Here is a proof using resultants, taken mostly from https://mathoverflow.net/questions/189181//189344#189344 . (For a short summary, see one of my comments to the OP.)
$\newcommand{\KK}{\mathbb{K}}
\newcommand{\LL}{\mathbb{L}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\ww}{\mathbf{w}}
\newcommand{\eps}{\varepsilon}
\newcommand{\Res}{\operatorname{Res}}
\newcommand{\Syl}{\operatorname{Syl}}
\newcommand{\adj}{\operatorname{adj}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\tilF}{\widetilde{F}}
\newcommand{\tilG}{\widetilde{G}}
\newcommand{\tilKK}{\widetilde{\KK}}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\zeroes}[1]{\underbrace{0,0,\ldots,0}_{#1 \text{ zeroes}}}$
We shall prove a more general statement:
Theorem 1. Let $\KK$ be a nontrivial commutative ring. Let $F$ and $G$ be two
polynomials in the polynomial ring $\KK \ive{T}$. Let
$d$ and $e$ be nonnegative integers such that $d+e > 0$ and $\deg F \leq d$ and $\deg G \leq e$.
Then, there exists a nonzero polynomial $P\in\KK \ive{X, Y}$ in two indeterminates $X$ and $Y$ such that $\deg_X P\leq e$
and $\deg_Y P\leq d$ and $P\tup{F, G} =0$.
Here and in the following, we are using the following notations:
"Ring" always means "associative ring with unity".
A ring $R$ is said to be nontrivial if $0 \neq 1$ in $R$.
If $R$ is any polynomial in the polynomial ring $\KK \ive{X, Y}$, then $\deg_X R$ denotes the degree of $R$ with respect to the variable $X$ (that is, it denotes the degree of $R$ when $R$ is considered as a polynomial in $\tup{\KK \ive{Y}} \ive{X}$), whereas $\deg_Y R$ denotes the degree of the polynomial $R$ with respect to the variable $Y$.
To prove Theorem 1, we recall the notion of the resultant of two polynomials over a
commutative ring:
Definition. Let $\KK$ be a commutative ring.
Let $P\in \KK \ive{T}$ and $Q\in\KK \ive{T}$ be two polynomials in the polynomial ring $\KK \ive{T}$.
Let $d\in\NN$ and $e\in\NN$ be such that $\deg P\leq d$ and $\deg Q\leq e$.
Thus, write the polynomials $P$ and $Q$ in the forms
\begin{align*}
P & =p_0 +p_1 T+p_2 T^2 +\cdots+p_d T^d \qquad\text{and}\\
Q & =q_0 +q_1 T+q_2 T^2 +\cdots+q_e T^e ,
\end{align*}
where $p_0 ,p_1 ,\ldots,p_d ,q_0 ,q_1 ,\ldots,q_e $ belong to $\KK$.
Then, we let $\Syl_{d,e} \tup{P, Q}$ be the matrix
\begin{align}
\left(
\begin{array}[c]{c}
\begin{array}[c]{ccccccccc}
p_0 & 0 & 0 & \cdots & 0 & q_0 & 0 & \cdots & 0\\
p_1 & p_0 & 0 & \cdots & 0 & q_1 & q_0 & \cdots & 0\\
\vdots & p_1 & p_0 & \cdots & 0 & \vdots & q_1 & \ddots & \vdots\\
\vdots & \vdots & p_1 & \ddots & \vdots & \vdots & \vdots & \ddots &
q_0 \\
p_d & \vdots & \vdots & \ddots & p_0 & \vdots & \vdots & \ddots & q_1 \\
0 & p_d & \vdots & \ddots & p_1 & q_e & \vdots & \ddots & \vdots\\
\vdots & \vdots & \ddots & \ddots & \vdots & 0 & q_e & \ddots & \vdots\\
0 & 0 & 0 & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & p_d & 0 & 0 & \cdots & q_e
\end{array}
\\
\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{e\text{ columns}}
\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{d\text{ columns}}
\end{array}
\right) \in\KK^{\tup{d+e} \times\tup{d+e}};
\end{align}
this is the $\tup{d+e} \times\tup{d+e}$-matrix whose first $e$ columns have the form
\begin{align}
\left( \zeroes{k},p_0 ,p_1 ,\ldots ,p_d ,\zeroes{e-1-k}\right) ^{T}
\qquad\text{for }k\in\left\{ 0,1,\ldots,e-1\right\} ,
\end{align}
and whose last $d$ columns have the form
\begin{align}
\left( \zeroes{\ell},q_0 ,q_1 ,\ldots,q_e ,\zeroes{d-1-\ell}\right) ^{T}
\qquad\text{for }\ell\in\left\{ 0,1,\ldots,d-1\right\} .
\end{align}
Furthermore, we define $\Res_{d,e}\tup{P, Q}$ to be the element
\begin{align}
\det \tup{ \Syl_{d,e}\tup{P, Q} } \in \KK .
