[proofplan] We prove both directions. For the forward direction, we show that any irreducible polynomial over an algebraically closed field must be linear by using the existence of a root and performing polynomial division. For the converse, we show that if every irreducible polynomial is linear, then every non-constant polynomial has a root by factoring it into irreducible factors. [/proofplan] [step:Show that irreducible polynomials over an algebraically closed field are linear] Suppose $k$ is algebraically closed and let $f \in k[x]$ be irreducible. Since $f$ is irreducible, it is non-constant, so $\deg f \geq 1$. Because $k$ is algebraically closed, $f$ has a root $\alpha \in k$. By the [Factor Theorem](/theorems/???), $f(x) = (x - \alpha) g(x)$ for some $g \in k[x]$. Since $f$ is irreducible in $k[x]$ and $x - \alpha$ is a non-unit factor, we must have $g \in k^\times$ (i.e., $g$ is a unit in $k[x]$). Therefore $\deg f = 1$, so $f$ is a scalar multiple of $x - \alpha$. Since irreducible polynomials are non-zero by definition, $f = c(x - \alpha)$ for some $c \in k^\times$. As irreducibility is unaffected by multiplication by units, every irreducible polynomial is associate to some $x - \alpha$ with $\alpha \in k$. [guided] We want to show that if $k$ is algebraically closed, then no irreducible polynomial can have degree $\geq 2$. The key idea is: an irreducible polynomial cannot be factored into non-trivial factors, but having a root forces a non-trivial factorisation for any polynomial of degree $\geq 2$. Suppose $k$ is algebraically closed and let $f \in k[x]$ be irreducible. Since $f$ is irreducible, it is in particular non-constant, so $\deg f \geq 1$. Because $k$ is algebraically closed, every non-constant polynomial over $k$ has a root in $k$. Let $\alpha \in k$ satisfy $f(\alpha) = 0$. By the [Factor Theorem](/theorems/???), we can write $f(x) = (x - \alpha) g(x)$ for some $g \in k[x]$. Now, $f$ is irreducible, meaning it cannot be written as a product of two non-unit elements of $k[x]$. The factor $x - \alpha$ has degree $1$, so it is not a unit (units in $k[x]$ are exactly the non-zero constants). Therefore $g$ must be a unit, i.e., $g \in k^\times$. This forces $\deg f = \deg(x - \alpha) + \deg g = 1 + 0 = 1$. So $f = c(x - \alpha)$ for some non-zero $c \in k$. Since multiplying by a unit does not change irreducibility, the irreducible polynomials in $k[x]$ are precisely the associates of the linear polynomials $x - \alpha$ for $\alpha \in k$. [/guided] [/step] [step:Show that if every irreducible polynomial is linear, then $k$ is algebraically closed] Conversely, suppose that the only irreducible polynomials in $k[x]$ are the linear polynomials $x - \alpha$ for $\alpha \in k$. Let $f \in k[x]$ be non-constant. Since $k[x]$ is a unique factorisation domain, $f$ factors as a product of irreducible polynomials: \begin{align*} f(x) = c \prod_{i=1}^{m} (x - \alpha_i) \end{align*} for some $c \in k^\times$, $m \geq 1$, and $\alpha_1, \dots, \alpha_m \in k$, where each factor $x - \alpha_i$ is irreducible by hypothesis. In particular, $f(\alpha_1) = 0$, so $f$ has a root in $k$. Since $f$ was an arbitrary non-constant polynomial, $k$ is algebraically closed. [guided] Now suppose every irreducible polynomial in $k[x]$ has the form $x - \alpha$ for some $\alpha \in k$. We must show that $k$ is algebraically closed, meaning every non-constant polynomial $f \in k[x]$ has a root in $k$. Since $k$ is a field, $k[x]$ is a Euclidean domain and hence a unique factorisation domain. Every non-constant polynomial $f \in k[x]$ therefore factors as a finite product of irreducible polynomials (unique up to order and units): \begin{align*} f(x) = c \prod_{i=1}^{m} p_i(x) \end{align*} where $c \in k^\times$ and each $p_i$ is irreducible. By our hypothesis, each $p_i$ is of the form $x - \alpha_i$ for some $\alpha_i \in k$. Since $f$ is non-constant, $m \geq 1$, so there is at least one linear factor. Evaluating at $\alpha_1$: \begin{align*} f(\alpha_1) = c \cdot 0 \cdot \prod_{i=2}^{m} (\alpha_1 - \alpha_i) = 0. \end{align*} Therefore $f$ has a root in $k$. Since $f$ was an arbitrary non-constant polynomial over $k$, this shows $k$ is algebraically closed. [/guided] [/step]