Are intersection multiplicities independent of the base field? Let $F, G$ be affine curves over some field $K$, i.e. elements of $K[x, y]$. The intersection multiplicity $\mu_P(F, G)$ of $F$ and $G$ in $P=(x_0, y_0)$ is defined as
$$
\mu_P(F, G) := \dim_K \mathcal{O}_P/(F, G)\;,
$$
where $\mathcal{O}_P$ is the local ring of $\mathbb{A}^2$ at $P$, i.e. the localisation of $K[x, y]$ in the maximal ideal $(x-x_0, y-y_0)$.
I am wondering if the quantity $\mu_P(F, G)$ is independent of the base field, i.e. if
$$
\dim_K \mathcal{O}_{P, K}/(F, G)=\dim_L \mathcal{O}_{P, L}/(F, G)\;,
$$
where we denoted $\mathcal{O}_{P, K}:=K[x, y]_{(x-x_0, y-y_0)}$ and analogously for $L$ for simplicity. Specifically, I am interested in the case $K=\mathbb{R}$ and $L=\mathbb{C}$.
My idea is to show that there is an isomorphism
$$
\mathbb{C}\otimes_{\mathbb{R}}\mathcal{O}_{P, \mathbb{R}}/(F, G)\cong \mathcal{O}_{P, \mathbb{C}}/(F, G)\;,
$$
of $\mathbb{C}$-vector spaces, from which the result would follow as scalar extension preserves dimensions. My guess is that such an isomorphism is given by
$$
\begin{split}
\varphi: \mathbb{C}\otimes_{\mathbb{R}}\mathcal{O}_{P, \mathbb{R}}/(F, G)&\rightarrow \mathcal{O}_{P, \mathbb{C}}/(F, G) \\
a\otimes f&\mapsto af
\end{split}
$$
with inverse
$$
\begin{split}
\psi: \mathcal{O}_{P, \mathbb{C}}/(F, G)&\rightarrow \mathbb{C}\otimes_{\mathbb{R}}\mathcal{O}_{P, \mathbb{R}}/(F, G) \\
\sum_{i, j} a_{ij}x^iy^j&\mapsto \sum_{i, j} a_{ij}\otimes x^iy^j\;.
\end{split}
$$
(Note that we can WLOG assume $F$ and $G$ to be irreducible since
$$
\mu_P(R, ST)=\mu_P(R, S)+\mu_P(R, T)
$$
for any $R, S, T\in K[x, y]$ and then we can furthermore assume that $F$ and $G$ are coprime. In such a case, I can show that any element of $\mathcal{O}_P/(F, G)$ has a polynomial representative, so it is indeed enough to define the above map on polynomials only.)
Indeed, I can show that $\varphi$ and $\psi$ are mutually inverse $\mathbb{C}$-linear maps. I can also show that $\varphi$ is well-defined, but I am struggling with the well-definedness of $\psi$: While I do not have any problems showing that $\psi(AF+BG)=0$ for $A, B\in \mathbb{C}[x, y]$, I am a bit clueless as to how to show
$$
\psi\left(\frac{AF+BG}{C}\right)=0
$$
for $C\not\in (x-x_0, y-y_0)$ with the property that $C\mid AF+BG$ in $\mathbb{C}[x, y]$.
It seems that this should be the right way to do this, but I cannot figure out this last step, does anyone have an idea how to finish the argument? Or, alternatively, how to prove the statement in a different way?
 A: Assume that $L/K$ is finite Galois with Galois group $G$.
Let us first show that $\mathcal O_{P,K} \otimes_K L \cong \mathcal O_{P,L}$, i.e. let's solve the case $F=G=0$. Let $S=k[X,Y] \setminus (x-x_0,y-y_0)$. Then we have
$$\mathcal O_{P,K} \otimes_K L \cong \mathcal O_{P,K} \otimes_{K[x,y]} K[x,y] \otimes_K L \cong \mathcal O_{P,K} \otimes_{K[x,y]} L[x,y] \cong S^{-1} L[x,y]$$
This is almost $\mathcal O_{P,L}$, but not quite: $\mathcal O_{P,L}$ is the localization at $T:=L[x,y] \setminus (x-x_0,y-y_0)$ not at $S$. We do have $S \subset T$, so that we obtain a map $S^{-1}L[x,y] \to T^{-1} L[x,y]$. We want to show that this map is an isomorphism, for this it suffices to show that any element of $T$ is a unit in $S^{-1}L[x,y]$. $G$ acts coefficient-wise on $L$. Let $f \in T$, then we get that $\prod_{\sigma \in G}\sigma(f) \in S$ (use Galois theory to show that it is in $K[x,y]$ and evalute at $(x_0,y_0)$ to get that it is in $S$). Thus $f$ is a unit in $S^{-1}L[x,y]$, as that ring contains $$\frac{\prod_{\sigma \in G \setminus \{\mathrm{id}\}}\sigma(f)}{\prod_{\sigma \in G}\sigma(f)}=\frac{1}{f}$$
This finishes the proof that $\mathcal O_{P,K} \otimes_K L\cong \mathcal O_{P,L}$
We can deduce the general result from this. Consider the exact sequence of $K$ vector spaces
$$(\mathcal O_{P,K})^2\to \mathcal O_{P,K} \to \mathcal O_{P,K}/(F,G) \to 0$$
where the map $(\mathcal O_{P,K})^2\to \mathcal O_{P,K}$ is given by $(a,b) \mapsto aF+bG$
Now we tensor with $L$ over $K$ and obtain the exact sequence
$$\mathcal O_{P,L}^2 \to \mathcal O_{P,L} \to \mathcal O_{P,K}/(F,G) \otimes_K L \to 0$$
where again  the map $(\mathcal O_{P,L})^2\to \mathcal O_{P,L}$ is given by $(a,b) \mapsto aF+bG$. From the exactness of this sequence, we obtain
$$\mathcal O_{P,L}/(F,G) \cong \mathcal O_{P,K}/(F,G) \otimes_K L$$ as desired.
