Let $R$ be a closed integral domain with its fraction field $F$. Let $K$ be a finite separable extension field of $F$, and let $A$ be the integral closure of $R$ in $K$.


It is well known that the trace map $Tr: K \to F$ is non-trivial and hence is surjective because of separable extension. If we restrict $Tr$ to $A$, then it is obvious that $Tr$ maps $A$ into $R$ because $R$ is integral closed.



> 
> Question: Is the restriction $Tr: A \to R$ also surjective?
> 
> 
> 



Surjectivity of trace map
No, the trace map need not be surjective on the level of rings. This is one of the difficulties of "wild ramification". For instance, let $k$ be a field of characteristic $p$, let $R$ be $k[x]$, and let $A$ be the $R$-algebra, 
$$ A = R[y]/\langle y^p +xy-1 \rangle = k[x,y]/\langle y^p + xy-1 \rangle.$$  

Using the Jacobian criterion, $A$ is a regular ring, hence integrally closed in its fraction field. The trace map, $$\text{Trace}\_{A/R}:A \to R,$$ is an $R$-module homomorphism. If it were surjective, it would be surjective after tensoring with $R/xR$. However, formation of the trace map is compatible with base change: $\text{Trace}\_{A/R}\otimes \text{Id}\_{R/xR}$ equals $\text{Trace}\_{(A/xA)/(R/xR)} $. Of course, $R/xR$ equals $k$, and 
$$ A/xA = (R/xR)[y]/\langle (y-1)^p \rangle = k[y]/\langle (y-1)^p \rangle.$$ It is straightforward to compute that the trace map for this ring extension is identically zero. Thus the image of $\text{Trace}\_{A/R}$ is contained in the proper ideal $xR$. In fact, the image is equal to $xR$.


