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\begin{document}

For a polynomial $f \in K[x_1, \cdots, x_n]$, its \textbf{Jacobian Ideal} is defined as
\[ \mathsf{J}(f) := \langle \frac{\partial f}{\partial x_1}, \cdots, \frac{\partial f}{\partial x_n} \rangle, \]
and the \textbf{Mapping-Jacobian Ideal} of $f$ is defined as
\[ \mathsf{MJ}(f) := \langle y_1 - \frac{\partial f}{\partial x_1}, \cdots, y_n - \frac{\partial f}{\partial x_n} \rangle, \]
where $y_1, \cdots, y_n$ are newly introduced distinct variables. Note that singular points of $f$ are those also vanishing on all its partial derivatives, and hence
\[ \Sing(f) = V(f, \frac{\partial f}{\partial x_1}, \cdots, \frac{\partial f}{\partial x_n}) = V(f, \mathsf{J}(f)).\]
\vspace{5ex}

\begin{lemma}
Isomorphic varieties have the same dimension.
\end{lemma}



\noindent \textbf{The Join of Two Varieties.}
\begin{definition}
Given any two disjoint projective varieties $X, Y \in \PP^n$, the join of $X$ and $Y$, denoted by $Join(X,Y)$, is defined as the union of the lines joining $X$ to $Y$:
\[ \join(X,Y) = \bigcup_{x \in X, y \in Y} \overline{x,y}.\]
\end{definition}
It is easy to see that this join $Join(X,Y)$ is a subvariety of the Grassmannian $\G(1,n)$ which is proved to be a projective variety, and hence the union of lines $Join(X,Y)$ is also a subvariety of $\PP^n$.

\begin{lemma}[{\cite{HarAG}}]\label{lm:joinGdeg}
Let $X,Y \in \PP^n$ be any two disjoint projective varieties,
\[ \gdeg(\join(X,Y)) = \gdeg(X) \cdot \gdeg(Y).\]
\end{lemma}
\begin{proof}
It suffices to prove this in the special case where $X$ and $Y$ live in complementary linear subspaces $\PP^m$ and $\PP^{n-m-1} \subset \PP^n$, because any join may be realized as the regular projection of such a join.

Let $\dim(X) = k$ and $\dim(Y) = l$. We take $\Lambda_X$ be a $(m-k)$-dimensional general plane intersecting $X$ transversely. Similarly, let $\Lambda_Y$ intersect $Y$ transversely. Let $\Lambda^* = \join(\Lambda_X, \Lambda_Y)$ be the subspace spanned by $\Lambda_X$ and $\Lambda_Y$. Now consider
\[
\Lambda^* \cap \join(X,Y).
\]
Note that a point in $\join(X,Y)$ gives a point in $X$ and a point in $Y$. Then a point in $\Lambda^* \cap \join(X,Y)$ should consist of a $X$-component in $\Lambda_X$ and a $Y$-component in $\Lambda_Y$. Thus this intersection should have all lines passing through a point in $X \cap \Lambda_X$ and a point in $Y \cap \Lambda_Y$.

Moreover, if $\Lambda_X$ and $\Lambda_Y$ intersect $X$ and $Y$ transversely, then the intersection $\Lambda^* \cap \join(X,Y)$ is generically transverse. Hence it intersects this set of lines in precisely
\[
| \Lambda_X \cap X | \cdot | \Lambda_Y \cap Y | = \gdeg(X) \cdot \gdeg(Y).
\]
So $\gdeg(\join(X,Y)) = \gdeg(X) \cdot \gdeg(Y)$.
\end{proof}


