[proofplan] We argue by contradiction: an isomorphism $\Phi : \mathbb{P}^1_k \times \mathbb{P}^1_k \to \mathbb{P}^2_k$ would convert the first projection $\pi_1 : \mathbb{P}^1_k \times \mathbb{P}^1_k \to \mathbb{P}^1_k$ into a morphism $\pi_1 \circ \Phi^{-1} : \mathbb{P}^2_k \to \mathbb{P}^1_k$. By [All Morphisms from $\mathbb{P}^n_k$ to $\mathbb{P}^1_k$ Are Constant](/theorems/2158) applied with $n = 2$, this morphism is constant. But composing with the isomorphism $\Phi$ shows $\pi_1$ must also be constant, contradicting the fact that $\pi_1$ is non-constant (it surjects onto $\mathbb{P}^1_k$ by varying the first factor while fixing the second). The birationality is shown by exhibiting matching standard affine charts on both sides, both of which are $\mathbb{A}^2_k$ with function field $k(s, t)$. [/proofplan] [step:Set up the projection $\pi_1$ and verify it is a non-constant morphism] Define the first projection \begin{align*} \pi_1 : \mathbb{P}^1_k \times \mathbb{P}^1_k &\to \mathbb{P}^1_k \\ (p, q) &\mapsto p. \end{align*} We verify two properties of $\pi_1$: \textbf{$\pi_1$ is a morphism of projective varieties.} Under the [Segre embedding](/theorems/2147) \begin{align*} \Sigma_{1,1} : \mathbb{P}^1_k \times \mathbb{P}^1_k &\to \mathbb{P}^3_k \\ ([x_0 : x_1], [y_0 : y_1]) &\mapsto [x_0 y_0 : x_0 y_1 : x_1 y_0 : x_1 y_1], \end{align*} the product $\mathbb{P}^1_k \times \mathbb{P}^1_k$ is identified with the Segre variety $S_{1,1} \subset \mathbb{P}^3_k$, a projective variety. The projection $\pi_1$ corresponds, on $S_{1,1}$, to "reading off the first column" of the $2 \times 2$ matrix of homogeneous coordinates: writing $Z_{ij} = x_i y_j$ for the four Segre coordinates of $S_{1,1}$, on the chart $\{Z_{00} \neq 0\} \cap S_{1,1}$ (where $y_0 \neq 0$), \begin{align*} \pi_1([Z_{00} : Z_{01} : Z_{10} : Z_{11}]) = [Z_{00} : Z_{10}] = [x_0 y_0 : x_1 y_0] = [x_0 : x_1], \end{align*} where the last equality is by cancelling the common scalar $y_0 \in k^\times$ in projective coordinates. On the other chart $\{Z_{01} \neq 0\} \cap S_{1,1}$ (where $y_1 \neq 0$), \begin{align*} \pi_1([Z]) = [Z_{01} : Z_{11}] = [x_0 y_1 : x_1 y_1] = [x_0 : x_1]. \end{align*} The two chart formulas agree on overlap (both equal $[x_0 : x_1]$ on points where both $y_0, y_1 \neq 0$), and the charts $\{Z_{00} \neq 0\}, \{Z_{01} \neq 0\}$ cover $S_{1,1}$ (every point has at least one of $y_0, y_1$ non-zero). Hence $\pi_1$ is a well-defined morphism on $S_{1,1}$. The standard verification that $\pi_1$ is a morphism of projective varieties is detailed in the proof of [Products of Projective Varieties are Projective](/theorems/2147). \textbf{$\pi_1$ is non-constant.} The image of $\pi_1$ is all of $\mathbb{P}^1_k$: for any $p \in \mathbb{P}^1_k$, picking any $q \in \mathbb{P}^1_k$ (e.g. $q = [1 : 0]$), we have $\pi_1((p, q)) = p$. Hence $\pi_1$ is surjective. In particular, $\pi_1$ takes at least two distinct values (since $|\mathbb{P}^1_k| \geq 2$ — for example $[0 : 1]$ and $[1 : 0]$ are distinct points), so $\pi_1$ is not constant. [/step] [step:Suppose for contradiction that $\Phi : \mathbb{P}^1_k \times \mathbb{P}^1_k \to \mathbb{P}^2_k$ is an isomorphism] Suppose there exists an isomorphism of projective varieties \begin{align*} \Phi : \mathbb{P}^1_k \times \mathbb{P}^1_k \xrightarrow{\sim} \mathbb{P}^2_k. \end{align*} By definition of isomorphism, $\Phi$ is a bijective morphism with morphism inverse $\Phi^{-1} : \mathbb{P}^2_k \to \mathbb{P}^1_k \times \mathbb{P}^1_k$. In particular, both $\Phi$ and $\Phi^{-1}$ are morphisms of projective varieties. Define the composition \begin{align*} \psi : \mathbb{P}^2_k \xrightarrow{\Phi^{-1}} \mathbb{P}^1_k \times \mathbb{P}^1_k \xrightarrow{\pi_1} \mathbb{P}^1_k. \end{align*} The composition of morphisms is a morphism (a basic property of the category of projective varieties: morphism composition is well-defined, since the composition of regular maps is again regular). Hence \begin{align*} \psi : \mathbb{P}^2_k \to \mathbb{P}^1_k \end{align*} is a morphism of projective varieties. [/step] [step:Apply the constancy theorem and derive a contradiction] By [All Morphisms from $\mathbb{P}^n_k$ to $\mathbb{P}^1_k$ Are Constant](/theorems/2158) applied with $n = 2$, every morphism $\mathbb{P}^2_k \to \mathbb{P}^1_k$ is constant. The hypotheses are: \begin{itemize} \item $k$ algebraically closed — given. \item $n \geq 2$ — verified, $n = 2$. \end{itemize} Hence $\psi : \mathbb{P}^2_k \to \mathbb{P}^1_k$ is constant: there exists $p_0 \in \mathbb{P}^1_k$ such that $\psi(r) = p_0$ for every $r \in \mathbb{P}^2_k$. Now we transport this constancy back to $\pi_1$ via $\Phi$. For every $(p, q) \in \mathbb{P}^1_k \times \mathbb{P}^1_k$, set $r = \Phi((p, q)) \in \mathbb{P}^2_k$. Then \begin{align*} \pi_1((p, q)) = \pi_1(\Phi^{-1}(\Phi((p, q)))) = \pi_1(\Phi^{-1}(r)) = \psi(r) = p_0. \end{align*} Hence $\pi_1((p, q)) = p_0$ for every $(p, q) \in \mathbb{P}^1_k \times \mathbb{P}^1_k$ — i.e. $\pi_1$ is constant. This contradicts Step 1, where $\pi_1$ was shown to be non-constant. The assumption "there exists an isomorphism $\Phi$" is therefore false: no such isomorphism exists, completing the proof of $\mathbb{P}^1_k \times \mathbb{P}^1_k \not\cong \mathbb{P}^2_k$. [guided] The proof is a clean transport-of-structure argument: an isomorphism $\Phi$ would let us push the projection $\pi_1$ across to a non-constant morphism out of $\mathbb{P}^2_k$, but the constancy theorem forbids non-constant morphisms out of $\mathbb{P}^2_k$ to $\mathbb{P}^1_k$. \textbf{The strategy in one sentence.} If two varieties are isomorphic, they have the same morphism structure — in particular, the same set of morphisms to any third variety. So an obstruction to morphisms out of one is automatically an obstruction to morphisms out of the other, after composing with the isomorphism. \textbf{Why the projection $\pi_1$?} The projections are the simplest non-constant morphisms out of $\mathbb{P}^1_k \times \mathbb{P}^1_k$. Their existence is what distinguishes products from indecomposable projective varieties. By "exporting" $\pi_1$ across the hypothetical isomorphism, we get a non-constant morphism out of $\mathbb{P}^2_k$ — but the latter cannot exist. \textbf{Why doesn't $\mathbb{P}^2_k$ have non-constant morphisms to $\mathbb{P}^1_k$?} The deeper reason is rigidity: a morphism $\mathbb{P}^2_k \to \mathbb{P}^1_k$ would correspond to a pair of homogeneous polynomials of equal degree with no common zero in $\mathbb{P}^2_k$, but [Projective Hypersurfaces Always Intersect](/theorems/2157) forces them to share a zero. Geometrically, $\mathbb{P}^2_k$ is "too connected" — any two curves in $\mathbb{P}^2_k$ meet by Bezout's theorem (a refinement of [Projective Hypersurfaces Always Intersect](/theorems/2157)). The product $\mathbb{P}^1_k \times \mathbb{P}^1_k$ is less connected: the projections give it a global "splitting" that is incompatible with being a 2-dimensional projective space. \textbf{Birationality is real, isomorphism is not.