= Solution
The usual <field norm> $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.
First a <unique factorization domain> is an <integrally closed domain>. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.
Let $m_b(T)=T^r+a_{r-1}T^{r-1}+\cdots+a_0\in K[T]$ be the monic <minimal polynomial> of $b$. Because $b$ is an <integral element> over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The <integral elements> form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.
Multiplication by $b$ on $K(b)$, in the basis $1,b,\ldots,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1)^ra_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting <direct sum> of $e$ copies of $K(b)$, so its determinant over $K$ is
$$
\boxed{N_{L/K}(b)=\bigl((-1)^ra_0\bigr)^e\in A.}
$$
In particular $N_{K(b)/K}(b)=(-1)^ra_0\in A$. The argument requires no separability; in fact it holds for every <integrally closed domain>, not just a <unique factorization domain>.
For the connection with dimension, let $B$ be the coordinate ring of an irreducible <affine variety>. By <Noether normalization> there is a finite integral inclusion $A=k[t_1,\ldots,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation gives
$$
a_0=-b\bigl(b^{r-1}+a_{r-1}b^{r-2}+\cdots+a_1\bigr)\in bB\cap A.
$$
Thus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the <field norm> is a power of this $a_0$, so it too belongs to $bB\cap A$.
Under the <finite morphism> to $\mathbb A^d$, the image of $V_B(b)$ is therefore contained in the proper <Zariski-closed set> $V_A(a_0)$ of dimension at most $d-1$. <Finite morphisms> preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves \b[proper closed subvarieties have strictly smaller dimension]. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.
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