\end{align}
The matrix $\Syl_{d,e}\tup{P, Q}$ is called the Sylvester matrix of $P$ and $Q$ in degrees $d$ and $e$.
Its determinant $\Res_{d,e}\tup{P, Q}$ is called the resultant of $P$ and $Q$ in degrees $d$ and $e$.
It is common to apply this definition to the case when $d=\deg P$ and $e=\deg Q$; in this case, we simply call $\Res_{d,e}\tup{P, Q}$ the resultant of $P$ and $Q$, and denote it by $\Res \tup{P, Q}$.
Here, we take $\NN$ to mean the set $\left\{0,1,2,\ldots\right\}$ of all nonnegative integers.
One of the main properties of resultants is the following:
Theorem 2. Let $\KK$ be a commutative ring.
Let $P\in \KK \ive{T}$ and $Q\in\KK \ive{T}$ be two polynomials in the polynomial ring $\KK \ive{T}$.
Let $d\in\NN$ and $e\in\NN$ be such that $d+e > 0$ and $\deg P\leq d$ and $\deg Q\leq e$.
Let $\LL$ be a commutative $\KK$-algebra, and let $w\in\LL$ satisfy $P\tup{w} =0$ and $Q\tup{w} = 0$.
Then, $\Res_{d,e}\tup{P, Q} =0$ in $\LL$.
Proof of Theorem 2 (sketched). Recall that $\Res_{d,e}\tup{P, Q} =\det \tup{ \Syl_{d,e}\tup{P, Q} }$ (by the definition of $\Res_{d,e}\tup{P, Q}$).
Write the polynomials $P$ and $Q$ in the forms
\begin{align*}
P & =p_0 +p_1 T+p_2 T^2 +\cdots+p_d T^d \qquad\text{and}\\
Q & =q_0 +q_1 T+q_2 T^2 +\cdots+q_e T^e ,
\end{align*}
where $p_0 ,p_1 ,\ldots,p_d ,q_0 ,q_1 ,\ldots,q_e $ belong to
$\KK$.
(We can do this, since $\deg P \leq d$ and $\deg Q \leq e$.)
From $p_0 +p_1 T+p_2 T^2 +\cdots+p_d T^d = P$,
we obtain
$p_0 + p_1 w + p_2 w^2 + \cdots + p_d w^d = P\left(w\right) = 0$.
Similarly,
$q_0 + q_1 w + q_2 w^2 + \cdots + q_e w^e = 0$.
Let $A$ be the matrix $\Syl_{d,e}\tup{P, Q}
\in\KK^{\tup{d+e} \times\tup{d+e} }$, regarded as a
matrix in $\LL ^{\tup{d+e} \times\tup{d+e} }$ (by
applying the canonical $\KK$-algebra homomorphism $\KK
\rightarrow\LL$ to all its entries).
Let $\ww$ be the row vector $\left( w^{0},w^{1},\ldots,w^{d+e-1}
\right) \in\LL ^{1\times\tup{d+e} }$. Let $\mathbf{0}$ denote
the zero vector in $\LL ^{1\times\tup{d+e} }$.
Now, it is easy to see that $\ww A=\mathbf{0}$. (Indeed, for each
$k\in\left\{ 1,2,\ldots,d+e\right\} $, we have
\begin{align*}
& \ww\left( \text{the }k\text{-th column of }A\right) \\
& =
\begin{cases}
p_0 w^{k-1}+p_1 w^k +p_2 w^{k+1}+\cdots+p_d w^{k-1+d}, & \text{if }k\leq e;\\
q_0 w^{k-e-1}+q_1 w^{k-e}+q_2 w^{k-e+1}+\cdots+q_e w^{k-1}, & \text{if }k>e
\end{cases}
\\
& =
\begin{cases}
w^{k-1}\left( p_0 +p_1 w+p_2 w^2 +\cdots+p_d w^d \right) , & \text{if }k\leq e;\\
w^{k-e-1}\left( q_0 +q_1 w+q_2 w^2 +\cdots+q_e w^e\right) , & \text{if }k>e
\end{cases}
\\
& =
\begin{cases}
w^{k-1}0, & \text{if }k\leq e;\\
w^{k-e-1}0, & \text{if }k>e
\end{cases}
\\
& \qquad\left(
\begin{array}[c]{c}
\text{since }p_0 +p_1 w+p_2 w^2 +\cdots+p_d w^d =0\\
\text{and }q_0 +q_1 w+q_2 w^2 +\cdots+q_e w^e =0
\end{array}
\right) \\
& =0.
\end{align*}
But this means precisely that $\ww A=\mathbf{0}$.)
But $A$ is a square matrix over a commutative ring; thus, the adjugate $\adj A$ of $A$ satisfies $A\cdot\adj A=\det A\cdot I_{d+e}$ (where $I_{d+e}$ denotes the identity matrix of size $d+e$).