} Both varieties have the same dimension (two), and both have the same function field $k(s, t)$ — the birational equivalence reflects the fact that they look the same on a dense open subset. But the global structure differs: $\mathbb{P}^2_k$ has Picard group $\mathbb{Z}$ (line bundles classified by a single integer, the degree), while $\mathbb{P}^1_k \times \mathbb{P}^1_k$ has Picard group $\mathbb{Z} \oplus \mathbb{Z}$ (one integer per factor). This is the deeper reason for non-isomorphism, and the morphism-counting argument above is one concrete manifestation. \textbf{Where does the algebraic-closedness of $k$ enter?} Through the Nullstellensatz, used implicitly in [Projective Hypersurfaces Always Intersect](/theorems/2157), which feeds into [All Morphisms from $\mathbb{P}^n_k$ to $\mathbb{P}^1_k$ Are Constant](/theorems/2158). Without algebraic closedness, the projective varieties might have fewer geometric points than algebra suggests, and the no-common-zero condition for morphisms could fail to translate cleanly to ideal conditions. [/guided] [/step] [step:Verify that $\mathbb{P}^1_k \times \mathbb{P}^1_k$ and $\mathbb{P}^2_k$ are nonetheless birational] We exhibit a $k$-isomorphism of function fields, which by [Birational Equivalence via Function Fields](/theorems/2144) implies birational equivalence. \textbf{The function field of $\mathbb{P}^2_k$.} The standard affine chart $U_0 = \{[1 : a_1 : a_2] : (a_1, a_2) \in \mathbb{A}^2_k\}$ of $\mathbb{P}^2_k$ is isomorphic to $\mathbb{A}^2_k$ by [Affine Cover of Projective Space](/theorems/2129). The function field of $\mathbb{A}^2_k$ is $\operatorname{Frac}(k[s, t]) = k(s, t)$ (rational functions in two variables). Since the function field of an irreducible variety equals the function field of any non-empty open subset, $k(\mathbb{P}^2_k) = k(s, t)$. \textbf{The function field of $\mathbb{P}^1_k \times \mathbb{P}^1_k$.} The product of standard affine charts $U_0 \times U_0 \subset \mathbb{P}^1_k \times \mathbb{P}^1_k$ is isomorphic to $\mathbb{A}^1_k \times \mathbb{A}^1_k = \mathbb{A}^2_k$. Specifically, on this chart we have coordinates $s, t$ where $s = a_1/a_0$ for the first factor $[a_0 : a_1] \in \mathbb{P}^1_k$ (with $a_0 \neq 0$ on $U_0$) and $t = b_1/b_0$ for the second factor $[b_0 : b_1]$ (with $b_0 \neq 0$ on $U_0$). Hence the function field of $\mathbb{P}^1_k \times \mathbb{P}^1_k$ is $k(\mathbb{A}^2_k) = k(s, t)$. \textbf{Conclusion.} Both varieties have function field $k$-isomorphic to $k(s, t)$. By [Birational Equivalence via Function Fields](/theorems/2144), \begin{align*} \mathbb{P}^1_k \times \mathbb{P}^1_k \text{ is birational to } \mathbb{P}^2_k. \end{align*} This birationality is realised explicitly by the rational maps \begin{align*} \mathbb{P}^1_k \times \mathbb{P}^1_k \dashrightarrow \mathbb{P}^2_k, \qquad ([a_0 : a_1], [b_0 : b_1]) \mapsto [a_0 b_0 : a_1 b_0 : a_0 b_1], \end{align*} which is defined on the open dense subset $\{a_0 \neq 0, b_0 \neq 0\}$ and inverted by the rational map \begin{align*} \mathbb{P}^2_k \dashrightarrow \mathbb{P}^1_k \times \mathbb{P}^1_k, \qquad [c_0 : c_1 : c_2] \mapsto ([c_0 : c_1], [c_0 : c_2]), \end{align*} defined on the open dense subset $\{c_0 \neq 0\}$. These rational maps are inverses on the common open subset, witnessing the birational equivalence — but neither extends to a global morphism, consistent with the non-isomorphism established in Steps 1–3. [/step]