Ufyukyu

Question

The problem statement is "Let $F in mathbb{C}[t]$ have degree at most $D geq 1$, and let $G in mathbb{C}[t]$ have degree $E geq 1$.

Show that there is a $P not = 0$ in $mathbb{C}[X,Y]$ with degree at most $E$ in $X$ and $D$ in $Y$ such that $P(F,G) = 0$."

I've tried to work this out by supposing $$P = sum_{i=0}^{E} sum_{j=0}^{D}c_{ij}X^iY^j$$, and noticing that I have control over the choice of $(1+E)(1+D)$ coefficients. I was hoping to be able to use this information to create a homogeneous system of linear equations to give a non-trivial solution for the $c_{ij}$ that would force $P(F,G) = 0$.

Is this a viable approach? If so, what would my next step be? I asked my professor, and the hint he gave me was to think about the resultant of $F$ and $G$. I know how to construct the matrix whose determinant is the resultant of $F$ and $G$, and I know the resultant is $0$ if $F$ and $G$ share a common factor, but I don't know how that helps us with this problem

Thanks in advance for any comments, hints, or solutions!

Your approach, I fear, only gives weaker bounds (degrees at most $2E$ and $2D$, not $E$ and $D$). — Sep 30 at 21:52
See mathoverflow.net/a/189344 for the proof using resultants. The idea is to take the resultant of the two polynomials $Fleft(Tright) - X$ and $Gleft(Tright) - Y$ in the indeterminate $T$ over the ring $mathbb{C}left[X,Yright]$. This resultant is nonzero as a polynomial, but will become $0$ when $Fleft(Uright)$ and $Gleft(Uright)$ (with $U$ being yet another indeterminate) are substituted for $X$ and $Y$, since then the two polynomials will have the common root $U = T$. — Sep 30 at 21:58
Would you like to post this as an answer so you can receive the bounty on the problem? This answered everything I had in mind. — Oct 8 at 0:19

score 2 · Accepted Answer · 2018-11-19 19:30:15Z

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.)

$newcommand{KK}{mathbb{K}}
newcommand{LL}{mathbb{L}}
newcommand{NN}{mathbb{N}}
newcommand{ww}{mathbf{w}}
newcommand{eps}{varepsilon}
newcommand{Res}{operatorname{Res}}
newcommand{Syl}{operatorname{Syl}}
newcommand{adj}{operatorname{adj}}
newcommand{id}{operatorname{id}}
newcommand{tilF}{widetilde{F}}
newcommand{tilG}{widetilde{G}}
newcommand{ive}[1]{left[ #1 right]}
newcommand{tup}[1]{left( #1 right)}
newcommand{zeroes}[1]{underbrace{0,0,ldots,0}_{#1 text{ zeroes}}}$
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 $PinKK ive{X, Y}$ in two indeterminates $X$ and $Y$ such that $deg_X Pleq e$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$.

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 $Pin KK ive{T}$ and $QinKK ive{T}$ be two polynomials in the polynomial ring $KK ive{T}$.
Let $dinNN$ and $einNN$ be such that $deg Pleq d$ and $deg Qleq e$.
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 qquadtext{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{equation}
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{ }_{etext{ columns}}
underbrace{ }_{dtext{ columns}}
end{array}
right) inKK^{tup{d+e} timestup{d+e}};
end{equation}
this is the $tup{d+e} timestup{d+e}$-matrix whose first $e$ columns have the form
begin{equation}
left( zeroes{k},p_0 ,p_1 ,ldots ,p_d ,zeroes{e-1-k}right) ^{T}
qquadtext{for }kinleft{ 0,1,ldots,e-1right} ,
end{equation}
and whose last $d$ columns have the form
begin{equation}
left( zeroes{ell},q_0 ,q_1 ,ldots,q_e ,zeroes{d-1-ell}right) ^{T}
qquadtext{for }ellinleft{ 0,1,ldots,d-1right} .
end{equation}
Furthermore, we define $Res_{d,e}tup{P, Q}$ to be the element
begin{equation}
det tup{ Syl_{d,e}tup{P, Q} } in KK .
end{equation}
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 $left{0,1,2,ldotsright}$ of all nonnegative integers.