In affine case, we can also find a variety analogous to that in projective case.
\begin{definition}
Given any two affine varieties $X \subset \A^n$ and $Y \subset \A^m$, the product of them is defined as
\[ \Prod(X,Y) = X \times Y \subset \A^{n+m}.\]
\end{definition}
By definition, $\Prod(X,Y)$ is also an affine variety because $\Prod(X,Y) = V(S_1,S_2)$ with $X = V(S_1)$ and $Y = V(S_2)$. Again using the proof technique similar to that in~\Cref{lm:joinGdeg}, we derive the following

\begin{lemma}
Let $X \in \A^m$ and $Y \in \A^{n-m}$ be any two disjoint affine varieties,
\[ \gdeg(\Prod(X,Y)) = \gdeg(X) \cdot \gdeg(Y).\]
\end{lemma}
\begin{proof}
Let $\dim(X) = k, \dim(Y) = l$. Let $\Lambda_X$ be a $(m-k)$-dimensional general plane intersecting $X$ transversely such that $\gdeg(X) = | \Lambda_X \cap X | $ and $\Lambda_Y$ be a $(n-m-l)$-dimensional general plane intersecting $Y$ transversely such that $\gdeg(Y) = | \Lambda_Y \cap Y|$. Take $\Lambda^*$  be the general plane spanned by $\Lambda_X$ and $\Lambda_Y$. So the dimension of $\Lambda^*$ is $n-k-l$.

Now we claim that $|\Lambda^* \cap \Prod(X,Y)|$ gives the geometric degree of $\Prod(X,Y)$. Via a similar argument in~\Cref{lm:joinGdeg}, we have
\[
|\Lambda^* \cap \Prod(X,Y)| = | \Lambda_X \cap X | \cdot | \Lambda_Y \cap Y| = \gdeg(X) \cdot \gdeg(Y).
\]
It is left to argue that $\gdeg(\Prod(X,Y)) = |\Lambda^* \cap \Prod(X,Y)|$, that is, the intersection of $\Lambda^*$ and $\Prod(X,Y)$ is indeed maximum. However, it is easy to see that any $(n-k-l)$-dimensional general plane ${\Lambda^*}'$ can be decomposed into $\Lambda'_X$ of dimension $(m-k)$ and $\Lambda'_Y$ of dimension $(n-m-l)$, and extra intersections between ${\Lambda^*}'$ with $\Prod(X,Y)$ will give extra intersection points in $\Lambda'_X \cap X$ and $\Lambda'_Y \cap Y$. Therefore, by contradiction, $\gdeg(\Prod(X,Y)) = |\Lambda^* \cap \Prod(X,Y)|$.
\end{proof}


\begin{lemma}\label{lm:unionGdeg}
Let $X$ and $Y$ be of dimensions $r_X, r_Y$, respectively, and assume that they do not have a common irreducible component. If $r_X = r_Y$, then
\[ \gdeg(X \cup Y) = \gdeg(X) + \gdeg(Y),\]
and if $r_X > r_Y$,
\[ \gdeg(X \cup Y) = \gdeg(X).\]
\end{lemma}
\begin{proof}
%Let the set $G_X$ be the $Gr\ddot{o}bner$ basis of $I(X)$ and the set $G_Y$ be that of $I(Y)$. We have $X = V_a(G_X) \subset \A^n$ and $Y = V_a(G_Y)$. Also let $G^h_X$ and $G^h_Y$ be the homogenization of $G_X$ and $G_Y$, respectively. Then the projective closure $\overline{X}$ of $X$ is the zero locus of the ideal generated by $G^h_X$, i.e. $\overline{X} = V_p(G^h_X) \subset \PP^n$. Analogously, $\overline{Y} = V_p(G^h_Y)$.
%
%Now
%\[
%\overline{X} \cup \overline{Y} =
%\]
%\[
%\overline{X \cup Y} = \overline{X} \cup \overline{Y}
%\]

Let $\overline{X}$ be the projective closure of $X$. Then $\overline{X} = V_p(I^h_X)$, where $I^h_X$ is the homogenization of the ideal $I_X = I(X)$.

