[proofplan] We show that $\sigma$ is an isomorphism by interpreting it as a $K$-linear map between $K$-vector spaces of the same finite dimension. First, we verify that $\sigma$ is $K$-linear by using the defining property of a $K$-homomorphism. Second, we establish injectivity: the kernel of $\sigma$ is an ideal of the field $L$, and since $\sigma$ is a nonzero ring homomorphism, this kernel must be $\{0\}$. Finally, the rank-nullity theorem for finite-dimensional vector spaces implies that an injective linear map between spaces of the same dimension is surjective, completing the proof. [/proofplan] [step:Verify that $\sigma$ is $K$-linear by expanding the homomorphism conditions] The map $\sigma: L \to L'$ is a ring homomorphism satisfying $\sigma|_K = \operatorname{id}_K$. We verify $K$-linearity. Let $\lambda \in K$ and $x, y \in L$. **Additivity.** Since $\sigma$ is a ring homomorphism, $\sigma(x + y) = \sigma(x) + \sigma(y)$. **Scalar compatibility.** Since $\sigma$ is a ring homomorphism that fixes $K$ pointwise: \begin{align*} \sigma(\lambda x) = \sigma(\lambda) \cdot \sigma(x) = \lambda \cdot \sigma(x). \end{align*} The second equality uses $\sigma(\lambda) = \lambda$, which holds because $\lambda \in K$ and $\sigma$ fixes $K$. Hence $\sigma: L \to L'$ is a $K$-linear map between the $K$-vector spaces $L$ and $L'$. [guided] A $K$-homomorphism $\sigma: L \to L'$ is, by definition, a ring homomorphism that restricts to the identity on $K$. We need to check that this makes $\sigma$ a $K$-linear map when we view $L$ and $L'$ as $K$-vector spaces. Why do we care about $K$-linearity? The entire argument hinges on applying a dimension-counting result from linear algebra. That result applies to linear maps between vector spaces — not to arbitrary ring homomorphisms. So we must first confirm that $\sigma$ fits the hypotheses of the linear algebra theorem. Let $\lambda \in K$ and $x, y \in L$. We verify the two axioms of $K$-linearity. **Additivity.** Since $\sigma$ is a ring homomorphism, it preserves addition: \begin{align*} \sigma(x + y) = \sigma(x) + \sigma(y). \end{align*} **Scalar compatibility.** We compute: \begin{align*} \sigma(\lambda x) = \sigma(\lambda) \cdot \sigma(x) = \lambda \cdot \sigma(x). \end{align*} The first equality holds because $\sigma$ is a ring homomorphism (it preserves multiplication). The second equality holds because $\lambda \in K$ and $\sigma$ fixes every element of $K$, so $\sigma(\lambda) = \lambda$. These two properties together show that $\sigma: L \to L'$ is a $K$-linear map. Note that the scalar compatibility step is precisely where the condition "$\sigma$ fixes $K$" is consumed — a ring homomorphism that does not fix $K$ would not be $K$-linear. [/guided] [/step] [step:Establish injectivity of $\sigma$ from the ideal structure of fields] The kernel $\ker \sigma = \{x \in L : \sigma(x) = 0\}$ is an ideal of $L$. Since $\sigma$ is a ring homomorphism with $\sigma(1_L) = 1_{L'}$, we have $1_L \notin \ker \sigma$, so $\ker \sigma \neq L$. A field has exactly two ideals: $\{0\}$ and itself. Since $\ker \sigma$ is a proper ideal of the field $L$, we conclude $\ker \sigma = \{0\}$. Therefore $\sigma$ is injective. [guided] We now show that $\sigma$ is injective. The argument uses a fundamental fact about the ideal structure of fields: a field $F$ has no proper nonzero ideals. The only ideals of $F$ are $\{0\}$ and $F$ itself. Why is this true? Suppose $I \subset F$ is a nonzero ideal, and let $a \in I$ with $a \neq 0$. Since $F$ is a field, $a$ has a multiplicative inverse $a^{-1} \in F$. By the ideal property, $a^{-1} \cdot a = 1_F \in I$. Once an ideal contains $1_F$, it contains every element $f = f \cdot 1_F$, so $I = F$. Now we apply this. The kernel $\ker \sigma = \{x \in L : \sigma(x) = 0\}$ is an ideal of $L$ (this is a standard fact about ring homomorphisms: the preimage of $\{0\}$ is closed under addition and absorbs multiplication). Since $\sigma$ is a ring homomorphism between fields, $\sigma(1_L) = 1_{L'} \neq 0$, so $1_L \notin \ker \sigma$, and therefore $\ker \sigma \neq L$. Since $L$ is a field and $\ker \sigma$ is a proper ideal of $L$, the classification of ideals in a field forces $\ker \sigma = \{0\}$. Hence $\sigma$ is injective. This is a key structural property of field homomorphisms: every nonzero ring homomorphism from a field is automatically injective. There is no need to check injectivity separately — it comes for free from the field axioms. [/guided] [/step] [step:Apply the rank-nullity theorem to conclude surjectivity] We have established that $\sigma: L \to L'$ is an injective $K$-linear map. By hypothesis, both $L$ and $L'$ are finite-dimensional $K$-vector spaces with \begin{align*} \dim_K L = [L : K] = [L' : K] = \dim_K L'. \end{align*} Denote this common dimension by $n := [L : K]$. By the rank-nullity theorem applied to the $K$-linear map $\sigma: L \to L'$: \begin{align*} \dim_K L = \dim_K (\ker \sigma) + \dim_K (\operatorname{im} \sigma). \end{align*} Since $\sigma$ is injective, $\ker \sigma = \{0\}$, so $\dim_K(\ker \sigma) = 0$. Therefore: \begin{align*} \dim_K(\operatorname{im} \sigma) = \dim_K L = n. \end{align*} The image $\operatorname{im} \sigma$ is a $K$-linear subspace of $L'$ with $\dim_K(\operatorname{im} \sigma) = n = \dim_K L'$. A subspace of a finite-dimensional vector space whose dimension equals that of the ambient space is the entire space, so $\operatorname{im} \sigma = L'$. Hence $\sigma$ is surjective. Combined with injectivity, $\sigma: L \to L'$ is a bijective $K$-homomorphism — that is, a $K$-isomorphism. [guided] We now bring together the two ingredients: $\sigma$ is $K$-linear (from the first step) and injective (from the second step). The final step is purely linear algebra. Both $L$ and $L'$ are $K$-vector spaces. The hypothesis $[L : K] = [L' : K]$ tells us they have the same finite dimension over $K$; call this common dimension $n$. The rank-nullity theorem states that for any linear map $T: V \to W$ between finite-dimensional vector spaces: \begin{align*} \dim V = \dim(\ker T) + \dim(\operatorname{im} T). \end{align*} We apply this with $T = \sigma$, $V = L$, and $W = L'$. Since $\ker \sigma = \{0\}$, we have $\dim_K(\ker \sigma) = 0$, and the rank-nullity theorem gives: \begin{align*} n = \dim_K L = 0 + \dim_K(\operatorname{im} \sigma) = \dim_K(\operatorname{im} \sigma). \end{align*} So the image of $\sigma$ is an $n$-dimensional $K$-subspace of $L'$, which itself has dimension $n$. Since any subspace of a finite-dimensional vector space $W$ with $\dim(\text{subspace}) = \dim W$ must equal $W$ (a basis for the subspace is a set of $\dim W$ linearly independent vectors in $W$, hence a basis for $W$), we conclude $\operatorname{im} \sigma = L'$. Therefore $\sigma$ is surjective, and hence bijective. A bijective ring homomorphism is an isomorphism, so $\sigma: L \xrightarrow{\sim} L'$ is a $K$-isomorphism. It is worth noting where each hypothesis is consumed. The condition $[L : K] = [L' : K]$ is used in this final step to match dimensions. Without it — say if $[L : K] < [L' : K]$ — the image of $\sigma$ would be a proper subspace of $L'$, and $\sigma$ would be an injective but non-surjective embedding. The condition that $L/K$ and $L'/K$ are *finite* extensions is essential for the rank-nullity argument: in infinite-dimensional vector spaces, an injective linear map need not be surjective (consider, for instance, the right-shift operator on a sequence space). [/guided] [/step]