One of the main properties of resultants is the following:

Theorem 2. Let $KK$ be a commutative ring.
Let $Pin KK ive{T}$ and $QinKK ive{T}$ be two polynomials in the polynomial ring $KK ive{T}$.
Let $dinNN$ and $einNN$ be such that $d+e > 0$ and $deg Pleq d$ and $deg Qleq e$.
Let $LL$ be a commutative $KK$-algebra, and let $winLL$ satisfy $Ptup{w} =0$ and $Qtup{w} = 0$.
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 qquadtext{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
$p_0 + p_1 w + p_2 w^2 + cdots + p_d w^d = Pleft(wright) = 0$.
Similarly,
$q_0 + q_1 w + q_2 w^2 + cdots + q_e w^e = 0$.

Let $A$ be the matrix $Syl_{d,e}tup{P, Q}
inKK^{tup{d+e} timestup{d+e} }$, regarded as a
matrix in $LL ^{tup{d+e} timestup{d+e} }$ (by
applying the canonical $KK$-algebra homomorphism $KK
rightarrowLL$ to all its entries).

Let $ww$ be the row vector $left( w^{0},w^{1},ldots,w^{d+e-1}
right) inLL ^{1timestup{d+e} }$. Let $mathbf{0}$ denote
the zero vector in $LL ^{1timestup{d+e} }$.

Now, it is easy to see that $ww A=mathbf{0}$. (Indeed, for each
$kinleft{ 1,2,ldots,d+eright} $, we have
begin{align*}
& wwleft( text{the }ktext{-th column of }Aright) \
& =
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 }kleq 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 }kleq e;\
w^{k-e-1}left( q_0 +q_1 w+q_2 w^2 +cdots+q_e w^eright) , & text{if }k>e
end{cases}
\
& =
begin{cases}
w^{k-1}0, & text{if }kleq e;\
w^{k-e-1}0, & text{if }k>e
end{cases}
\
& qquadleft(
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 $Acdotadj A=det
Acdot I_{d+e}$ (where $I_{d+e}$ denotes the identity matrix of size $d+e$).
Hence, $wwunderbrace{Acdotadj A}_{=det Acdot
I_{d+e}}=wwdet Acdot I_{d+e}=det Acdotww$. Comparing this
with $underbrace{ww A}_{=mathbf{0}}cdotadj A
=mathbf{0}cdotadj A=mathbf{0}$, we obtain
$det Acdotww=mathbf{0}$.

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 $det Acdotww$ is $det A cdot 1 = det A$.
Hence, from $det Acdotww=mathbf{0}$, we conclude that $det A=0$.
Comparing this with
begin{equation}
detunderbrace{A}_{=Syl_{d,e}tup{P, Q}} =det tup{ Syl_{d,e}tup{P, Q} }
=Res_{d,e}tup{P, Q} ,
end{equation}
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:

Antoine Chambert-Loir, Résultants (minor errata).

Svante Janson, Resultant and discriminant of polynomials.

Gerald Myerson, On resultants, Proc. Amer. Math. Soc. 89 (1983), 419--420.

Some of these sources use the matrix $left( Syl_{d,e}tup{P, Q} right) ^{T}$ 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.

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 $g_0 ,g_1 ,ldots,g_e inKK$.
Assume that $g_e^d neq 0$.
Then, there exists a nonzero polynomial $PinKK ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $deg_X Pleq e$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$.

Proof of Lemma 3 (sketched). Let $widetilde{KK}$ be the
commutative ring $KK ive{X, Y}$. Define two polynomials
$tilF inwidetilde{KK}ive{T}$ and $tilG inwidetilde{KK}ive{T}$ by
begin{equation}
tilF =F-X=Ftup{T} -Xqquadtext{and}qquadtilG
=G-Y=Gtup{T} -Y.
end{equation}
Note that $X$ and $Y$ have degree $0$ when considered as polynomials in
$widetilde{KK}ive{T}$ (since $X$ and $Y$ belong to the
ring $widetilde{KK}$). Thus, these new polynomials $tilF = F - X$
and $tilG = G - Y$ have degrees $degtilF leq d$ (because
$deg X = 0 leq d$ and $deg F leq d$) and
$degtilG leq e$ (similarly).
Hence, the resultant $Res_{d,e}tup{tilF, tilG} in
widetilde{KK}$ of these polynomials $tilF$ and
$tilG$ in degrees $d$ and $e$ is well-defined. Let us denote this
resultant $Res_{d,e}left( tilF
,tilG right)$ by $P$. Hence,
begin{equation}
P=Res_{d,e}left( tilF ,tilG
right) inwidetilde{KK}=KK ive{X, Y} .
end{equation}

Our next goal is to show that $P$ is a nonzero polynomial and satisfies
$deg_X Pleq e$ and $deg_Y Pleq d$ and $Ptup{F, G} =0$. Once
this is shown, Lemma 3 will obviously follow.