To prove this lemma, we use the definition that $\gdeg(X) := \gdeg(\overline{X})$. We can simply argue that $\gdeg(X \cup Y) = \gdeg(\overline{X \cup Y}) = \gdeg(\overline{X} \cup \overline{Y})$. This is because $(I_X \cdot I_Y)^h = I^h_X \cdot I^h_Y$ (which is easy to see) and then 
\[
\overline{X \cup Y} = V_p((I_X \cdot I_Y)^h) = V_p(I^h_X \cdot I^h_Y) = V_p(I^h_X) \cup V_p(I^h_Y) = \overline{X} \cup \overline{Y}.
\]
Now this lemma becomes to prove $\gdeg(\overline{X} \cup \overline{Y}) = \gdeg(\overline{X}) + \gdeg(\overline{Y})$ if $r_X = r_Y$ or $\gdeg(\overline{X} \cup \overline{Y}) = \gdeg(\overline{X})$ if $r_X > r_Y$. This can be proved using Hilbert polynomial, and the details can be found in~\cite{HarAG}.
\end{proof}
\vspace{5ex}


\noindent \textbf{Measuring ``Entangleness.''}
\begin{lemma}
Given $f \in K[x_1, \cdots, x_n]$ and $g \in K[w_1, \cdots, w_m]$ be two polynomials of disjoint variables, then
\[ \Sing(fg) = \Sing(f) \cup \Sing(g).\]
\end{lemma}
\begin{proof}
Note that when $f$ and $g$ are of disjoint variables
\[ \Sing(fg) = V(fg, g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}, f \frac{\partial g}{\partial w_1}, \cdots, f \frac{\partial g}{\partial w_m}).\]
We first show that
\[ \Sing(fg) \subset (\Sing(f) \cup \Sing(g)).\]
If a point $P \in \Sing(fg)$, at least $P \in V(f)$ or $V(g)$. If $P \in V(f)$, $f(P) = 0$.
\[ (g\frac{\partial f}{\partial x_i})(P) = g(P) \frac{\partial f}{\partial x_i}(P) = 0 \]
implies $g(P) = 0$ or else $\frac{\partial f}{\partial x_i}(P) = 0$ for all $i$. That is, either
\[ P \in (V(f) \cap V(g))\ \ \ \mathrm{or}\ \ \ P \in \Sing(f).\]
Since $f$ and $g$ are of disjoint variables, $V(f) \cap V(g)) \subset \Sing(f)$. Similarly, if $P \in V(g)$, we have $P \in \Sing(g)$. Therefore, $P \in (\Sing(f) \cup \Sing(g))$.

Conversely, we show
\[ (\Sing(f) \cup \Sing(g)) \subset \Sing(fg).\]
If $P \in \Sing(f)$, we have
\[ f(P) = \frac{\partial f}{\partial x_i}(P) = 0.\]
With these,
\[ (fg)(P) = g(P) \frac{\partial f}{\partial x_i}(P) = f(P) \frac{\partial g}{\partial w_i}(P) = 0.\]
So $P \in \Sing(fg)$. Similarly, if $P \in \Sing(g)$ then $P \in \Sing(fg)$.

Putting everything together, we have
\[  \Sing(fg) = \Sing(f) \cup \Sing(g).\]
\end{proof}


%
%\begin{lemma}
%Given $f \in K[x_0, x_1, \cdots, x_n]$ and $F = f \in K[x_0, x_1, \cdots, x_n, x_{n+1}]$ and let $X = V(f) \subset \mathbb{P}^n$ and $X' = V(F) \subset \mathbb{P}^{n+1}$, then
%\[ \gdeg(X') = \gdeg(X).\]
%\end{lemma}
%\begin{proof}
%
%\end{proof}



\begin{lemma}
Given $f(X) \in K[x_1, \cdots, x_n]$, by extending the affine space and introducing new variable $t$, then
\[
\gdeg \left( g(W)t - 1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n} \right)\ \ =\ \ \gdeg \left( g(W)t - 1, \MJ(f) \right).
\]
\end{lemma}