Hence, $\ww\underbrace{A\cdot\adj A}_{=\det A\cdot I_{d+e}}=\ww\det A\cdot I_{d+e}=\det A\cdot\ww$.
Comparing this with $\underbrace{\ww A}_{=\mathbf{0}}\cdot\adj A =\mathbf{0}\cdot\adj A=\mathbf{0}$, we obtain
$\det A\cdot\ww=\mathbf{0}$.
But $d+e > 0$; thus, the row vector $\ww$ has a well-defined first entry.
This first entry is $w^0 = 1$.
Hence, the first entry of the row vector $\det A\cdot\ww$ is $\det A \cdot 1 = \det A$.
Hence, from $\det A\cdot\ww=\mathbf{0}$, we conclude that $\det A=0$.
Comparing this with
\begin{align}
\det\underbrace{A}_{=\Syl_{d,e}\tup{P, Q}} =\det \tup{ \Syl_{d,e}\tup{P, Q} }
=\Res_{d,e}\tup{P, Q} ,
\end{align}
we obtain $\Res_{d,e}\tup{P, Q} =0$ (in $\LL$). This proves Theorem 2. $\blacksquare$
Theorem 2 (which I have proven in detail to stress how the proof uses nothing
about $\LL$ other than its commutativity) was just the meek tip of the
resultant iceberg. Here are some further sources with deeper results:
Some of these sources use the matrix $\tup{\Syl_{d,e}\tup{P, Q}}^T$ instead of our $\Syl_{d,e}\tup{P, Q}$, but of course this matrix has the same determinant as $\Syl_{d,e}\tup{P, Q}$, so that their definition of a resultant is the same as mine.
Some other among these sources use a different matrix, in which the coefficients $p_0, p_1, \ldots, p_d$ occur in the opposite order (i.e., from bottom to top rather than from top to bottom) and similarly for the coefficients $q_0, q_1, \ldots, q_e$; this leads to a notion of resultant that differs from mine in sign (because their matrix is obtained from mine by a permutation of rows and columns).
We are not yet ready to prove Theorem 1 directly. Instead, let us prove a
weaker version of Theorem 1:
Lemma 3. Let $\KK$ be a commutative ring. Let $F$ and $G$ be two
polynomials in the polynomial ring $\KK \ive{T}$. Let
$d$ and $e$ be nonnegative integers such that $d+e > 0$ and $\deg F \leq d$ and $\deg G \leq e$.
Write the polynomial $G$ in the form $G=g_0 +g_1 T+g_2 T^2 +\cdots
+g_e T^e $, where $g_0 ,g_1 ,\ldots,g_e \in\KK$.
Assume that $g_e^d \neq 0$.
Then, there exists a nonzero polynomial $P\in\KK \ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $\deg_X P\leq e$
and $\deg_Y P\leq d$ and $P\tup{F, G} =0$.
Proof of Lemma 3 (sketched). Let $\tilKK$ be the
commutative ring $\KK \ive{X, Y}$. Define two polynomials
$\tilF \in\tilKK\ive{T}$ and $\tilG \in\tilKK\ive{T}$ by
\begin{align}
\tilF = F-X = F\tup{T}-X \qquad\text{and}\qquad
\tilG = G-Y = G\tup{T}-Y.
\end{align}
Note that $X$ and $Y$ have degree $0$ when considered as polynomials in
$\tilKK\ive{T}$ (since $X$ and $Y$ belong to the
ring $\tilKK$). Thus, these new polynomials $\tilF = F - X$
and $\tilG = G - Y$ have degrees $\deg\tilF \leq d$ (because
$\deg X = 0 \leq d$ and $\deg F \leq d$) and
$\deg\tilG \leq e$ (similarly).
Hence, the resultant $\Res_{d,e}\tup{\tilF, \tilG} \in \tilKK$ of these polynomials $\tilF$ and $\tilG$ in degrees $d$ and $e$ is well-defined. Let us denote this resultant $\Res_{d,e}\tup{\tilF, \tilG}$ by $P$. Hence,
\begin{align}
P = \Res_{d,e}{\tilF, \tilG} \in \tilKK =\KK \ive{X, Y} .
\end{align}
Our next goal is to show that $P$ is a nonzero polynomial and satisfies
$\deg_X P\leq e$ and $\deg_Y P\leq d$ and $P\tup{F, G} =0$. Once
this is shown, Lemma 3 will obviously follow.
We have
\begin{align}
P = \Res_{d,e} \tup{\tilF, \tilG} = \det\tup{\Syl_{d,e}\tup{\tilF, \tilG}}
\label{darij1.pf.t1.P=det}
\tag{1}
\end{align}
(by the definition of $\Res_{d,e}\tup{\tilF, \tilG}$).
Write the polynomial $F$ in the form $F=f_0 +f_1 T+f_2 T^2 +\cdots
+f_d T^d $, where $f_0 ,f_1 ,\ldots,f_d \in\KK$. (This can be
done, since $\deg F \leq d$.)