We have
begin{equation}
P=Res_{d,e}left( tilF ,tilG
right) =detleft( Syl_{d,e}left( tilF
,tilG right) right)
label{darij1.pf.t1.P=det}
tag{1}
end{equation}
(by the definition of $Res_{d,e}left(
tilF ,tilG right)$).

Write the polynomial $F$ in the form $F=f_0 +f_1 T+f_2 T^2 +cdots
+f_d T^d $, where $f_0 ,f_1 ,ldots,f_d inKK$. (This can be
done, since $deg F leq d$.)

Recall that $g_e ^d neq 0$. Thus, $left( -1right) ^e g_e^d neq 0$.

For each $pinNN$, we let $S_{p}$ be the group of all permutations of
the set $left{ 1,2,ldots,pright} $.

Now,
begin{align*}
tilF & =F-X=left( f_0 +f_1 T+f_2 T^2 +cdots+f_d
T^d right) -X\
& qquadleft( 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 $tilF inwidetilde{KK}ive{T}$ (since
$f_0 -Xinwidetilde{KK}$). Similarly, $g_0 -Y,g_1 ,g_2
,ldots,g_e $ are the coefficients of the polynomial $tilG
inwidetilde{KK}ive{T}$. 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}
_{etext{ columns}}
underbrace{qquad qquad qquad qquad qquad qquad qquad}
_{dtext{ columns}}
end{array}
right) \
&inwidetilde{KK}^{tup{d+e} timesleft(
d+eright) }.
end{align}
Now, let us use this explicit form of $Syl_{d,e} tup{tilF, tilG}$
to compute $detleft( Syl_{d,e}tup{tilF, tilG} right)$ using
the Leibniz formula.
The Leibniz formula yields
begin{equation}
detleft( Syl_{d,e} tup{ tilF, tilG } right)
=sum_{sigmain S_{d+e}}a_{sigma},
label{darij1.pf.t1.det=sum}
tag{2}
end{equation}
where for each permutation $sigmain S_{d+e}$, the addend $a_{sigma}$ is a
product of entries of $Syl_{d,e} tup{tilF, tilG}$, possibly with a minus sign. More
precisely,
begin{equation}
a_{sigma}=tup{-1}^{sigma}prod_{i=1}^{d+e}left( text{the }
left( i,sigmaleft( iright) right) text{-th entry of }
Syl_{d,e}tup{tilF, tilG}
right)
end{equation}
for each $sigmain S_{d+e}$ (where $tup{-1}^{sigma}$ denotes the
sign of the permutation $sigma$).

Now, eqref{darij1.pf.t1.P=det} becomes
begin{equation}
P=detleft( Syl_{d,e}tup{tilF, tilG} right)
=sum_{sigmain S_{d+e}}a_{sigma}
label{darij1.pf.t1.P=sum}
tag{3}
end{equation}
(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 $sigmain S_{d+e}$, 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 $sigmain S_{d+e}$, the
product $a_{sigma}$ has degree $leq ecdot1=e$ with respect to the variable
$X$. Hence, the sum $sum_{sigmain S_{d+e}}a_{sigma}$ of all these products
$a_{sigma}$ also has degree $leq e$ with respect to the variable $X$. In
other words, $deg_X left( sum_{sigmain S_{d+e}}a_{sigma}right) leq
e$. In view of eqref{darij1.pf.t1.P=sum}, this rewrites as $deg_X Pleq e$.
Similarly, $deg_Y Pleq d$ (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 $widetilde{KK}$ as polynomials in the
variable $X$ over the ring $KK ive{Y} $. For each
permutation $sigmain S_{d+e}$, 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 $sigmain S_{d+e}$ does not satisfy $left(
sigmaleft( iright) =itext{ for each }iinleft{ 1,2,ldots,eright}
right)$. 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 $sigmain S_{d+e}$ satisfies $left(
sigmaleft( iright) =itext{ for each }iinleft{ 1,2,ldots,eright}
right)$, but is not the identity map $idin S_{d+e}$. Thus,
there must exist at least one $iinleft{ 1,2,ldots,d+eright} $ such
that $sigmaleft( iright) <i$. Consider such an $i$, and notice that it
must satisfy $i>e$ and $sigmaleft( iright) >e$; hence, the $left(
i,sigmaleft( iright) right)$-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 $sigmain S_{d+e}$ is the identity map
$idin S_{d+e}$. Thus, the product $a_{sigma}$ is $left(
f_0 -Xright) ^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 $sum_{sigmain S_{d+e}}a_{sigma}$ is $tup{-1}^e g_e^d
neq0$. Therefore, $sum_{sigmain S_{d+e}}a_{sigma}neq0$. In view of
eqref{darij1.pf.t1.P=sum}, this rewrites as $Pneq0$. In other words, the
polynomial $P$ is nonzero.