\begin{proof}
Without loss of generality, assume working in $\A^{2n + m + 1}$ with
\[ f \in K[x_1, \cdots, x_n, y_1, \cdots, y_n, w_1, \cdots, w_m, t]. \]
Note that
\begin{multline*}
V\left( g(W)t -1, g(W)(y_1 - \frac{\partial f}{\partial x_1}), \cdots, g(W)(y_n - g\frac{\partial f}{\partial x_n}) \right) \\
= V \left( g(W)t - 1, \MJ(f) \right).
\end{multline*}
Hence it suffices to prove
\begin{multline*}
\gdeg \left( g(W)t - 1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n} \right) \\
= \gdeg \left( g(W)t - 1,  g(W)(y_1 - \frac{\partial f}{\partial x_1}), \cdots, g(W)(y_n - \frac{\partial f}{\partial x_n}) \right).
\end{multline*}
Let
\begin{align*}
X &= V \left( g(W)t - 1, g(W)(y_1 - \frac{\partial f}{\partial x_1}), \cdots, g(W)(y_n - \frac{\partial f}{\partial x_n}) \right), \\
Y &= V \left( g(W)t - 1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n} \right).
\end{align*}
Now define $\phi$ to be the following morphism from $X$ to $Y$:
\[
y_i \mapsto y_i \cdot g(W),
\]
and $\phi^{-1} : Y \rightarrow X$ to be:
\[
y_i \mapsto \frac{y_i}{g(W)}.
\]

\end{proof}


\begin{comment}
\begin{lemma}
Given $f(X) \in K[x_1, \cdots, x_n]$ and $g(W) \in K[w_1, \cdots, w_m]$ be two polynomials of disjoint variables, by extending the affine space and introducing new variables $s,t$, then
\begin{multline*}
\gdeg(f(X)s - 1, g(W)t - 1, \MJ(fg)) \\
= \gdeg(f(X)s - 1, g(W)t - 1, \MJ(f), \MJ(g)).
\end{multline*}
\end{lemma}
\begin{proof}
Without loss of generality, assume working in $\A^{2n + 2m + 2}$ with
\[ f,g \in K[x_1, \cdots, x_n, y_1, \cdots, y_n, w_1, \cdots, w_m, z_1, \cdots, z_m, s, t]. \]
Note that
\begin{multline*}
V(f(X)s - 1, g(W)t -1, \MJ(f)g(W), \MJ(g)f(X)) \\
= V(f(X)s-1, g(W)t - 1, \MJ(f), \MJ(g)).
\end{multline*}
Hence it suffices to prove
\begin{multline*}
\gdeg(f(X)s - 1, g(W)t - 1, \MJ(fg)) \\
= \gdeg(f(X)s - 1, g(W)t - 1, \MJ(f)g(W), \MJ(g)f(X)).
\end{multline*}
Let
\begin{align*}
X &= V(f(X)s - 1, g(W)t - 1, \MJ(f)g(W), \MJ(g)f(X)), \\
Y &= (f(X)s - 1, g(W)t - 1, \MJ(fg))
\end{align*}
and define $\phi$ to be the following morphism from $X$ to $Y$:
\end{proof}
\end{comment}



