[proofplan] We exhibit the image of the Segre map as exactly the locus where all $2 \times 2$ minors of the $(n+1) \times (m+1)$ matrix $[Z_{ij}]$ vanish. This locus is the set of rank-$\le 1$ matrices in $\mathbb{P}^N_k$. Forward: a Segre image $[x_i y_j]$ is the outer product $x y^\top$, which has rank $\le 1$, so it satisfies all $2 \times 2$ minor relations. Reverse: a nonzero matrix with all $2 \times 2$ minors zero has rank exactly one, hence factors as $x y^\top$ for some nonzero $x \in k^{n+1}$, $y \in k^{m+1}$ — and the projective classes $[x]$, $[y]$ are uniquely determined up to the inverse-scaling ambiguity $(x, y) \sim (\lambda x, \lambda^{-1} y)$, so they give a unique preimage in $\mathbb{P}^n_k \times \mathbb{P}^m_k$. Injectivity follows from the same uniqueness statement. [/proofplan] [step:Verify that the Segre map is well-defined] Let $p = [x] = [x_0 : \cdots : x_n] \in \mathbb{P}^n_k$ and $q = [y] = [y_0 : \cdots : y_m] \in \mathbb{P}^m_k$ with $x \in k^{n+1} \setminus \{0\}$, $y \in k^{m+1} \setminus \{0\}$. The proposed image is the tuple $(x_i y_j)_{i,j}$. We verify two things. \textbf{Nonzero.} Pick indices $i_*, j_*$ with $x_{i_*} \neq 0$ and $y_{j_*} \neq 0$ (possible because $x, y \neq 0$). Then $x_{i_*} y_{j_*} \neq 0$ in $k$, so the tuple $(x_i y_j)$ is a nonzero element of $k^{N+1}$ and defines a valid point of $\mathbb{P}^N_k$. \textbf{Independence of representatives.} Replacing $x$ with $\lambda x$ for $\lambda \in k^\times$ and $y$ with $\mu y$ for $\mu \in k^\times$ replaces every entry $x_i y_j$ with $\lambda \mu \, x_i y_j$, which is a global rescaling of the projective tuple by $\lambda \mu \in k^\times$ — hence preserves the projective class $[x_i y_j]$. So $\Sigma_{n,m}$ depends only on the projective classes $[x], [y]$, and is well-defined as a map $\mathbb{P}^n_k \times \mathbb{P}^m_k \to \mathbb{P}^N_k$. [/step] [step:Show the image lies in the vanishing locus of the $2 \times 2$ minors] For $0 \le i, k \le n$ and $0 \le j, l \le m$, define the homogeneous quadratic polynomials \begin{align*} Q_{ijkl}(Z) := Z_{ij} Z_{kl} - Z_{il} Z_{kj} \in k[Z_{ij} : 0 \le i \le n,\, 0 \le j \le m]. \end{align*} These are the $2 \times 2$ minors of the $(n+1) \times (m+1)$ matrix $(Z_{ij})$. Let $\mathfrak{a} \subset k[Z]$ denote the homogeneous ideal generated by all $Q_{ijkl}$, and let $\Sigma := V(\mathfrak{a}) \subset \mathbb{P}^N_k$ denote its zero locus. For any $(p, q) = ([x], [y]) \in \mathbb{P}^n_k \times \mathbb{P}^m_k$ and any indices $i, j, k, l$, \begin{align*} Q_{ijkl}(\Sigma_{n,m}(p, q)) = (x_i y_j)(x_k y_l) - (x_i y_l)(x_k y_j) = x_i x_k y_j y_l - x_i x_k y_l y_j = 0 \end{align*} in $k$. Hence $\Sigma_{n,m}(p, q) \in \Sigma$, so \begin{align*} \Sigma_{n,m}(\mathbb{P}^n_k \times \mathbb{P}^m_k) \subset \Sigma. \end{align*} [/step] [step:Show that points of $\Sigma$ correspond to nonzero rank-one matrices] We translate $\Sigma$ into matrix language. Identify the homogeneous coordinates $(Z_{ij})$ of $\mathbb{P}^N_k$ with the entries of an $(n+1) \times (m+1)$ matrix $M = (M_{ij})$ where $M_{ij} = Z_{ij}$. The vanishing of $Q_{ijkl} = Z_{ij} Z_{kl} - Z_{il} Z_{kj}$ for all $i, j, k, l$ is precisely the vanishing of all $2 \times 2$ minors of $M$. [claim:A nonzero $(n+1) \times (m+1)$ matrix over $k$ has all $2 \times 2$ minors equal to zero if and only if it has rank exactly one] [proof] \textbf{($\Rightarrow$)} A matrix with rank zero is the zero matrix, contradicting