Product of smooth varities is smooth Let $X$ and $Y$ be smooth varieties, then $X\times Y$ is a smooth variety.
Here is my definition of smooth: 
It is enough to show that if $X$ is smooth at $p$ and $Y$ is smooth at $q$ then the product is smooth at $(p,q)$. The smoothness of $X$ at $P$ gives us a $U,\phi,(f_1,..,f_a)$ where $a=n-d$ with the Jacobi having maximal rank. Simillary the smoothness of $Y$ gives a $V,\psi,(g_1,...g_b)$ with $b=m-d'$ such that the Jacobi has maximal rank.
By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety.
I think $\phi\times\psi$ should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank.
 A: If the base field $k$ is algebraically closed and you are willing to use the criterion on the cotangent bundle (Hartshorne Prop III.10.4)
it seems the result follows from the following argument. Let $X,Y \rightarrow Spec(k)$ be smooth varieties of dimension $d,e$. It follows from III.10.4 that
$\Omega^1_{X/k}, \Omega^1_{Y/k}$ are locally trivial of rank $d$ and $e$. There is the following formula (HH Ex II.8.3) for the cotangent bundle of the product:
$$C1.\text{    }p^*\Omega^1_{X\times_k Y/k}\cong p^*\Omega^1_{X/k}\oplus q^*\Omega^1_{Y/k}.$$
It follows $\Omega^1_{X\times_k Y/k}$ has rank $d+e$. Again by III.10.4 it follows $X\times_k Y$ is smooth since $dim(X\times_k Y)=d+e$. This result is mentioned in the relative setting in the same book HH III.10.1d: If $X,Y$ are smooth over $S$ of relative dimension $d,e$ it follow $X\times_S Y$ is smooth of relative dimension $d+e$.
If $X,Y$ are schemes of finite type over $S$, where $S$ a smooth scheme of finite type over $k$, the above argument applies since the formula $C1$ holds in general - hence the above argument holds in the relative setting if you work over an algebraically closed field $k$.
Example. It should be possible to write this out explicitly in terms ot the jacobian matrix as you try above. This may be a good exercise. If you choose open subschemes $U:=Spec(A), V:=Spec(B)$ with $p\in U, q\in V$ you may choose $A:=k[x_1,..,x_n]/(f_1,..,f_a)$ and $B:=k[y_1,..,y_m]/(g_1,..,g_b)$. It follows
$$ \Omega^1_{A/k}\cong A\{dx_1,..,x_n\}/A\{df_1,..,df_a\}.$$
Where
$$ df(x_1,..,x_n):=\frac{\partial f}{\partial x_1}dx_1+\cdots +\frac{\partial f}{\partial x_n}dx_n.$$
Formula $C1$ gives
$$ \Omega^1_{A\otimes_k B/k}\cong \Omega^1_{A/k}\otimes_k B \oplus A\otimes_k \Omega^1_{B/k}.$$
It may be this approach can  be used to prove smoothness at $(p,q)$. I believe this is done in the "Red book of varieties and schemes" - at least the book gives some hints.
Question: "By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety. I think $\phi \times \psi$
should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank."
If $I\subseteq A, J\subseteq B$ are ideals in $k$-algebras $A,B$ it follows
$$ A/I\otimes_k B/J\cong A\otimes_k B/(I\otimes B + A\otimes J).$$
There is an isomorphism $U\times_k V \cong Spec(A\otimes_k B)$ by HH Theorem II.3.3. It seems this proves your open set $U\times_k V \cong Spec(A\otimes_k B)$
has (as you guessed) the following algebra of functions
$$A\otimes_k B\cong k[x_i,y_j](f_u, g_v).$$
Note: It is not always true that products of varieties are varieties. If $X:=Y:=Spec(\mathbb{C})$ and $k:=\mathbb{R}$,  it follows
$$X\times_k Y \cong Spec(\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C})$$
and
$$\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}\oplus \mathbb{C}$$
is not an integral domain. Hence the product of two integral schemes over a field need not be integral.
You find discussions and questions on varieties/schemes that are "geometrically integral" here:
https://math.stackexchange.com/search?q=geometrically+integral
A: $\phi\times\psi$ does indeed give an isomorphism of $U\times V$ with $Z(f_1,\cdots,f_a)\times Z(g_1,\cdots,g_b) \subset \Bbb A^n\times\Bbb A^m$, which is exactly $Z(f_1,\ldots,f_a,g_1,\ldots,g_b)\subset \Bbb A^{n+m}$. We can check the first part by noting that $\phi^{-1}\times\psi^{-1}$ is an inverse to $\phi\times\psi$, and the second part is true because $(x_1,\ldots,x_n)$ satisfies all the $f_i$ and $(y_1,\ldots,y_m)$ satisfies all the $g_j$ iff $(x_1,\ldots,x_n,y_1,\ldots,y_m)$ satisfies all the $f_i$ and the $g_j$.
To check smoothness, it's instructive to write down the Jacobian matrix:
$$\begin{pmatrix}
 \frac{\partial f_1}{\partial x_1} & \ldots & \frac{\partial f_1}{x_n} & \frac{\partial f_1}{\partial y_1} & \ldots & \frac{\partial f_1}{\partial y_m} \\ 
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 \frac{\partial f_a}{\partial x_1} & \ldots & \frac{\partial f_a}{x_n} & \frac{\partial f_a}{\partial y_1} & \ldots & \frac{\partial f_a}{\partial y_m} \\
 \frac{\partial g_1}{\partial x_1} & \ldots & \frac{\partial g_1}{x_n} & \frac{\partial g_1}{\partial y_1} & \ldots & \frac{\partial g_1}{\partial y_m} \\ 
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 \frac{\partial g_b}{\partial x_1} & \ldots & \frac{\partial g_b}{x_n} & \frac{\partial g_b}{\partial y_1} & \ldots & \frac{\partial g_b}{\partial y_m} \\
\end{pmatrix}$$
but $\frac{\partial f_i}{y_j}=0$ and $\frac{\partial g_i}{x_j}=0$ because no $f$ contains any $y$ and no $g$ contains any $x$. So our Jacobian matrix is of the form $$\begin{pmatrix} J_{f,x} & 0 \\ 0 & J_{g,y} \end{pmatrix}$$ where $J_{f,x}$ is the Jacobian of the $f$s with respect to the $x$s and similarly with $J_{g,y}$. But this exactly means that the big Jacobian is of maximal rank since the small ones are.
