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  "body": "\n## Proof\n\nWe prove a stronger claim, which allows a change of base field, by strong [[induction|0ASJVEFA9J]] on the degree $[L : K]$.\n\n1. For a ring isomorphism $\\tau \\colon K \\to K'$ of fields and $h = \\sum_i c_i x^i \\in K[x]$, write $\\tau h = \\sum_i \\tau(c_i) x^i$. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], $h \\mapsto \\tau h$ is a ring isomorphism $K[x] \\to K'[x]$, with inverse $h \\mapsto \\tau^{-1} h$.\n2. Claim: let $\\tau \\colon K \\to K'$ be a [[ring isomorphism|0B6QC0C0G1]] of [[fields|0QA8CHH8MG]], and $f \\in K[x]$ nonzero. Let $L$ be a [[splitting field|02NZZ1J0FH]] of $f$ over $K$, and $L'$ one of $\\tau f$ over $K'$. Then some ring isomorphism $\\sigma \\colon L \\to L'$ extends $\\tau$.\n3. The statement is the case $K' = K$ and $\\tau = \\id$, so it suffices to prove the claim. First, $[L : K]$ is finite. Indeed, $L = K(a_1, \\dots, a_n)$ for roots $a_i$ of $f$, which are algebraic over $K$.\n4. Adjoining $a_1, \\dots, a_n$ one at a time gives a chain of fields from $K$ to $L$. Each step is finite by the [[degree of a simple extension|0N5QW9K7MH]], so $[L : K]$ is finite by the [[tower law|0A7KJ68XHT]].\n5. Write $f = c (x - a_1) \\cdots (x - a_n)$ in $L[x]$ and $\\tau f = c' (x - b_1) \\cdots (x - b_n)$ in $L'[x]$, with $L = K(a_1, \\dots, a_n)$ and $L' = K'(b_1, \\dots, b_n)$, as in the definitions of [[splitting|0SHMC00RJV]] and [[generated field|0MA5B744K3]]. Let $d = [L : K]$.\n6. Case $d = 1$. Then $L = K$, so all $a_i \\in K$. Applying the ring homomorphism of Step 1 gives $\\tau f = \\tau(c) (x - \\tau(a_1)) \\cdots (x - \\tau(a_n))$ in $K'[x]$.\n7. Comparing with Step 5 by unique factorisation in $L'[x]$, which is a [[Euclidean domain|0DQG39K2NH]], hence a [[PID|0QKVE1GD9A]] and a [[UFD|055HT85T3S]], each $b_j$ is some $\\tau(a_i) \\in K'$. So $L' = K'$, and $\\sigma = \\tau$ works.\n8. Case $d \u003e 1$, assuming the claim whenever the degree is less than $d$. If all $a_i$ were in $K$, then $L = K$ and $d = 1$. So some root $a = a_i$ is not in $K$.\n9. Let $m$ be the [[minimal polynomial|0GK1WB9R98]] of $a$ over $K$. It is [[irreducible|0NFEG48RDR]], it [[divides|06K7SARZKR]] $f$, and $\\deg m = [K(a) : K] \u003e 1$ since $a \\notin K$.\n10. By Step 1, $\\tau m$ is monic, irreducible in $K'[x]$, of degree $\\deg m$, and $\\tau m$ divides $\\tau f$. So $\\tau m$ divides $\\prod_j (x - b_j)$ in $L'[x]$, as $c' \\ne 0$.\n11. Let $\\pi$ be a monic divisor of $\\tau m$ in $L'[x]$ of least degree $\\ge 1$. A factorisation of $\\pi$ into two nonunits would give a monic divisor of smaller [[degree|0T233AZ1GE]] $\\ge 1$.\n12. So $\\pi$ is irreducible. By Step 7, $L'[x]$ is a UFD, so $\\pi$ is [[prime|0QRCFZKXH8]].\n13. By the definition of a [[prime element|0MNHXBSX3M]], $\\pi$ divides some factor $x - b_j$. Comparing degrees, $\\pi = x - b_j$. So $b = b_j$ is a root of $\\tau m$.\n14. Since $\\tau m$ is monic, irreducible and $\\tau m(b) = 0$, divisibility gives that $\\tau m$ is the minimal polynomial of $b$ over $K'$.\n15. Let $\\psi \\colon K[x] \\to K'(b)$, $h \\mapsto (\\tau h)(b)$. It is a ring homomorphism, and it is surjective because $K'(b) = \\set{ u(b) : u \\in K'[x] }$ by the [[structure of simple extensions|0ZN39QDTP9]].\n16. Its kernel is the set of $h$ with $\\tau m \\mid \\tau h$, by Step 14 and divisibility. By Step 1, this is the set of $h$ with $m \\mid h$, that is, $(m)$.\n17. By the [[first isomorphism theorem|0ZMYX7GMH0]], $K[x]/(m) \\cong K'(b)$ through $h + (m) \\mapsto (\\tau h)(b)$. Also $K[x]/(m) \\cong K(a)$ through $h + (m) \\mapsto h(a)$, by the structure of simple extensions.\n18. Composing, $\\sigma_1 \\colon K(a) \\to K'(b)$, $h(a) \\mapsto (\\tau h)(b)$, is a ring isomorphism. On constants it equals $\\tau$, so it extends $\\tau$, and $\\sigma_1 f = \\tau f$.\n19. The field $L$ is a splitting field of $f$ over $K(a)$: $f$ splits over $L$, and $L$ is generated over $K(a)$ by $a_1, \\dots, a_n$, since $K(a) \\subseteq L$.\n20. In the same way, $L'$ is a splitting field of $\\sigma_1 f = \\tau f$ over $K'(b)$, because $b$ is one of the $b_j$.\n21. By the tower law, $[L : K(a)] = d / [K(a) : K] \u003c d$. The induction hypothesis, for $\\sigma_1$, gives a ring isomorphism $\\sigma \\colon L \\to L'$ extending $\\sigma_1$.\n22. Then $\\sigma$ extends $\\tau$, which proves the claim. With $\\tau = \\id_K$, we get $\\sigma \\colon L \\to L'$ with $\\sigma(c) = c$ for all $c \\in K$.\n\n## Idea\n\nBuild the isomorphism one root at a time. A root $a$ of $f$ outside $K$ has a minimal polynomial $m$, and its image $\\tau m$ has a root $b$ among the roots of $\\tau f$. Both $K(a)$ and $K'(b)$ are copies of $K[x]/(m)$, so $\\tau$ extends to them. Over these larger fields the degree is smaller, and induction finishes.\n",
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