I am reading Milne's Etale Cohomology and ran across this problem which has so-far eluded me.

According to Milne, in any category with fiber products, we say that a morphism $f:Y \to X$ is a strict epimorphism if the sequence of sets $$\operatorname{Hom}(X,Z) \overset{f^*}{\to} \operatorname{Hom}(Y,Z) \underset{p_2^*}{\overset{p_1^*}{\rightrightarrows}} \operatorname{Hom}(Y \times_X Y,Z) $$ is exact for all objects $Z$, that is, an equaliser in the category.

Now fix our category to be schemes over a field $k$, and consider the morphism $\operatorname{Spec}(k[t]) \to \operatorname{Spec}(k[t^3,t^5])$. The question is to show that this is an epimorphism that is not strict.

The map on underlying rings is injective, and easily seen to be surjective on spectra, so this is certainly an epimorphism in Sch/$k$, yet I am seemingly unable to produce a scheme $Z$ which makes the above diagram fail.

One thought was to regard $\operatorname{Spec}(k[t^3,t^5])$ as a cuspidal curve $y^3-x^5$ in affine 2-space, and then blow up at the origin. The exceptional divisor of the strict transform would then be a triple point (nilpotent), and so hopefully this would mess things up appropriately. Yet I cannot seem to find a section. The blow up comes with a natural map down to the original curve, but I am struggling to find one upwards.


Take $Z=\mathrm{Spec}\, k[x]$, and consider the map $\varphi:Y\to Z$ induced by the ring map $k[x]\to k[t]$, $x\mapsto t^7$. Then $\varphi$ is not in the image of $\mathrm{Hom}(X,Z)=\mathrm{Hom}(k[x],k[t^3,t^5])$ because $t^7\not\in k[t^3,t^5]$. However, I claim that $\varphi$ is equalized by the two maps to $\mathrm{Hom}(Y\times_X Y,Z)$. This is equivalent to the condition that $t^7\otimes 1=1\otimes t^7$ in $k[t]\otimes_{k[t^3,t^5]}k[t]$.

For notational convenience, write $u=t\otimes 1$ and $v=1\otimes t$. We then need to show that $u^7=v^7$. By construction of the tensor product, we have $u^3=v^3$ and $u^5=v^5$. We conclude that $$ u^7-v^7=(u^2+v^2)\big(u^5-v^5\big)-u^2v^2\big(u^3-v^3\big)=0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.