[proofplan] A field is perfect when every irreducible polynomial over it is separable. We split the proof into two cases according to the characteristic. In characteristic zero, every irreducible polynomial has nonzero formal derivative, so it is automatically separable. In characteristic $p > 0$, an irreducible polynomial fails to be separable precisely when it lies in $k[x^p]$, and we show this pathology is impossible if and only if the Frobenius is surjective. Conversely, if the Frobenius is not surjective, we exhibit an explicit inseparable irreducible polynomial. [/proofplan] [step:Recall the separability criterion for irreducible polynomials] By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), an irreducible polynomial $f \in k[x]$ is inseparable if and only if its formal derivative $f'(x)$ vanishes identically. This occurs if and only if $\operatorname{char}(k) = p > 0$ and $f(x) = g(x^p)$ for some $g \in k[x]$. A field $k$ is perfect if and only if every irreducible polynomial in $k[x]$ is separable. We must therefore show that every irreducible polynomial in $k[x]$ is separable if and only if one of conditions (1) or (2) holds. [guided] The definition of a perfect field is: every algebraic extension of $k$ is separable. Equivalently, every irreducible polynomial in $k[x]$ is separable (since an algebraic extension is separable if and only if the minimal polynomial of every element is separable). How can an irreducible polynomial fail to be separable? By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), an irreducible $f \in k[x]$ is inseparable if and only if $f'(x) = 0$ identically. Since differentiation reduces degree by one, the only way for $f'$ to vanish identically is if the coefficient $n \cdot a_n$ of every monomial $a_n x^n$ with $n \geq 1$ equals zero. In characteristic zero this forces all such $a_n = 0$, making $f$ constant -- a contradiction. In characteristic $p > 0$, the coefficient $n \cdot a_n$ can vanish with $a_n \neq 0$ precisely when $p \mid n$. So $f'(x) = 0$ if and only if $f(x) = g(x^p)$ for some $g \in k[x]$. This criterion is the fulcrum of the entire proof: it tells us that inseparable irreducible polynomials can only exist in characteristic $p > 0$, and when they do, they have the very specific form $g(x^p)$. [/guided] [/step] [step:Show that characteristic zero implies perfection] Suppose $\operatorname{char}(k) = 0$. Let $f \in k[x]$ be irreducible with $\deg(f) = n \geq 1$. Write $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ with $a_n \neq 0$. Then \begin{align*} f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \cdots + a_1. \end{align*} Since $\operatorname{char}(k) = 0$, the integer $n$ is nonzero in $k$, so $n a_n \neq 0$. Therefore $f'(x) \neq 0$, hence $\deg(f') = n - 1 \geq 0$. By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), $f$ is separable. Since $f$ was an arbitrary irreducible polynomial, $k$ is perfect. [guided] Suppose $\operatorname{char}(k) = 0$. We want to show every irreducible polynomial is separable. Take any irreducible $f \in k[x]$ of degree $n \geq 1$, and write $f(x) = a_n x^n + \cdots + a_0$ with $a_n \neq 0$. The formal derivative is \begin{align*} f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \cdots + a_1. \end{align*} In characteristic zero, the leading coefficient $n a_n$ of $f'$ is nonzero: $n \neq 0$ in $k$ (since no positive integer maps to zero under the characteristic-zero ring homomorphism $\mathbb{Z} \to k$), and $a_n \neq 0$ by assumption. So $f'$ is a nonzero polynomial of degree $n - 1$, which means $f'$ does not vanish identically. By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), $f$ is separable. Since $f$ was arbitrary, every irreducible polynomial over $k$ is separable, so $k$ is perfect. This is why characteristic zero is the "easy" case: the formal derivative of a polynomial of degree $n$ automatically has degree $n - 1$, so it can never vanish identically. There is no room for inseparability. [/guided] [/step] [step:Show that surjective Frobenius implies perfection in characteristic $p$] Suppose $\operatorname{char}(k) = p > 0$ and $\varphi_p: k \to k$ is surjective. Let $f \in k[x]$ be irreducible. We claim $f$ is separable. Suppose for contradiction that $f$ is