[proofplan] We prove the equivalence by showing that the four conditions force each other through the standard structure of polynomials over a field. Algebraic closedness gives a root of each nonconstant polynomial, and induction on degree then gives complete splitting. Complete splitting forces irreducible polynomials to be linear, while linearity of all irreducibles gives roots by taking an irreducible factor. Finally, irreducible polynomials of degree greater than $1$ are ruled out either by minimal polynomials in finite extensions or by the quotient field $k[x]/(f)$. [/proofplan] [step:Show that algebraic closedness forces every nonconstant polynomial to split] Assume condition 1. We prove condition 2 by induction on $n=\deg f$. If $n=1$, then $f$ has the form $f=c(x-a)$ for some $c\in k^\times$ and $a\in k$, so $f$ splits into linear factors. Assume now that $n\geq 2$ and that every polynomial in $k[x]$ of degree less than $n$ splits into linear factors. Let $f\in k[x]$ be a polynomial with $\deg f=n$. Since $k$ is algebraically closed, there exists $a\in k$ such that $f(a)=0$. By polynomial division by the monic linear polynomial $x-a$, there exists $g\in k[x]$ such that \begin{align*} f=(x-a)g. \end{align*} Because $\deg f=n$ and $\deg(x-a)=1$, we have $\deg g=n-1$. By the induction hypothesis, there exist $c\in k^\times$ and $a_2,\dots,a_n\in k$ such that \begin{align*} g=c\prod_{j=2}^{n}(x-a_j). \end{align*} Therefore \begin{align*} f=c(x-a)\prod_{j=2}^{n}(x-a_j). \end{align*} Renaming $a$ as $a_1$ gives the required splitting. Thus condition 2 holds. [/step] [step:Show that a split irreducible polynomial must be linear] Assume condition 2. Let $p\in k[x]$ be irreducible. Since irreducible polynomials are nonconstant, condition 2 gives an integer $n\geq 1$, a scalar $c\in k^\times$, and elements $a_1,\dots,a_n\in k$ such that \begin{align*} p=c\prod_{j=1}^{n}(x-a_j). \end{align*} If $n\geq 2$, then \begin{align*} p=\bigl(c(x-a_1)\bigr)\prod_{j=2}^{n}(x-a_j), \end{align*} where both factors have positive degree. This contradicts irreducibility of $p$. Hence $n=1$, and therefore $\deg p=1$. Thus condition 3 holds. [/step] [step:Recover algebraic closedness from linearity of irreducible polynomials] Assume condition 3. Let $f\in k[x]$ be nonconstant. Since $k[x]$ is a [Euclidean domain](/page/Euclidean%20Domain) with degree function $\deg:k[x]\setminus\{0\}\to \mathbb{N}\cup\{0\}$, every nonconstant polynomial has an irreducible factor: among all nonconstant divisors of $f$, choose one of minimal positive degree; if it factored into two nonconstant polynomials, one factor would be a nonconstant divisor of smaller positive degree. Let $q\in k[x]$ be an irreducible factor of $f$. By condition 3, $\deg q=1$, so there exist $c\in k^\times$ and $a\in k$ such that \begin{align*} q=c(x-a). \end{align*} Since $q$ divides $f$, there exists $h\in k[x]$ such that $f=qh$. Evaluating at $a$ gives \begin{align*} f(a)=q(a)h(a)=0. \end{align*} Thus every nonconstant polynomial in $k[x]$ has a root in $k$, which is exactly condition 1. [guided] Assume condition 3, meaning that every irreducible polynomial in $k[x]$ has degree $1$. To prove that $k$ is algebraically closed under the definition used in the statement, we must show that every nonconstant polynomial $f\in k[x]$ has a root in $k$. The first point is to find an irreducible factor of $f$. The ring $k[x]$ has a degree function \begin{align*} \deg:k[x]\setminus\{0\}\to \mathbb{N}\cup\{0\}. \end{align*} Consider the collection of all nonconstant divisors of $f$. This collection is nonempty because $f$ divides itself. Choose a divisor $q\in k[x]$ of minimal positive degree among these divisors. If $q$ were reducible, then there would exist nonconstant polynomials $u,v\in k[x]$ such that \begin{align*} q=uv. \end{align*} Both $u$ and $v$ would have positive degree strictly smaller than $\deg q$, and $u$ would still divide $f$ because $f=qh$ for some $h\in k[x]$. That contradicts the minimality of $\deg q$. Therefore $q$ is irreducible. Now condition 3 applies to this particular irreducible factor $q$. Hence $\deg q=1$, so there exist $c\in k^\times$ and $a\in k$ with \begin{align*} q=c(x-a). \end{align*} Because $q$ divides $f$, there exists $h\in k[x]$ such that \begin{align*} f=qh. \end{align*} Evaluate this identity at $a$. Since $q(a)=c(a-a)=0$, we get \begin{align*} f(a)=q(a)h(a)=0. \end{align*} Thus $a\in k$ is a root of $f$. Since $f$ was an arbitrary nonconstant polynomial in $k[x]$, every nonconstant polynomial over $k$ has a root in $k$. By the definition fixed in the statement, $k$ is algebraically closed. [/guided] [/step] [step:Use minimal polynomials to show that finite algebraic extensions are trivial] Assume condition 3. Let $K/k$ be a finite algebraic [field extension](/page/Field%20Extension). We prove that every element of $K$ lies in $k$. Let $\alpha\in K$. Define the evaluation homomorphism \begin{align*} \operatorname{ev}_{\alpha}:k[x]\to K,\quad h\mapsto h(\alpha). \end{align*} Since $\alpha$ is algebraic over $k$, the kernel of $\operatorname{ev}_{\alpha}$ contains a nonzero polynomial. Let $m_{\alpha,k}\in k[x]$ be the monic polynomial of smallest positive degree in this kernel. We show that $m_{\alpha,k}$ is irreducible. If \begin{align*} m_{\alpha,k}=uv \end{align*} with $u,v\in k[x]$ both nonconstant, then \begin{align*} 0=m_{\alpha,k}(\alpha)=u(\alpha)v(\alpha). \end{align*} Because $K$ is a field, either $u(\alpha)=0$ or $v(\alpha)=0$, contradicting the minimality of $\deg m_{\alpha,k}$. Thus $m_{\alpha,k}$ is irreducible. By condition 3, $\deg m_{\alpha,k}=1$. Since $m_{\alpha,k}$ is monic, there exists $a\in k$ such that \begin{align*} m_{\alpha,k}=x-a. \end{align*} The identity $m_{\alpha,k}(\alpha)=0$ gives $\alpha=a\in k$. Since $\alpha\in K$ was arbitrary, $K\subset k$. The extension already has $k\subset K$, so $K=k$. Thus condition 4 holds. [/step] [step:Construct a finite extension from a nonlinear irreducible polynomial] Assume condition 4. Let $f\in k[x]$ be irreducible. We prove that $\deg f=1$. Suppose, for contradiction, that $n=\deg f>1$. Because $f$ is irreducible in the [principal ideal domain](/page/Principal%20Ideal%20Domain) $k[x]$, the ideal $(f)\trianglelefteq k[x]$ is maximal. Hence \begin{align*} K:=k[x]/(f) \end{align*} is a field. Define the natural embedding \begin{align*} \iota:k\to K,\quad a\mapsto a+(f). \end{align*} This map is injective because $(f)$ contains no nonzero constant polynomial. We therefore regard $K/\iota(k)$ as a field extension. Let $\alpha:=x+(f)\in K$. Every class in $K$ has a representative of degree less than $n$ by polynomial division by the monic associate of $f$. Hence $K$ is spanned over $\iota(k)$ by \begin{align*} 1+(f),\alpha,\alpha^2,\dots,\alpha^{n-1}. \end{align*} Thus $K/\iota(k)$ is finite. Finite field extensions are algebraic, so $K/\iota(k)$ is a finite algebraic extension. By condition 4, $K=\iota(k)$. Since $K=\iota(k)$, there exists $a\in k$ such that \begin{align*} \alpha=\iota(a). \end{align*} Equivalently, \begin{align*} x-a\in(f). \end{align*} Thus $f$ divides $x-a$ in $k[x]$, which is impossible when $\deg f=n>1$. This contradiction shows that $\deg f=1$. Therefore condition 3 holds. [guided] Assume condition 4: every finite algebraic extension of $k$ is equal to $k$. We want to prove that every irreducible polynomial in $k[x]$ has degree $1$. Let $f\in k[x]$ be irreducible, and suppose for contradiction that \begin{align*} n:=\deg f>1. \end{align*} The standard way to force a root of $f$ into an extension field is to pass to the quotient by the ideal generated by $f$. Since $k[x]$ is a principal ideal domain and $f$ is irreducible, the ideal $(f)\trianglelefteq k[x]$ is maximal. Therefore the [quotient ring](/page/Quotient%20Ring) \begin{align*} K:=k[x]/(f) \end{align*} is a field. We must make precise how $k$ sits inside this quotient field. Define \begin{align*} \iota:k\to K,\quad a\mapsto a+(f). \end{align*} This is a field homomorphism. It is injective because if $\iota(a)=0$ for some $a\in k$, then $a\in(f)$. A nonzero constant polynomial cannot be a multiple of the nonconstant polynomial $f$, so $a=0$. Thus $K$ is a field extension of the embedded copy $\iota(k)$. Now define the element \begin{align*} \alpha:=x+(f)\in K. \end{align*} This element is the class of the polynomial variable. The quotient construction makes $\alpha$ a root of $f$: if \begin{align*} f=b_0+b_1x+\cdots+b_nx^n \end{align*} with $b_0,\dots,b_n\in k$, then in $K$ we have \begin{align*} b_0+(f)+b_1\alpha+\cdots+b_n\alpha^n=f+(f)=0. \end{align*} Next we verify that the extension is finite. By polynomial division by a monic associate of $f$, every polynomial $h\in k[x]$ can be written as \begin{align*} h=qf+r \end{align*} with $q,r\in k[x]$ and either $r=0$ or $\deg r<n$. Passing to the quotient gives \begin{align*} h+(f)=r+(f). \end{align*} Therefore every element of $K$ is an $\iota(k)$-linear combination of \begin{align*} 1+(f),\alpha,\alpha^2,\dots,\alpha^{n-1}. \end{align*} So $K/\iota(k)$ is finite-dimensional, hence a finite field extension. Every finite field extension is algebraic because each element of a finite-dimensional [vector space](/page/Vector%20Space) over the base field satisfies a nonzero polynomial relation among $1,\beta,\beta^2,\dots,\beta^m$ for sufficiently large $m$. Thus $K/\iota(k)$ is a finite algebraic extension. Condition 4 now applies, so $K=\iota(k)$. In particular, the element $\alpha\in K$ must equal $\iota(a)$ for some $a\in k$. This means \begin{align*} x+(f)=a+(f). \end{align*} Equivalently, \begin{align*} x-a\in(f). \end{align*} Thus $f$ divides the linear polynomial $x-a$ in $k[x]$. This is impossible because $\deg f=n>1$. The contradiction proves that no irreducible polynomial in $k[x]$ can have degree greater than $1$, so every irreducible polynomial in $k[x]$ has degree $1$. [/guided] [/step] [step:Combine the implications] We have proved condition 1 implies condition 2, condition 2 implies condition 3, condition 3 implies condition 1, condition 3 implies condition 4, and condition 4 implies condition 3. Hence conditions 1, 2, 3, and 4 are equivalent. [/step]