Recall that $g_e ^d \neq 0$. Thus, $\left( -1\right) ^e g_e^d \neq 0$.
For each $p\in\NN$, we let $S_{p}$ be the group of all permutations of
the set $\left\{ 1,2,\ldots,p\right\} $.
Now,
\begin{align*}
\tilF & =F-X=\left( f_0 +f_1 T+f_2 T^2 +\cdots+f_d
T^d \right) -X\\
& \qquad\left( \text{since }F=f_0 +f_1 T+f_2 T^2 +\cdots+f_d
T^d \right) \\
& =\tup{f_0 - X} +f_1 T+f_2 T^2 +\cdots+f_d T^d .
\end{align*}
Thus, $f_0 -X,f_1 ,f_2 ,\ldots,f_d $ are the coefficients of the
polynomial $\tilF \in\tilKK\ive{T}$ (since
$f_0 -X\in\tilKK$). Similarly, $g_0 -Y,g_1 ,g_2
,\ldots,g_e $ are the coefficients of the polynomial $\tilG
\in\tilKK\ive{T}$. Hence, the definition of the
matrix $\Syl_{d,e}\left( \tilF ,\tilG
\right)$ yields
\begin{align}
&\Syl_{d,e}\tup{\tilF, \tilG} \\
&=\left(
\begin{array}[c]{c}
\begin{array}[c]{ccccccccc}
f_0 -X & 0 & 0 & \cdots & 0 & g_0 -Y & 0 & \cdots & 0\\
f_1 & f_0 -X & 0 & \cdots & 0 & g_1 & g_0 -Y & \cdots & 0\\
\vdots & f_1 & f_0-X & \cdots & 0 & \vdots & g_1 & \ddots & \vdots\\
\vdots & \vdots & f_1 & \ddots & \vdots & \vdots & \vdots & \ddots &
g_0 -Y\\
f_d & \vdots & \vdots & \ddots & f_0 -X & \vdots & \vdots & \ddots &
g_1 \\
0 & f_d & \vdots & \ddots & f_1 & g_e & \vdots & \ddots & \vdots\\
\vdots & \vdots & \ddots & \ddots & \vdots & 0 & g_e & \ddots & \vdots\\
0 & 0 & 0 & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & f_d & 0 & 0 & \cdots & g_e
\end{array}
\\
\underbrace{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}
_{e\text{ columns}}
\underbrace{\qquad \qquad \qquad \qquad \qquad \qquad \qquad}
_{d\text{ columns}}
\end{array}
\right) \\
&\in\tilKK^{\tup{d+e} \times\left(
d+e\right) }.
\end{align}
Now, let us use this explicit form of $\Syl_{d,e} \tup{\tilF, \tilG}$
to compute $\det\left( \Syl_{d,e}\tup{\tilF, \tilG} \right)$ using
the Leibniz formula.
The Leibniz formula yields
\begin{align}
\det\left( \Syl_{d,e} \tup{ \tilF, \tilG } \right)
=\sum_{\sigma\in S_{d+e}}a_{\sigma},
\label{darij1.pf.t1.det=sum}
\tag{2}
\end{align}
where for each permutation $\sigma\in S_{d+e}$, the addend $a_{\sigma}$ is a
product of entries of $\Syl_{d,e} \tup{\tilF, \tilG}$, possibly with a minus sign. More
precisely,
\begin{align}
a_{\sigma}=\tup{-1}^{\sigma}\prod_{i=1}^{d+e}\left( \text{the }
\left( i,\sigma\left( i\right) \right) \text{-th entry of }
\Syl_{d,e}\tup{\tilF, \tilG}
\right)
\end{align}
for each $\sigma\in S_{d+e}$ (where $\tup{-1}^{\sigma}$ denotes the
sign of the permutation $\sigma$).
Now, \eqref{darij1.pf.t1.P=det} becomes
\begin{align}
P=\det\left( \Syl_{d,e}\tup{\tilF, \tilG} \right)
=\sum_{\sigma\in S_{d+e}}a_{\sigma}
\label{darij1.pf.t1.P=sum}
\tag{3}
\end{align}
(by \eqref{darij1.pf.t1.det=sum}).