Finally, it remains to prove that $Ptup{F, G} =0$. In order to do
this, we let $LL$ be the polynomial ring $KK ive{U}
$ in a new indeterminate $U$. We let $varphi:KK ive{X, Y}
rightarrowLL$ be the unique $KK$-algebra homomorphism that
sends $X$ to $Ftup{U}$ and sends $Y$ to $Gtup{U}$.
(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
$widetilde{KK}$ to $LL$ (since $KK ive{X, Y}
=widetilde{KK}$). Thus, $LL$ becomes a $widetilde{KK
}$-algebra via this homomorphism $varphi$.

Now, recall that the polynomial $tilF inwidetilde{KK}ive{T}$
was defined by $tilF =F-X$. Hence, $tilF left(
Uright) =Ftup{U} -varphitup{X}$. (Indeed, when we
regard $X$ as an element of $widetilde{KK}ive{T}$, the
polynomial $X$ is simply a constant, and thus evaluating it at $U$ yields the
canonical image of $X$ in $LL$, which is $varphitup{X}$.)
But $varphitup{X} =Ftup{U}$ (by the definition of
$varphi$).
Hence, $tilF tup{U} =Ftup{U} -varphitup{X} =0$ (since $varphitup{X} =Ftup{U}$). Similarly, $tilG tup{U} =0$.

Thus, the element $UinLL$ satisfies $tilF tup{U}
=0$ and $tilG tup{U} =0$. Hence, Theorem 2 (applied to
$widetilde{KK}$, $tilF$, $tilG$ and $U$ instead of $KK$, $P$, $Q$ and $w$) yields that
$Res_{d,e}tup{tilF, tilG} = 0$ in $LL$.
In other words, $varphitup{ Res_{d,e}tup{tilF, tilG} } =0$. In view of
$Res_{d,e}tup{tilF, tilG} =P$, this rewrites as $varphitup{P} =0$.

But recall that $varphi$ is the $KK$-algebra homomorphism that sends
$X$ to $Ftup{U}$ and sends $Y$ to $Gtup{U}$. Hence, it
sends any polynomial $QinKK ive{X, Y}$ to $Q tup{ Ftup{U}, Gtup{U} }$. Applying this to $Q=P$, we
conclude that it sends $P$ to $P tup{ Ftup{U}, Gtup{U} }$.
In other words, $varphitup{P} =P tup{ Ftup{U}, Gtup{U} }$;
hence, $P tup{ Ftup{U}, Gtup{U} } =varphitup{P} =0$.

Now, $Ftup{U}$ and $Gtup{U}$ are polynomials in the
indeterminate $U$ over $KK$. If we rename the indeterminate $U$ as
$T$, then these polynomials $Ftup{U}$ and $Gtup{U}$
become $Ftup{T}$ and $Gtup{T}$, and therefore the
polynomial $Pleft( Ftup{U} ,Gtup{U} right)$ becomes
$Ptup{ Ftup{T}, Gtup{T} }$.
Hence, $Ptup{ Ftup{T}, Gtup{T} } =0$ (since $Ptup{ Ftup{U}, Gtup{U} } =0$).
In other words, $P tup{F, G} =0$ (since $Ftup{T} =F$ and $Gtup{T} =G$).
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 $X - f in KKive{X, Y}$.
Then, $Q$ is nonzero and satisfies
$deg_X Q = 1 leq e$ (since $e geq 1$) and
$deg_Y Q = 0 leq d$ and
$Qleft(F, Gright) = F - f = 0$ (since $F = f$).
Hence, there exists a nonzero polynomial $PinKK ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $deg_X Pleq e$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$
(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 $PinKK ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $deg_X Pleq eps$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$.