nonzero. A matrix with rank $\ge 2$ contains two linearly independent columns (or rows), and the $2 \times 2$ submatrix formed by any two rows that distinguish them has nonzero determinant — i.e. some $2 \times 2$ minor is nonzero. So rank is exactly $1$. \textbf{($\Leftarrow$)} A rank-$1$ matrix $M$ has all columns proportional: there exist $x \in k^{n+1}$ (a fixed nonzero column scaled out) and $y \in k^{m+1} \setminus \{0\}$ with $j$-th entry $y_j$ such that the $j$-th column of $M$ is $y_j x$. Writing this entrywise, $M_{ij} = x_i y_j$, so any $2 \times 2$ minor is \begin{align*} M_{ij} M_{kl} - M_{il} M_{kj} = (x_i y_j)(x_k y_l) - (x_i y_l)(x_k y_j) = 0. \end{align*} [/proof] [/claim] In other words, $\Sigma$ is exactly the projective set of nonzero matrices of rank $\le 1$, and since the zero matrix is excluded from projective space, $\Sigma$ is the projective set of rank-exactly-one matrices. [/step] [step:Show that every rank-one matrix factors as $x y^\top$ uniquely up to inverse scaling] Let $M = (M_{ij}) \in \Sigma$, viewed as a nonzero rank-$1$ matrix in $k^{(n+1) \times (m+1)} / k^\times$. We exhibit a factorisation $M = x y^\top$, i.e. $M_{ij} = x_i y_j$ for all $i, j$, with $x \in k^{n+1} \setminus \{0\}$ and $y \in k^{m+1} \setminus \{0\}$. Pick indices $i_*, j_*$ with $M_{i_* j_*} \neq 0$ (exist because $M \neq 0$). Define \begin{align*} x_i &:= M_{i j_*} / M_{i_* j_*} \in k \quad \text{for } 0 \le i \le n, \\ y_j &:= M_{i_* j} \in k \quad \text{for } 0 \le j \le m. \end{align*} Then $x_{i_*} = 1$ and $y_{j_*} = M_{i_* j_*} \neq 0$, so neither vector is zero. Moreover, for all $i, j$, \begin{align*} x_i \cdot y_j = \frac{M_{i j_*}}{M_{i_* j_*}} \cdot M_{i_* j} = \frac{M_{i j_*} M_{i_* j}}{M_{i_* j_*}}. \end{align*} Apply the $2 \times 2$ minor relation $Q_{i j i_* j_*}(M) = 0$, i.e. \begin{align*} M_{ij} M_{i_* j_*} = M_{i j_*} M_{i_* j}, \qquad \text{equivalently} \qquad M_{ij} = \frac{M_{i j_*} M_{i_* j}}{M_{i_* j_*}}. \end{align*} Hence $M_{ij} = x_i y_j$, establishing the factorisation $M = x y^\top$. [guided] The geometric content is the following: a rank-one $(n+1) \times (m+1)$ matrix is, by definition, an outer product of two vectors. The proof above is the explicit reconstruction of these vectors from the matrix entries. Let us trace why the formulas work. \textbf{Reconstructing the columns.} Any rank-one matrix $M$ has all its columns proportional: there is a fixed direction (the column space) and a vector of column-coefficients (the row space). The pivot row is the row index $i_*$ where some entry is nonzero; this distinguishes a "reference column direction" via $x = M[\, \cdot \, , j_*] / M_{i_* j_*}$ (the $j_*$-th column of $M$ rescaled to have $1$ in slot $i_*$). The "row coefficient" then must be $y_j = M_{i_* j}$, the $j$-th entry of the pivot row. \textbf{Why the minor relation closes the argument.} If we just guessed $x$ and $y$ this way, we might worry: what if $M_{ij}$ doesn't equal $x_i y_j$ for some non-pivot pair $(i, j)$? The minor equation $Q_{i j i_* j_*}(M) = 0$ — which says $M_{ij} M_{i_* j_*} = M_{i j_*} M_{i_* j}$ — is exactly what we need. Rearranging: \begin{align*} M_{ij} = \frac{M_{i j_*} M_{i_* j}}{M_{i_* j_*}} = \frac{M_{i j_*}}{M_{i_* j_*}} \cdot M_{i_* j} = x_i \cdot y_j. \end{align*} So the minor relations exactly enforce the rank-one factorisation: \emph{the algebraic equations defining $\Sigma$ are the algebraic content of "rank one"}. \textbf{Uniqueness up to inverse scaling.