All entries of the matrix $\Syl_{d,e}\tup{\tilF, \tilG}$ are polynomials in the two
indeterminates $X$ and $Y$; but only $d+e$ of these entries are non-constant
polynomials (since all of $f_0 ,f_1 ,\ldots,f_d ,g_0 ,g_1 ,\ldots,g_e $
belong to $\KK$). More precisely, only $e$ entries of
$\Syl_{d,e}\tup{\tilF, \tilG}$
have non-zero degree with respect to the variable $X$ (namely, the first $e$
entries of the diagonal of $\Syl_{d,e}\tup{\tilF, \tilG}$), and these $e$ entries have degree $1$
with respect to this variable. Thus, for each $\sigma\in S_{d+e}$, the product
$a_{\sigma}$ contains at most $e$ many factors that have degree $1$ with
respect to the variable $X$, while all its remaining factors have degree $0$
with respect to this variable. Therefore, for each $\sigma\in S_{d+e}$, the
product $a_{\sigma}$ has degree $\leq e\cdot1=e$ with respect to the variable
$X$. Hence, the sum $\sum_{\sigma\in S_{d+e}}a_{\sigma}$ of all these products
$a_{\sigma}$ also has degree $\leq e$ with respect to the variable $X$. In
other words, $\deg_X \left( \sum_{\sigma\in S_{d+e}}a_{\sigma}\right) \leq
e$. In view of \eqref{darij1.pf.t1.P=sum}, this rewrites as $\deg_X P\leq e$.
Similarly, $\deg_Y P\leq d$ (since only $d$ entries of the matrix
$\Syl_{d,e}\tup{\tilF, \tilG}
$ have non-zero degree with respect to the variable $Y$, and these $d$ entries
have degree $1$ with respect to this variable).
Next, we shall show that the polynomial $P$ is nonzero. Indeed, let us
consider all elements of $\tilKK$ as polynomials in the
variable $X$ over the ring $\KK \ive{Y} $. For each
permutation $\sigma\in S_{d+e}$, the product $a_{\sigma}$ (thus considered)
has degree $\leq e$ (as we have previously shown). Let us now compute the
coefficient of $X^e $ in this product $a_{\sigma}$. There are three possible cases:
Case 1: The permutation $\sigma\in S_{d+e}$ does not satisfy $\left(
\sigma\left( i\right) =i\text{ for each }i\in\left\{ 1,2,\ldots,e\right\}
\right)$. Thus, the product $a_{\sigma}$ has strictly fewer than $e$
factors that have degree $1$ with respect to the variable $X$, while all its
remaining factors have degree $0$ with respect to this variable. Thus, the
whole product $a_{\sigma}$ has degree $<e$ with respect to the variable $X$.
Hence, the coefficient of $X^e $ in this product $a_{\sigma}$ is $0$.
Case 2: The permutation $\sigma\in S_{d+e}$ satisfies $\left(
\sigma\left( i\right) =i\text{ for each }i\in\left\{ 1,2,\ldots,e\right\}
\right)$, but is not the identity map $\id\in S_{d+e}$. Thus,
there must exist at least one $i\in\left\{ 1,2,\ldots,d+e\right\} $ such
that $\sigma\left( i\right) <i$. Consider such an $i$, and notice that it
must satisfy $i>e$ and $\sigma\left( i\right) >e$; hence, the $\left(
i,\sigma\left( i\right) \right)$-th entry of
$\Syl_{d,e}\tup{\tilF, \tilG}$ is $0$. Thus, the
whole product $a_{\sigma}$ is $0$ (since the latter entry is a factor in this
product). Thus, the coefficient of $X^e $ in this product $a_{\sigma}$ is $0$.
Case 3: The permutation $\sigma\in S_{d+e}$ is the identity map
$\id\in S_{d+e}$. Thus, the product $a_{\sigma}$ is $\left(
f_0 -X\right) ^e g_e^d $ (since $\tup{-1}^{\id
}=1$). Hence, the coefficient of $X^e $ in this product $a_{\sigma}$ is
$\tup{-1}^e g_e^d $.
Summarizing, we thus conclude that the coefficient of $X^e $ in the product
$a_{\sigma}$ is $0$ unless $\sigma=\id$, in which case it is
$\tup{-1}^e g_e^d $. Hence, the coefficient of $X^e $ in the
sum $\sum_{\sigma\in S_{d+e}}a_{\sigma}$ is $\tup{-1}^e g_e^d
\neq0$. Therefore, $\sum_{\sigma\in S_{d+e}}a_{\sigma}\neq0$. In view of
\eqref{darij1.pf.t1.P=sum}, this rewrites as $P\neq0$. In other words, the
polynomial $P$ is nonzero.
Finally, it remains to prove that $P\tup{F, G} =0$. In order to do
this, we let $\LL$ be the polynomial ring $\KK \ive{U}
$ in a new indeterminate $U$. We let $\varphi:\KK \ive{X, Y}
\rightarrow\LL$ be the unique $\KK$-algebra homomorphism that
sends $X$ to $F\tup{U}$ and sends $Y$ to $G\tup{U}$.
(This is well-defined by the universal property of the polynomial ring $\KK \ive{X, Y}$.)
Note that $\varphi$ is a $\KK$-algebra homomorphism from $\tilKK$ to $\LL$ (since $\KK \ive{X, Y} = \tilKK$). Thus, $\LL$ becomes a $\tilKK$-algebra via this homomorphism $\varphi$.