inseparable. By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), $f(x) = g(x^p)$ for some $g \in k[x]$. Write \begin{align*} g(x) = b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0 \end{align*} with $b_i \in k$. Since $\varphi_p$ is surjective, for each $i$ there exists $c_i \in k$ with $c_i^p = b_i$. Then \begin{align*} f(x) = g(x^p) = \sum_{i=0}^{m} b_i x^{pi} = \sum_{i=0}^{m} c_i^p x^{pi} = \left( \sum_{i=0}^{m} c_i x^i \right)^p, \end{align*} where the last equality uses the Frobenius endomorphism on the polynomial ring: in characteristic $p$, $(a + b)^p = a^p + b^p$ for all $a, b$ in any commutative ring of characteristic $p$, applied inductively to the sum $\sum c_i x^i$. Setting $h(x) = \sum_{i=0}^{m} c_i x^i$, we obtain $f(x) = h(x)^p$. Since $\deg(h) = m$ and $\deg(f) = pm$, and $m \geq 1$ (as $\deg(f) \geq 1$ and $p \geq 2$ would require $m \geq 1$; more precisely, $\deg(f) = pm \geq p > 1$ so $h$ is nonconstant), $f$ factors nontrivially. This contradicts the irreducibility of $f$. Therefore $f$ must be separable. [guided] Suppose $\operatorname{char}(k) = p > 0$ and the Frobenius $\varphi_p: k \to k$, $x \mapsto x^p$, is surjective. We want to show every irreducible polynomial over $k$ is separable. Suppose for contradiction that some irreducible $f \in k[x]$ is inseparable. By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), this means $f(x) = g(x^p)$ for some $g \in k[x]$. Write $g(x) = \sum_{i=0}^{m} b_i x^i$ with $b_i \in k$. Now we use the surjectivity of $\varphi_p$: for each coefficient $b_i$, there exists $c_i \in k$ such that $c_i^p = b_i$. Substituting into $f$: \begin{align*} f(x) = g(x^p) = \sum_{i=0}^{m} b_i (x^p)^i = \sum_{i=0}^{m} c_i^p \cdot x^{pi}. \end{align*} The key algebraic fact is that in characteristic $p$, the Frobenius identity $(a + b)^p = a^p + b^p$ holds in any commutative ring of characteristic $p$. Applying this identity inductively to the finite sum $\sum_{i=0}^{m} c_i x^i$, we obtain \begin{align*} \left( \sum_{i=0}^{m} c_i x^i \right)^p = \sum_{i=0}^{m} c_i^p x^{pi}. \end{align*} So $f(x) = h(x)^p$ where $h(x) = \sum_{i=0}^{m} c_i x^i$. Since $\deg(f) = pm \geq p \geq 2$, we have $m \geq 1$, so $h$ is nonconstant. But then $f = h^p$ is a nontrivial factorisation of $f$ (as $1 \leq \deg(h) = m < pm = \deg(f)$), contradicting the irreducibility of $f$. The contradiction shows that no irreducible polynomial over $k$ can be inseparable. Therefore $k$ is perfect. The idea here is that surjectivity of the Frobenius lets us "extract $p$-th roots of coefficients," which in turn lets us factor any polynomial of the form $g(x^p)$ as a perfect $p$-th power. So the only polynomials that could be inseparable are automatically reducible -- they cannot be both irreducible and inseparable. [/guided] [/step] [step:Show that non-surjective Frobenius yields an inseparable irreducible polynomial] Suppose $\operatorname{char}(k) = p > 0$ and $\varphi_p$ is not surjective. Then there exists $a \in k$ such that $a \notin k^p := \{c^p : c \in k\}$, i.e., $a$ has no $p$-th root in $k$. Consider the polynomial \begin{align*} f(x) = x^p - a \in k[x]. \end{align*} [claim:$f(x) = x^p - a$ is irreducible in $k[x]$] Let $\alpha \in \overline{k}$ be a root of $f$, so that $\alpha^p = a$. In $\overline{k}[x]$, using the Frobenius identity, $f$ factors as \begin{align*} x^p - a = x^p - \alpha^p = (x - \alpha)^p. \end{align*} So $\alpha$ is the unique root of $f$ in $\overline{k}$, with multiplicity $p$. Let $g \in k[x]$ be the minimal polynomial of $\alpha$ over $k$, and let $d = \deg(g)$, so $g \mid f$ in $k[x]$. Since $f = (x - \alpha)^p$ in $\overline{k}[x]$, every factor of $f$ in $\overline{k}[x]$ has the form $(x - \alpha)^j$ for some $1 \leq j \leq p$. In particular, $g(x) = (x - \alpha)^d$ in $\overline{k}[x]$. Expanding $(x - \alpha)^d$ by the binomial theorem, the constant term of $g$ is $(-\alpha)^d = (-1)^d \alpha^d$, and the coefficient of $x^{d-1}$ is $\binom{d}{1}(-\alpha)^{d-1} = -d\alpha^{d-1}$. Since $g \in k[x]$, these coefficients lie in $k$. We show $d = p$. Suppose $d < p$. Since $p$ is prime and $1 \leq d < p$, we have $\gcd(d, p) = 1$. The coefficient of $x^{d-1}$ in $g$ is $-d \alpha^{d-1}$. Since $\operatorname{char}(k) = p$ and $\gcd(d, p) = 1$, the integer $d$ is invertible