[proofplan] We prove irreducibility directly from the action of the standard generators $e$ and $f$ on the basis vectors. Given a nonzero submodule $W \subset V$, choose a nonzero vector $w = a v_1 + b v_2$ in $W$. If the coefficient of $v_1$ is nonzero, applying $f$ gives $v_2 \in W$, and then applying $e$ gives $v_1 \in W$. If the coefficient of $v_1$ is zero, then $w$ already gives $v_2 \in W$, and applying $e$ again gives $v_1 \in W$. Thus every nonzero submodule contains a basis of $V$. [/proofplan] [step:Record the action of the standard generators on the basis vectors] By the standard matrix action of $\mathfrak{sl}_2(k)$ on $V = k^2$, the elements $e$ and $f$ act on the basis vectors $v_1 = (1,0)$ and $v_2 = (0,1)$ by \begin{align*} e v_1 &= 0, & e v_2 &= v_1, & f v_1 &= v_2, & f v_2 &= 0. \end{align*} If $W \subset V$ is an $\mathfrak{sl}_2(k)$-submodule, then $eW \subset W$ and $fW \subset W$. [/step] [step:Show that a nonzero submodule contains $v_2$] Let $W \subset V$ be a nonzero $\mathfrak{sl}_2(k)$-submodule. Choose $w \in W$ with $w \ne 0$. Since $(v_1,v_2)$ is a basis of $V$, there exist unique scalars $a,b \in k$ such that \begin{align*} w = a v_1 + b v_2. \end{align*} If $a \ne 0$, then applying $f$ gives \begin{align*} f w = f(a v_1 + b v_2) = a f v_1 + b f v_2 = a v_2. \end{align*} Because $W$ is stable under $f$, we have $f w \in W$. Since $a \ne 0$ and $k$ is a field, $a^{-1} \in k$, so \begin{align*} v_2 = a^{-1} f w \in W. \end{align*} If $a = 0$, then $w = b v_2$. Since $w \ne 0$, we have $b \ne 0$, and therefore \begin{align*} v_2 = b^{-1} w \in W. \end{align*} Thus in every case $v_2 \in W$. [/step] [step:Apply $e$ to obtain the remaining basis vector] Since $v_2 \in W$ and $W$ is stable under the action of $e$, we have \begin{align*} v_1 = e v_2 \in W. \end{align*} Therefore both basis vectors $v_1$ and $v_2$ belong to $W$. [/step] [step:Conclude that the only nonzero submodule is the whole module] Because $v_1,v_2 \in W$ and $W$ is a $k$-linear subspace of $V$, every vector $c v_1 + d v_2$ with $c,d \in k$ belongs to $W$. Hence \begin{align*} V = \operatorname{span}_k\{v_1,v_2\} \subset W. \end{align*} The reverse inclusion $W \subset V$ holds by definition of submodule, so $W = V$. Thus every nonzero $\mathfrak{sl}_2(k)$-submodule of $V$ is equal to $V$, and the standard two-dimensional $\mathfrak{sl}_2(k)$-module is irreducible. [/step]