Now, recall that the polynomial $\tilF \in\tilKK\ive{T}$ was defined by $\tilF =F-X$. Hence, $\tilF \tup{U} = F\tup{U} - \varphi\tup{X}$. (Indeed, when we regard $X$ as an element of $\tilKK\ive{T}$, the polynomial $X$ is simply a constant, and thus evaluating it at $U$ yields the canonical image of $X$ in $\LL$, which is $\varphi\tup{X}$.)
But $\varphi\tup{X} = F\tup{U}$ (by the definition of $\varphi$).
Hence, $\tilF \tup{U} = F\tup{U} - \varphi\tup{X} = 0$ (since $\varphi\tup{X} = F\tup{U}$). Similarly, $\tilG \tup{U} = 0$.
Thus, the element $U\in\LL$ satisfies $\tilF \tup{U} =0$ and $\tilG \tup{U} = 0$. Hence, Theorem 2 (applied to
$\tilKK$, $\tilF$, $\tilG$ and $U$ instead of $\KK$, $P$, $Q$ and $w$) yields that $\Res_{d,e}\tup{\tilF, \tilG} = 0$ in $\LL$. In other words, $\varphi\tup{ \Res_{d,e}\tup{\tilF, \tilG} } =0$. In view of $\Res_{d,e}\tup{\tilF, \tilG} =P$, this rewrites as $\varphi\tup{P} =0$.
But recall that $\varphi$ is the $\KK$-algebra homomorphism that sends
$X$ to $F\tup{U}$ and sends $Y$ to $G\tup{U}$. Hence, it
sends any polynomial $Q\in\KK \ive{X, Y}$ to $Q \tup{ F\tup{U}, G\tup{U} }$. Applying this to $Q=P$, we
conclude that it sends $P$ to $P \tup{ F\tup{U}, G\tup{U} }$.
In other words, $\varphi\tup{P} =P \tup{ F\tup{U}, G\tup{U} }$;
hence, $P \tup{ F\tup{U}, G\tup{U} } =\varphi\tup{P} =0$.
Now, $F\tup{U}$ and $G\tup{U}$ are polynomials in the
indeterminate $U$ over $\KK$. If we rename the indeterminate $U$ as
$T$, then these polynomials $F\tup{U}$ and $G\tup{U}$
become $F\tup{T}$ and $G\tup{T}$, and therefore the
polynomial $P\left( F\tup{U} ,G\tup{U} \right)$ becomes
$P\tup{ F\tup{T}, G\tup{T} }$.
Hence, $P\tup{ F\tup{T}, G\tup{T} } =0$ (since $P\tup{ F\tup{U}, G\tup{U} } =0$).
In other words, $P \tup{F, G} =0$ (since $F\tup{T} =F$ and $G\tup{T} =G$).
This completes the proof of Lemma 3. $\blacksquare$
Lemma 4. (a) Theorem 1 holds when $d = 0$.
(b) Theorem 1 holds when $e = 0$.
Proof of Lemma 4. (a) Assume that $d = 0$.
Thus, $d + e > 0$ rewrites as $e > 0$. Hence, $e \geq 1$.
But the polynomial $F$ is constant (since $\deg F \leq d = 0$).
In other words, $F = f$ for some $f \in \KK$. Consider this $f$.
Now, let $Q$ be the polynomial $X - f \in \KK\ive{X, Y}$.
Then, $Q$ is nonzero and satisfies
$\deg_X Q = 1 \leq e$ (since $e \geq 1$) and
$\deg_Y Q = 0 \leq d$ and
$Q\left(F, G\right) = F - f = 0$ (since $F = f$).
Hence, there exists a nonzero polynomial $P\in\KK \ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $\deg_X P\leq e$
and $\deg_Y P\leq d$ and $P\tup{F, G} =0$
(namely, $P = Q$). In other words, Theorem 1 holds (under
our assumption that $d = 0$). This proves Lemma 4 (a).
(b) The proof of Lemma 4 (b) is analogous to
our above proof of Lemma 4 (a). $\blacksquare$
Now, we can prove Theorem 1 at last:
Proof of Theorem 1. We shall prove Theorem 1 by induction on $e$.
The induction base is the case when $e = 0$; this case follows
from Lemma 4 (b).
For the induction step, we fix a positive integer $\eps$.
Assume (as the induction hypothesis) that Theorem 1 holds for $e = \eps - 1$.
We must now prove that Theorem 1 holds for $e = \eps$.
Let $\KK$ be a commutative ring. Let $F$ and $G$ be two
polynomials in the polynomial ring $\KK \ive{T}$. Let
$d$ be a nonnegative integer such that $d+\eps > 0$ and $\deg F \leq d$ and $\deg G \leq \eps$.
Our goal is now to prove that the claim of Theorem 1 holds for $e = \eps$.
In other words, our goal is to prove that there exists a nonzero polynomial $P\in\KK \ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $\deg_X P\leq \eps$
and $\deg_Y P\leq d$ and $P\tup{F, G} =0$.