in $k$, so $\alpha^{d-1} = (-d)^{-1} \cdot (-d\alpha^{d-1}) \in k$. Similarly, the constant term gives $\alpha^d \in k$ (up to a sign). Therefore $\alpha = \alpha^d \cdot (\alpha^{d-1})^{-1} \in k$ (the quotient is valid since $\alpha \neq 0$: if $\alpha = 0$ then $a = \alpha^p = 0 \in k^p$, contradicting $a \notin k^p$). But $\alpha \in k$ implies $a = \alpha^p \in k^p$, contradicting our assumption that $a \notin k^p$. Therefore $d = p$, so $g = f$ (both are monic of degree $p$ with $g \mid f$), and $f$ is irreducible. [/claim] [proof] Proved inline above. [/proof] Since $f(x) = x^p - a = (x - \alpha)^p$ in $\overline{k}[x]$, the polynomial $f$ has $\alpha$ as a root of multiplicity $p$. Thus $f$ is inseparable. We have exhibited an irreducible inseparable polynomial, so $k$ is not perfect. [guided] Now we handle the converse: if $\operatorname{char}(k) = p > 0$ and $\varphi_p$ is not surjective, we must exhibit an irreducible inseparable polynomial over $k$ to show $k$ is not perfect. Since $\varphi_p$ is not surjective, there exists $a \in k$ that is not a $p$-th power in $k$: $a \notin k^p$. Consider the polynomial $f(x) = x^p - a$. **Why is $f$ inseparable?** In the algebraic closure $\overline{k}$, let $\alpha$ be a root, so $\alpha^p = a$. Using the Frobenius identity $(x - \alpha)^p = x^p - \alpha^p$ (valid since $\operatorname{char}(k) = p$), we get \begin{align*} f(x) = x^p - a = x^p - \alpha^p = (x - \alpha)^p. \end{align*} So $\alpha$ is the only root, with multiplicity $p$. In particular, $f$ is inseparable. **Why is $f$ irreducible?** This is the substantive part. Let $g \in k[x]$ be the minimal polynomial of $\alpha$ over $k$, with $\deg(g) = d$. Since $g \mid f$ in $k[x]$ and $f = (x - \alpha)^p$ in $\overline{k}[x]$, we must have $g(x) = (x - \alpha)^d$ in $\overline{k}[x]$ for some $1 \leq d \leq p$. We need to show $d = p$. Suppose for contradiction that $d < p$. Since $p$ is prime and $1 \leq d < p$, we have $\gcd(d, p) = 1$. Now look at the coefficients of $g(x) = (x - \alpha)^d$, which must all lie in $k$: - The coefficient of $x^{d-1}$ is $\binom{d}{1}(-\alpha)^{d-1} = -d\alpha^{d-1}$. - The constant term is $(-\alpha)^d$. Since $g \in k[x]$, both $-d\alpha^{d-1} \in k$ and $(-\alpha)^d \in k$. Because $\gcd(d, p) = 1$, the integer $d$ is nonzero in $k$ (it is not divisible by $p$), so $d$ is invertible. Therefore $\alpha^{d-1} \in k$. Also $\alpha^d \in k$ (from the constant term). If $\alpha = 0$, then $a = \alpha^p = 0$, but $0 = 0^p \in k^p$, contradicting $a \notin k^p$. So $\alpha \neq 0$, and we can form $\alpha = \alpha^d / \alpha^{d-1} \in k$. But then $a = \alpha^p \in k^p$, again contradicting $a \notin k^p$. This contradiction shows $d = p$, so the minimal polynomial $g$ has degree $p = \deg(f)$. Since $g$ divides $f$ and both are monic of the same degree, $g = f$. Since $g$ is irreducible (as a minimal polynomial), $f$ is irreducible. So $f(x) = x^p - a$ is an irreducible polynomial that is inseparable (it has a single root of multiplicity $p$). Therefore $k$ is not perfect. To summarize the logic: when the Frobenius is not surjective, we can find $a \notin k^p$, and the polynomial $x^p - a$ is simultaneously irreducible (because $\alpha$ cannot lie in $k$ or in any proper intermediate extension) and inseparable (because it equals $(x - \alpha)^p$ over $\overline{k}$). [/guided] [/step] [step:Combine the cases to conclude the equivalence] We have established: - If $\operatorname{char}(k) = 0$, then every irreducible polynomial in $k[x]$ is separable, so $k$ is perfect. - If $\operatorname{char}(k) = p > 0$ and $\varphi_p$ is surjective, then every irreducible polynomial in $k[x]$ is separable, so $k$ is perfect. - If $\operatorname{char}(k) = p > 0$ and $\varphi_p$ is not surjective, then $x^p - a$ (for any $a \notin k^p$) is an irreducible inseparable polynomial, so $k$ is not perfect. The first two points show that conditions (1) or (2) imply $k$ is perfect. For the converse, suppose $k$ is perfect and $\operatorname{char}(k) = p > 0$. If $\varphi_p$ were not surjective, the third point would give an inseparable irreducible polynomial, contradicting perfection. So $\varphi_p$ must be surjective, establishing condition (2). This completes the proof of the biconditional. [/step]