\begin{theorem}
Let $f \in K[x_1, \cdots, x_n]$ and $g \in K[w_1, \cdots, w_m]$ be two polynomials of disjoint variables. The following are true:
\begin{enumerate}[label=(\alph*)]
    \item $\gdeg(\J(fg)) \leq \gdeg(f) \cdot \gdeg(g)$,
    \item {\color{red}$\gdeg(\MJ(fg)) \geq \gdeg(\MJ(f)) \cdot \gdeg(\MJ(g))$},
    \item $\gdeg(\Sing(fg)) = \gdeg(\Sing(f)) + \gdeg(\Sing(g))$.
\end{enumerate}
\end{theorem}
\begin{proof}
~~\\
\noindent \textsf{(a).} Since $f$ and $g$ are of disjoint variables,
\[
\gdeg(\J(fg)) = \gdeg(g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}, f \frac{\partial g}{\partial w_1}, \cdots, f \frac{\partial g}{\partial w_m}).
\]
Now via B$\mathbf{\acute{e}}$zout's Theorem,
\[
\gdeg(\J(fg)) \leq \gdeg(g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}) \cdot \gdeg(f \frac{\partial g}{\partial w_1}, \cdots, f \frac{\partial g}{\partial w_m}).
\]
Note that
\[ V(g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}) = V(\J(f)) \cup V(g),\]
and then by~\Cref{lm:unionGdeg}, we have
\[ \gdeg(g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}) = \gdeg(g) \]
because $V(g)$ is a hypersurface and thus $\dim(V(g)) > \dim(V(\J(f)))$. Similarly, we also get
\[ \gdeg(f \frac{\partial g}{\partial w_1}, \cdots, f \frac{\partial g}{\partial w_m}) = \gdeg(f).\]
Therefore,
\[ \gdeg(\J(fg)) \leq \gdeg(f) \cdot \gdeg(g).\]
~~\\
\noindent \textsf{(b).}
\end{proof}


\begin{theorem}
\begin{multline*}
\gdeg(\MJ(fg)) \leq \\
 \Big( \gdeg(g(W)s-1, \MJ(f)) + \deg(g(W)) \Big) \cdot \Big( \gdeg(f(X)t-1, \MJ(g)) + \deg(f(W)) \Big)
\end{multline*}
\end{theorem}
\begin{proof}
\begin{align*}
&\ V(y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n})\\
=&\ V(g(W)s - 1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n}) \cup V(g(W), y_1 = 0, \cdots, y_n = 0) \\
\end{align*}
By isomorphism,
\[ \dim(V(g(W)s-1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n})) = \dim(V(g(W)s-1, y_1 - \frac{\partial f}{\partial x_1}, \cdots, y_n - \frac{\partial f}{\partial x_n})),\]
and simple calculations gives
\[ \dim(V(g(W)s-1, y_1 - \frac{\partial f}{\partial x_1}, \cdots, y_n - \frac{\partial f}{\partial x_n})) = \dim(g(W), y_1 = 0, \cdots, y_n = 0).\]
Since
\[
\gdeg(g(W)s-1, y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n}) = \gdeg(g(W)s-1, y_1 - \frac{\partial f}{\partial x_1}, \cdots, y_n - \frac{\partial f}{\partial x_n}),
\]
and by~\Cref{lm:unionGdeg},
\begin{align*}
&\ \gdeg(y_1 - g\frac{\partial f}{\partial x_1}, \cdots, y_n - g\frac{\partial f}{\partial x_n})\\
=&\ \gdeg(g(W)s-1, y_1 - \frac{\partial f}{\partial x_1}, \cdots, y_n - \frac{\partial f}{\partial x_n}) + \gdeg(g(W), y_1 = 0, \cdots, y_n = 0) \\
=&\ \gdeg(g(W)s-1, \MJ(f)) + \deg(g(W)) \\
=&\ \gdeg(g(W)s-1)\cdot \gdeg(\MJ(f)) + \gdeg(g(W)).
\end{align*}
\end{proof}


%\begin{theorem}
%\begin{align*}
%\gdeg(\J(fg)) &\leq \gdeg(g \frac{\partial f}{\partial x_1}, \cdots, g \frac{\partial f}{\partial x_n}) \cdot \gdeg(f \frac{\partial g}{\partial w_1}, \cdots, f \frac{\partial g}{\partial w_m})\\
%&= \gdeg(g) \cdot \gdeg(f).
%\end{align*}
%\end{theorem}
%\begin{proof}
%
%\end{proof}

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