Rank properties of matrix valued in linear forms

Question

Let $R(X,Y) \in \text{Mat}_{d,d+1}(\mathbb{C}[X,Y]_{(1)})$ be a $d \times (d+1)$-matrix valued in linear forms $\mathbb{C}[X,Y]_{(1)}:= \{aX+bY \ \vert \ a,b \in \Bbb C \}$.
Let denote $v_j(X,Y)$ its $j$-th column $d$-vector, and $R_j(X,Y) \in \text{Mat}_{d,d}(\mathbb{C}[X,Y]_{(1)})$ obtained from $R(X,Y)$ by discarding the $j$-th column vector $v_j(X,Y)$.

Assume that that $R(X,Y)$ has "pointwise rank $d$" in the sense that for every $\lambda:= (\lambda_0,\lambda_1) \in \Bbb C^2$ the evaluated $R(\lambda):= R(X,Y)(\lambda)$ in $(X,Y)= (\lambda_0,\lambda_1)$ has rank $d$.

[ Assume that moreover $R(X,Y)$ cannot be transformed via $\Bbb C$-valued row- and column operations into matrix having at least one zero column (note that after evaluation at any specific $\lambda \in \Bbb C^2$ the matrix $R(\lambda) \in \text{Mat}_{d,d+1}(\Bbb{C})$ can always betransformed in such form, but only after having specialized $(X,Y)$ to specific $\lambda$) ; as Jack Huizenga remarked, this bared "assumption" is redundant as this is an immediate consequence of the "pointwise rank $d$ assumption" on $R(X,Y)$.]

Question: Does this "pointwise rank $d$ assumption" already imply that for every $j \in 1,..., d+1$ the determinant $\det R_j(X,Y) $ - which is by construction a $d$-homogenous polynomial in $X,Y$ - is not a zero polynomial? So the question is why this cannot happen here under above assumptions;
Note that as we are over $\Bbb C$ every $d$-homogeneous polynomial $\det R_j(X,Y) $ admits homogeneous zeroes - and my concern is to exclude that accidientally some $\det R_j(X,Y) $ might actually be zero polynomial.

Note that a priori that $R(\lambda)$ has rank $d$ a priori tells that only some maximal minor $R_j(\lambda)$ has rank $d$ and which one(s) depends on concrete $\lambda$. My feeling says the imposed conditions above make $R(X,Y)$ "generic" enough such that for every $j \in 1,..., d+1$ the determinant $\det R_j(X,Y) $ is non zero, but I cannot show it formally.

Geometric picture: This should serve as linear algebra explanation why the construction Dolgachev gave in CAG on page 150 is a non-degenerated curve.

Added after Jack Huizenga's answer: Is it possible to argue using only "elementary methods", ie only linear algebra & properties of polynomials without invoking tools from sheaf theory?
The claim also appears in Harris' First Course on AG (see Ex. 1.24, 1.25), so it suggests that this problem can be approached with "elementary" methods.

Answer 1

The assumption that $R(X,Y)$ is pointwise rank $d$ in fact implies your assumption about row and column operations. If after row and column operations you can get the final column to be zero, then the leading $d\times d$ matrix must have rank $d$ at every point. But its determinant is a polynomial with plenty of roots over $\mathbb{C}$.

We can also see that if $R(X,Y)$ has pointwise rank $d$ everywhere then each of the maximal minors of $R(X,Y)$ are nonzero. The matrix $R(X,Y)$ gives a map of vector bundles on $\mathbb{P}^1$

$$\mathcal{O}(-1)^{d} \to \mathcal{O}^{d+1}.$$

The assumption that the map has pointwise rank $d$ everywhere means that this map is injective and the cokernel is a line bundle, which must be $\mathcal{O}(d)$ by Chern class considerations. In the exact sequence

$$0\to \mathcal{O}(-1)^{d}\to \mathcal{O}^{d+1}\to \mathcal{O}(d)\to 0,$$

the entries of the matrix giving the map $\mathcal{O}^{d+1}\to \mathcal{O}(d)$ are essentially the minors of the matrix $R(X,Y)$. (This is basically Cramer's rule.) If one of them was zero, then the kernel of this map would have to have a factor of $\mathcal{O}$, but it doesn't. Thus each of the maximal minors is nonzero.