PomeLemma 5.43.8.11E6C3E4D36C Let $f:(\mathbb C^ m,0)\to (\mathbb C,0)$ be a nonconstant holomorphic germ, with $m\geq 1$, and let $e=\operatorname {ord}_0f$. For every quasi-subharmonic germ $v$ at zero,
\[ \nu (v\circ f,0)=e\nu (v,0). \]
Lemma 5.43.811E6C3E4D36C
Proof. A smooth correction of $v$ pulls back to a smooth function, so we may assume that $v$ is subharmonic. The Taylor expansion of $f$ gives $|f(z)|\leq M|z|^ e$ on a small ball. Choose a unit vector $\xi $ on which the leading homogeneous term is nonzero. Then
\[ f(\zeta \xi )=a\zeta ^ e+O(\zeta ^{e+1}),\qquad a\ne 0. \]
For all small $r$, the remainder has modulus at most $|a|r^ e/4$ on $|\zeta |=r$. If $|t|<|a|r^ e/4$, Rouché's theorem shows that $f(\zeta \xi )-t$ has $e$ zeros in $|\zeta |<r$. Consequently, with $c=|a|/4$,
\[ \{ |t|<cr^ e\} \subset f(B(0,r)) \subset \{ |t|<Mr^ e\} . \]
The suprema of $v$ on these three sets are ordered. Dividing by $\log r^2<0$, the two disk terms both tend to $e\nu (v,0)$ by the definition in section 5.34. The middle term therefore has the same limit, proving the formula. $\square$
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