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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?

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    $\begingroup$ This already has counterexamples in the simplest example of quadratic fields. Let $R = \mathbf Z$, $F = \mathbf Q$, and $K = \mathbf Q(\sqrt{d})$ where $d$ is a squarefree integer. If $d \not\equiv 1 \bmod 4$ then $A = \mathbf Z[\sqrt{d}]$ and the image of the trace map from $A$ to $R$ is $2\mathbf Z$. A concrete example is $A = \mathbf Z[i]$. In these examples $2$ is wildly ramified. $\endgroup$
    – KConrad
    Commented Apr 20, 2015 at 13:19

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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 $xR$.

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  • $\begingroup$ Thank you so much! It is a good counter-example. But what if the characteristic is 0? $\endgroup$
    – Hebe
    Commented Apr 20, 2015 at 9:45
  • $\begingroup$ @Hebe: "But what if the characteristic is $0$?" If the degree $d$ of $K/F$ is invertible in $R$, then the trace of $1/d$ equals $1$. Thus the trace map is surjective. $\endgroup$
    – Jason Starr
    Commented Apr 20, 2015 at 10:57
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    $\begingroup$ Dear all, I only just now realized that when the OP asks about "characteristic 0", probably that means that the fraction field should be characteristic 0 (i.e., mixed characteristic), rather than that the residue field should be characteristic 0 (equicharacteristic $0$). Others have given examples in mixed characteristic, so I will leave my answer as it is. $\endgroup$
    – Jason Starr
    Commented Apr 20, 2015 at 14:32
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In addition to Jason's counter-example, we have the case of wild extensions of local fields (cf. Algebraic Number Theory, Cassels and Fröhlich ed., Chap. I "Local fields", by A. Fröhlich). Let $L/K$ be a finite extension of local fields and let $k$ denote the residue field of $K$, $p$ the characteristic of $k$, $e$ the ramification index of $K/F$. Let ${\frak o}_K$ (resp. ${\frak o}_L$) the ring of integers of $K$ (resp. of $L$). One says that $L/K$ is tamely ramified if $p\not\vert e$ and if the residue field of $L$ is a separable extension of $k$ (loc. cit page 21). Then by Theorem 2 of loc. cit, page 21, one has ${\rm Tr}({\frak o}_L)={\frak o}_K$ iff $L/K$ is tamely ramified. A counter-example in characteristic $0$ is thus given by the "wild" quadratic extension ${\mathbb Q}_2 (\sqrt{2})/{\mathbb Q}_2$, where ${\mathbb Q}_2$ is the field of dyadic numbers.

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