Write the polynomial $G$ in the form $G=g_0 +g_1 T+g_2 T^2 +\cdots
+g_{\eps}T^{\eps}$, where $g_0 ,g_1 ,\ldots,g_{\eps}\in\KK$.
(This can be done, since $\deg G \leq \eps$.)
If $g_{\eps}^d \neq 0$, then our goal follows immediately
by applying Lemma 3 to $e = \eps$.
Thus, for the rest of this induction step, we WLOG assume that $g_{\eps}^d = 0$.
Hence, there exists a positive integer $m$ such that $g_{\eps}^m = 0$ (namely, $m = \eps$).
Thus, there exists a smallest such $m$.
Consider this smallest $m$.
Then, $g_{\eps}^m = 0$, but
\begin{align}
\text{every positive integer $\ell < m$ satisfies $g_{\eps}^{\ell} \neq 0$.}
\label{darij1.pf.t1.epsilon-ell}
\tag{4}
\end{align}
We claim that $g_{\eps}^{m-1} \neq 0$. Indeed, if $m-1$ is a
positive integer, then this follows from \eqref{darij1.pf.t1.epsilon-ell} (applied to $\ell = m-1$);
otherwise, it follows from the fact that $g_{\eps}^0 = 1 \neq 0$
(since the ring $\KK$ is nontrivial).
Now recall again that our goal is to prove that the claim of Theorem 1 holds for $e = \eps$.
If $d = 0$, then this goal follows from Lemma 4 (a).
Hence, for the rest of this induction step, we WLOG assume that $d \neq 0$.
Hence, $d > 0$ (since $d$ is a nonnegative integer).
We have $e \geq 1$ (since $e$ is a positive integer), thus
$e - 1 \geq 0$. Hence, $d + \left(e-1\right) \geq d > 0$.
Let $I$ be the subset $\left\{x \in \KK \mid g_{\eps}^{m-1} x = 0 \right\}$ of $\KK$.
Then, $I$ is an ideal of $\KK$ (namely, it is the
annihilator of the subset
$\left\{g_{\eps}^{m-1}\right\}$ of $\KK$);
thus, $\KK / I$ is a commutative $\KK$-algebra.
Denote this commutative $\KK$-algebra $\KK / I$ by $\LL$.
Let $\pi$ be the canonical projection $\KK \to \LL$.
Of course, $\pi$ is a surjective $\KK$-algebra homomorphism.
For any $a \in \KK$, we will denote the image of $a$ under $\pi$ by $\overline{a}$.
The $\KK$-algebra homomorphism $\pi : \KK \to \LL$ induces a canonical $\KK$-algebra homomorphism $\KK\ive{T} \to \LL\ive{T}$ (sending $T$ to $T$).
For any $a \in \KK\ive{T}$, we will denote the image of $a$ under the latter homomorphism by $\overline{a}$.
The $\KK$-algebra homomorphism $\pi : \KK \to \LL$ induces a canonical $\KK$-algebra homomorphism $\KK\ive{X, Y} \to \LL\ive{X, Y}$ (sending $X$ and $Y$ to $X$ and $Y$).
For any $a \in \KK\ive{X, Y}$, we will denote the image of $a$ under the latter homomorphism by $\overline{a}$.
We have $g_{\eps}^{m-1} g_{\eps} = g_{\eps}^m = 0$, so that $g_{\eps} \in I$ (by the definition of $I$);
hence, the residue class $\overline{g_{\eps}}$ of $g_{\eps}$ modulo the ideal $I$ is $0$.
We have $g_{\eps}^{m-1} \cdot 1 = g_{\eps}^{m-1} \neq 0$ in $\KK$,
and thus $1 \notin I$ (by the definition of $I$).
Hence, the ideal $I$ is not the whole ring $\KK$.
Thus, the quotient ring $\KK / I = \LL$ is nontrivial.
But $G=g_0 +g_1 T+g_2 T^2 +\cdots +g_{\eps}T^{\eps}$
and thus
\begin{align}
\overline{G}
&= \overline{g_0} + \overline{g_1} T + \overline{g_2} T^2 + \cdots + \overline{g_{\eps}} T^{\eps} \\
&= \left( \overline{g_0} + \overline{g_1} T + \overline{g_2} T^2 + \cdots + \overline{g_{\eps-1}} T^{\eps-1} \right)
+ \underbrace{\overline{g_{\eps}}}_{= 0} T^{\eps} \\
&= \overline{g_0} + \overline{g_1} T + \overline{g_2} T^2 + \cdots + \overline{g_{\eps-1}} T^{\eps-1} ,
\end{align}
so that $\deg \overline{G} \leq e-1$.
Also, $\deg \overline{F} \leq \deg F \leq d$.
But the induction hypothesis tells us that Theorem 1 holds for $e = \eps - 1$.
Hence, we can apply Theorem 1 to $\LL$, $\overline{F}$, $\overline{G}$ and $\eps - 1$
instead of $\KK$, $F$, $G$ and $e$.
