedding and we can take sums of coordinates through a linear projection, we have the result. (I'd recommend working through an example to see how this projection works for yourself - try projection from $[0:-1:1]$ to $V(x_2)$ in $\Bbb P^2$ and note that a point $[a:b:c]$ goes to $[a:b+c:0]$).
Best Answer
As already pointed out in the comments one elegant way to show this is by using that $\mathbb{A^1}$ is not Hausdorff, and thus the diagonal is not closed in the product topology. Another, more image-like proof is the following:
The Zariski topology on $\mathbb{A}^1$ is just the cofinite topology. Hence a basis of the product topology is given by the products $U \times V$ where $U, V \subseteq \mathbb{A}^1$ are both only missing finitely many points (if you try to visualize $\mathbb{A}^2$ as $\mathbb{R}^2$ then you can imagine these basis elements as the plane $\mathbb{R}^2$ where we remove finitely many lines parellel to the axes.)
Now take a look at the diagonal $\Delta = \{(x,x) \mid x \in \mathbb{A}^1\}$. Using the above visualization we can already “see” that the complement of $\Delta$ cannot be written as the union of the basis elements described above.
More formally let $U, V \subseteq \mathbb{A}^1$ be open. If $U \neq \emptyset$ or $V \neq \emptyset$, i.e. if $U \times V \neq \emptyset$, then there exist $x \in U \cap V$ because we are dealing with the cofinite topology and $\mathbb{A}^1$ is infinite. Thus $(x,x) \in U \times V$ and $(U \times V) \cap \Delta \neq \emptyset$. Therefore we cannot write $\Delta^C$, which is non-empty, as the union of basis elements $U \times V$ with $U,V \subseteq \mathbb{A}^1$ open.
So $\Delta$ is not closed in the product topology. But $\Delta$ is closed in the Zariski topology of $\mathbb{A}^2$, because it is the zero locus of $P(x,y) = x-y$.
(Notice however, that the product topology is coarser then the Zariski topology.)
PS: For visualization I also find it very useful to think about the Zariski topology on $\mathbb{R}^2$ and the product of the Zariski topology on $\mathbb{R} \times \mathbb{R}$. Instead of the diagonal an even more intuitive subset is then the unit circle $S^1 \subseteq \mathbb{R}^2$. This is certainly closed in the Zariski topology, as it is given as the zero locus of $x^2+y^2-1$. If $S^1$ were closed in the product topology on $\mathbb{R} \times \mathbb{R}$, then we could write $S^1$ by taking a finite number of lines, all parallel to the axes, and then intersecting this by some other lines, all of which are also parallel to the axes. This seems really absurd.