[proofplan]
We apply [Noether's Normalization Theorem](/theorems/2939) to $K$ viewed as a finitely generated $k$-algebra, obtaining a polynomial subalgebra $k[x_1, \ldots, x_d]$ over which $K$ is finite. Since $K$ is a field and $k[x_1, \ldots, x_d]$ sits inside $K$ as a subring over which $K$ is integral, the [Integrality and Fields](/theorems/2868) criterion forces $k[x_1, \ldots, x_d]$ to be a field. But a polynomial ring in $d \geq 1$ variables is never a field, so $d = 0$ and $K$ is finite over $k$.
[/proofplan]
[step:Apply Noether normalisation to obtain a polynomial subalgebra]
The field $K$ is a nonzero finitely generated $k$-algebra. By [Noether's Normalization Theorem](/theorems/2939), there exist elements $x_1, \ldots, x_d \in K$ that are algebraically independent over $k$ such that $K$ is finite (hence integral) over the polynomial subalgebra $A' = k[x_1, \ldots, x_d]$.
[/step]
[step:Show $k[x_1, \ldots, x_d]$ must be a field using integrality]
Since $K$ is a field and $A' = k[x_1, \ldots, x_d]$ is an integral domain with $A' \subset K$ an integral extension, we apply [Integrality and Fields](/theorems/2868): if $A \subset B$ is an integral extension of integral domains and $B$ is a field, then $A$ is a field. The hypotheses are satisfied: $A' \subset K$ is integral (since finite implies integral), $A'$ is an integral domain (as a polynomial ring over a field), and $K$ is a field. Therefore $A' = k[x_1, \ldots, x_d]$ is a field.
[guided]
Why does integrality force the subring to be a field? The statement we invoke is: if $A \subset B$ is an integral extension of integral domains and $B$ is a field, then $A$ is a field. The proof goes as follows: take any nonzero $a \in A$. Since $a \in B$ and $B$ is a field, $a^{-1}$ exists in $B$. Since $B$ is integral over $A$, $a^{-1}$ satisfies a monic polynomial over $A$:
\begin{align*}
(a^{-1})^n + c_1(a^{-1})^{n-1} + \cdots + c_n = 0, \quad c_i \in A.
\end{align*}
Multiplying through by $a^{n-1}$:
\begin{align*}
a^{-1} + c_1 + c_2 a + \cdots + c_n a^{n-1} = 0,
\end{align*}
so $a^{-1} = -(c_1 + c_2 a + \cdots + c_n a^{n-1}) \in A$. Hence every nonzero element of $A$ is invertible in $A$, so $A$ is a field.
[/guided]
[/step]
[step:Conclude $d = 0$ since a polynomial ring in positive variables is not a field]
Suppose for contradiction that $d \geq 1$. Then $A' = k[x_1, \ldots, x_d]$ contains the element $x_1$, which is nonzero (since $x_1, \ldots, x_d$ are algebraically independent over $k$, in particular $x_1$ is transcendental over $k$). If $A'$ were a field, then $x_1$ would have a multiplicative inverse $g \in k[x_1, \ldots, x_d]$, meaning $x_1 \cdot g = 1$. But in a polynomial ring over a field, the only units are the nonzero constants: if $x_1 \cdot g = 1$, then $\deg(x_1 \cdot g) = 1 + \deg(g) = \deg(1) = 0$, which is impossible since $\deg(g) \geq 0$.
Therefore $d = 0$, so $A' = k$ and $K$ is finite over $k$, meaning $\dim_k K < \infty$.
[guided]
The argument that $k[x_1, \ldots, x_d]$ is not a field when $d \geq 1$ uses a basic property of polynomial rings over fields: the degree function is multiplicative, $\deg(fg) = \deg(f) + \deg(g)$ (this holds because $k$ is a field, hence an integral domain, so leading coefficients do not cancel). If $x_1 g = 1$ for some $g \in k[x_1, \ldots, x_d]$, taking degrees gives $1 + \deg(g) = 0$, which is impossible for $\deg(g) \geq 0$. So $x_1$ is not a unit, and $k[x_1, \ldots, x_d]$ is not a field.
With $d = 0$ established, the normalisation says $K$ is finite over $k[\ ] = k$. "Finite" means $K$ is finitely generated as a $k$-module, i.e., $\dim_k K < \infty$.
[/guided]
[/step]