We thus conclude that there exists a nonzero polynomial $P\in \LL\ive{X, Y}$ in two indeterminates $X$ and $Y$ such that $\deg_X P \leq \eps - 1$ and $\deg_Y P \leq d$ and $P\left( \overline{F}, \overline{G} \right) =0$.
Consider this polynomial $P$, and denote it by $R$.
Thus, $R \in \LL \ive{X, Y}$ is a nonzero polynomial in two indeterminates $X$ and $Y$ and satisfies $\deg_X R \leq \eps - 1$ and $\deg_Y R \leq d$ and $R \left( \overline{F}, \overline{G} \right) =0$.
Clearly, there exists a polynomial $Q \in \KK\ive{X, Y}$ in two
indeterminates $X$ and $Y$ that satisfies $\deg_X Q = \deg_X R$ and
$\deg_Y Q = \deg_Y R$ and $\overline{Q} = R$.
(Indeed, we can construct such a $Q$ as follows: Write
$R$ in the form
$R = \sum\limits_{i = 0}^{\deg_X R} \sum\limits_{j = 0}^{\deg_Y R} r_{i, j} X^i Y^j$
for some coefficients $r_{i, j} \in \LL$.
For each pair $\left(i, j\right)$, pick some
$p_{i, j} \in \KK$ such that $\overline{p_{i, j}} = r_{i, j}$
(this can be done, since the homomorphism $\pi : \KK \to \LL$ is surjective).
Then, set $Q = \sum\limits_{i = 0}^{\deg_X R} \sum\limits_{j = 0}^{\deg_Y R} p_{i, j} X^i Y^j$.
It is clear that this polynomial $Q$ satisfies $\deg_X Q = \deg_X R$ and
$\deg_Y Q = \deg_Y R$ and $\overline{Q} = R$.)
We have $\overline{Q \left(F, G\right)}
= \underbrace{\overline{Q}}_{=R} \left( \overline{F}, \overline{G} \right)
= R \left( \overline{F}, \overline{G} \right) = 0$.
In other words, the polynomial $Q \left(F, G\right) \in \KK\ive{T}$
lies in the kernel of the canonical
$\KK$-algebra homomorphism $\KK\ive{T} \to \LL\ive{T}$.
This means that each coefficient of this
polynomial $Q \left(F, G\right) \in \KK\ive{T}$
lies in the kernel of the $\KK$-algebra homomorphism $\pi : \KK \to \LL$.
In other words, each coefficient of this
polynomial $Q \left(F, G\right) \in \KK\ive{T}$ lies in $I$
(since the kernel of the $\KK$-algebra homomorphism $\pi : \KK \to \LL$
is $I$).
Hence, each coefficient $c$ of this
polynomial $Q \left(F, G\right) \in \KK\ive{T}$
satisfies $g_{\eps}^{m-1} c = 0$ (by the definition of $I$).
Therefore, $g_{\eps}^{m-1} Q \left(F, G\right) = 0$.
On the other hand, $\overline{Q} = R$ is nonzero.
In other words, the polynomial $Q \in \KK\ive{X, Y}$ does not lie
in the kernel of the canonical
$\KK$-algebra homomorphism $\KK\ive{X, Y} \to \LL\ive{X, Y}$.
This means that not every coefficient of this
polynomial $Q \in \KK\ive{X, Y}$
lies in the kernel of the $\KK$-algebra homomorphism $\pi : \KK \to \LL$.
In other words, not every coefficient of this
polynomial $Q \in \KK\ive{X, Y}$ lies in $I$
(since the kernel of the $\KK$-algebra homomorphism $\pi : \KK \to \LL$
is $I$).
Hence, not every coefficient $c$ of this
polynomial $Q \in \KK\ive{X, Y}$
satisfies $g_{\eps}^{m-1} c = 0$ (by the definition of $I$).
Therefore, $g_{\eps}^{m-1} Q \neq 0$.
So $g_{\eps}^{m-1} Q \in \KK\ive{X, Y}$ is a nonzero polynomial
in two indeterminates $X$ and $Y$ and satisfies
$\deg_X \left( g_{\eps}^{m-1} Q \right) \leq \deg_X Q = \deg_X R \leq \eps - 1 \leq \eps$
and
$\deg_Y \left( g_{\eps}^{m-1} Q \right) \leq \deg_Y Q = \deg_Y R \leq d$
and $\left(g_{\eps}^{m-1} Q \right) \left(F, G\right) = g_{\eps}^{m-1} Q \left(F, G\right) = 0$.
Hence, there exists a nonzero polynomial $P\in\KK \ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $\deg_X P\leq \eps$
and $\deg_Y P\leq d$ and $P\tup{F, G} =0$
(namely, $P = g_{\eps}^{m-1} Q$).
We have thus reached our goal.
So we have proven that Theorem 1 holds for $e = \eps$.
This completes the induction step. Thus, Theorem 1 is proven by induction. $\blacksquare$