} If $x y^\top = x' (y')^\top$ in $k^{(n+1) \times (m+1)}$, with $x, x' \in k^{n+1} \setminus \{0\}$ and $y, y' \in k^{m+1} \setminus \{0\}$, pick $i$ with $x_i \neq 0$ and $j$ with $y_j \neq 0$. Comparing entries: $x_i y_j = x_i' y_j'$, so $\lambda := x_i'/x_i$ is well-defined and nonzero. Comparing the entire $i$-th row gives $x_i y = x_i' y'$, i.e. $y' = (x_i/x_i') y = \lambda^{-1} y$. Comparing the $j$-th column gives $y_j x = y_j' x'$, i.e. $x' = (y_j/y_j') x$, and substituting $y_j' = \lambda^{-1} y_j$ gives $x' = \lambda x$. So $(x', y') = (\lambda x, \lambda^{-1} y)$. \textbf{Why this is the right ambiguity.} On $\mathbb{P}^n_k \times \mathbb{P}^m_k$ we may rescale $x$ and $y$ \emph{independently} by any nonzero scalars: $(x, y) \sim (ax, by)$ for $a, b \in k^\times$. Under the Segre map this rescales $M$ by $ab$, so the projective class $[M]$ is preserved. The above uniqueness says that the only rescalings of $x$ and $y$ that leave $M$ \emph{itself} (not its projective class) unchanged are inverse rescalings $(\lambda, \lambda^{-1})$. Combined: the projective class $([x], [y]) \in \mathbb{P}^n_k \times \mathbb{P}^m_k$ is uniquely determined by the projective class $[M] \in \mathbb{P}^N_k$. [/guided] \textbf{Uniqueness up to inverse scaling.} Suppose also $M = x' (y')^\top$ with $x' \in k^{n+1} \setminus \{0\}$, $y' \in k^{m+1} \setminus \{0\}$. Pick $i$ with $x_i \neq 0$ and $j$ with $y_j \neq 0$ (necessarily $x_i' \neq 0$ and $y_j' \neq 0$ since $x_i y_j = x_i' y_j' \neq 0$). Set $\lambda := x_i'/x_i \in k^\times$. The $i$-th row of $x y^\top$ is $x_i y$, and of $x' (y')^\top$ is $x_i' y'$; equating gives $x_i y = x_i' y'$, so $y' = \lambda^{-1} y$. The $j$-th column comparison gives $x' = (y_j/y_j') x = \lambda x$. So $(x', y') = (\lambda x, \lambda^{-1} y)$, as claimed. [/step] [step:Conclude that $\Sigma_{n,m}$ is injective with image $\Sigma$] \textbf{Image equals $\Sigma$.} By Step 2, $\Sigma_{n,m}(\mathbb{P}^n_k \times \mathbb{P}^m_k) \subset \Sigma$. Conversely, given $[M] \in \Sigma$, by Step 4 there exist $x, y$ nonzero with $M = x y^\top$, equivalently $M_{ij} = x_i y_j$. The projective point $([x], [y]) \in \mathbb{P}^n_k \times \mathbb{P}^m_k$ satisfies $\Sigma_{n,m}([x], [y]) = [x_i y_j] = [M]$, so $[M]$ lies in the image. Hence \begin{align*} \Sigma_{n,m}(\mathbb{P}^n_k \times \mathbb{P}^m_k) = \Sigma = V(\{Z_{ij} Z_{kl} - Z_{il} Z_{kj}\}). \end{align*} \textbf{Injectivity.} Suppose $\Sigma_{n,m}([x], [y]) = \Sigma_{n,m}([x'], [y'])$ in $\mathbb{P}^N_k$, with representatives chosen so that $x y^\top = x' (y')^\top = M \in k^{(n+1) \times (m+1)}$ (after rescaling representatives, the projective equality of Segre images becomes an equality of matrices for some choice of scalar representatives; this is allowed because the Segre map is well-defined on representatives only up to global rescaling). By the uniqueness up to inverse scaling from Step 4, there exists $\lambda \in k^\times$ with $(x', y') = (\lambda x, \lambda^{-1} y)$. In $\mathbb{P}^n_k \times \mathbb{P}^m_k$, multiplying $x$ by $\lambda \in k^\times$ does not change $[x]$, and multiplying $y$ by $\lambda^{-1}$ does not change $[y]$. Hence $([x'], [y']) = ([x], [y])$, establishing injectivity. This completes the proof: $\Sigma_{n,m}$ is injective with image equal to the projective variety $V(\{Z_{ij} Z_{kl} - Z_{il} Z_{kj}\})$ cut out by the $2 \times 2$ minors of $[Z_{ij}]$. [/step]