Write the polynomial $G$ in the form $G=g_0 +g_1 T+g_2 T^2 +cdots
+g_{eps}T^{eps}$, where $g_0 ,g_1 ,ldots,g_{eps}inKK$.
(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, $d + left(e-1right) geq d > 0$.

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 $KKive{T} to LLive{T}$ (sending $T$ to $T$).
For any $a in KKive{T}$, 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 $KKive{X, Y} to LLive{X, Y}$ (sending $X$ and $Y$ to $X$ and $Y$).
For any $a in KKive{X, Y}$, 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 $Pin LLive{X, Y}$ in two indeterminates $X$ and $Y$ such that $deg_X P leq eps - 1$ and $deg_Y P leq d$ and $Pleft( overline{F}, overline{G} right) =0$.
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 $Q in KKive{X, Y}$ 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
$R = sumlimits_{i = 0}^{deg_X R} sumlimits_{j = 0}^{deg_Y R} r_{i, j} X^i Y^j$
for some coefficients $r_{i, j} in LL$.
For each pair $left(i, jright)$, 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 $Q = sumlimits_{i = 0}^{deg_X R} sumlimits_{j = 0}^{deg_Y R} p_{i, j} X^i Y^j$.
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 $overline{Q left(F, Gright)}
= underbrace{overline{Q}}_{=R} left( overline{F}, overline{G} right)
= R left( overline{F}, overline{G} right) = 0$.
In other words, the polynomial $Q left(F, Gright) in KKive{T}$
lies in the kernel of the canonical
$KK$-algebra homomorphism $KKive{T} to LLive{T}$.
This means that each coefficient of this
polynomial $Q left(F, Gright) in KKive{T}$
lies in the kernel of the $KK$-algebra homomorphism $pi : KK to LL$.
In other words, each coefficient of this
polynomial $Q left(F, Gright) in KKive{T}$ lies in $I$
(since the kernel of the $KK$-algebra homomorphism $pi : KK to LL$
is $I$).
Hence, each coefficient $c$ of this
polynomial $Q left(F, Gright) in KKive{T}$
satisfies $g_{eps}^{m-1} c = 0$ (by the definition of $I$).
Therefore, $g_{eps}^{m-1} Q left(F, Gright) = 0$.

On the other hand, $overline{Q} = R$ is nonzero.
In other words, the polynomial $Q in KKive{X, Y}$ does not lie
in the kernel of the canonical
$KK$-algebra homomorphism $KKive{X, Y} to LLive{X, Y}$.
This means that not every coefficient of this
polynomial $Q in KKive{X, Y}$
lies in the kernel of the $KK$-algebra homomorphism $pi : KK to LL$.
In other words, not every coefficient of this
polynomial $Q in KKive{X, Y}$ 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 $Q in KKive{X, Y}$
satisfies $g_{eps}^{m-1} c = 0$ (by the definition of $I$).
Therefore, $g_{eps}^{m-1} Q neq 0$.
So $g_{eps}^{m-1} Q in KKive{X, Y}$ 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 $left(g_{eps}^{m-1} Q right) left(F, Gright) = g_{eps}^{m-1} Q left(F, Gright) = 0$.
Hence, there exists a nonzero polynomial $PinKK ive{X, Y}$
in two indeterminates $X$ and $Y$ such that $deg_X Pleq eps$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$
(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. $blacksquare$

score 2 · Accepted Answer · 2018-11-19 19:30:15Z

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.)

$newcommand{KK}{mathbb{K}}
newcommand{LL}{mathbb{L}}
newcommand{NN}{mathbb{N}}
newcommand{ww}{mathbf{w}}
newcommand{eps}{varepsilon}
newcommand{Res}{operatorname{Res}}
newcommand{Syl}{operatorname{Syl}}
newcommand{adj}{operatorname{adj}}
newcommand{id}{operatorname{id}}
newcommand{tilF}{widetilde{F}}
newcommand{tilG}{widetilde{G}}
newcommand{ive}[1]{left[ #1 right]}
newcommand{tup}[1]{left( #1 right)}
newcommand{zeroes}[1]{underbrace{0,0,ldots,0}_{#1 text{ zeroes}}}$
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 $PinKK ive{X, Y}$ in two indeterminates $X$ and $Y$ such that $deg_X Pleq e$
and $deg_Y Pleq d$ and $Ptup{F, G} =0$.

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 $Y$.