The gradient test for a reduced hypersurface

The polynomial ring $R$ is a UFD by [F7], so write $f=u{q}_1\cdots q_m$ with $u\in k^{\times }$ and pairwise nonassociate irreducibles $q_i$; each $q_i$ is prime. If $g^r\in (f)$ for some $r\ge 1$, every $q_i$ divides $g^r$ and hence divides $g$. Since the $q_i$ are distinct prime factors, their product divides $g$, so $g\in (f)$ and $(f)$ is radical. By [F8], $I(X)=I(V(f))=\sqrt{(f)}=(f)$, so $X$ is reduced and its affine scheme is $\mathrm{Spec}(R\mathrm{/}(f))$ with the actual equation ideal. This finite factorization argument makes no choice; AC is used here only for the Nullstellensatz identification.

The affine space ${\mathbb{A}}_k^n$ is irreducible because $R$ is a domain by [F7], and [F5] and [F6] give $\dim {\mathbb{A}}_k^n=n$. The polynomial $f$ is a nonzero nonunit of its coordinate ring, so [F4] gives that $X$ is nonempty and each irreducible component $X_i$ has dimension $n-1$. For the fixed closed point $a$, [F3] therefore gives $\dim {\mathcal{O}}_{X,a}={\max }_{a\in X_i}\dim X_i=n-1$. Component dimensions come from [F4], and AC identifies their maximum with local dimension through [F3].

By [F1] applied to the actual ideal $(f)$ and its one-element generating list, the intrinsic tangent space at $a$ is the kernel of the single row $df(a)=({\mathrm{\partial }}_{1}f(a),\dots ,{\mathrm{\partial }}_{n}f(a)):k^n⟶k$. If some coefficient $c_j={\mathrm{\partial }}_{j}f(a)$ is nonzero, the equation ${\sum }_i{\mathrm{\partial }}_{i}f(a)v_i=0$ determines $v_j=-c_j^{-1}{\sum }_{i\mathrm{\ne }j}{\mathrm{\partial }}_{i}f(a)v_i$, so the other $n-1$ coordinates vary freely and ${\dim }_k{T}_{a}X=n-1$. If every partial vanishes, the kernel is all of $k^n$ and has dimension $n$. These alternatives include every characteristic because [F1] uses formal polynomial derivatives without a characteristic restriction.

The point $a$ is closed in the reduced classical finite-type space $X$. By [F2], it is regular exactly when ${\dim }_k{T}_{a}X={\max }_{a\in X_i}\dim X_i$. Step 1.2 identifies this maximum as $n-1$, and step 1.3 shows that equality holds exactly when some partial derivative of $f$ is nonzero. Since the singular locus is the complement of the regular locus by [F2], $a$ is singular exactly when all partial derivatives vanish. Since $f(a)=0$, this proves both inclusions in the displayed equality. The choice use is inherited through [F2]–[F4], as recorded in [F9].

The non-squarefree distinction is visible in one variable: in $k[t]\mathrm{/}(t^2)$ at $0$ the actual Jacobian row is $(2t){\mathrm{\mid }}_0=0$ in every characteristic, so [F1] gives tangent space $k$, whereas for the radical ideal $(t)$ the row is $(1)$ and the tangent space is zero. Thus radicalizing a non-squarefree equation changes its tangent computation; for the squarefree $f$ here, step 1.1 proves radicalization is redundant. For $n=1$ and $f=t$, the unique point has nonzero derivative, tangent dimension zero and local dimension zero, so it is regular. For the reducible squarefree example $f=xy$ in $k[x,y]$, at the origin the gradient $(y,x)$ vanishes, the tangent dimension is two and the local dimension is one; away from the origin on either axis the gradient is nonzero and tangent dimension is one, so those points are regular. The zero tangent vector lies in every Jacobian kernel. Here $n\ge 1$ and nonconstant nonzero $f$ make the principal-subvariety theorem applicable; [F4] makes the hypersurface nonempty, so the empty case has no instance. Steps 1.3–2.1 establish both implications, and no arbitrary choice is made beyond the declared AC uses.