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I'm trying to solve an exercise in Commutative Algebra in which I need the kernel of the following homomorphism:

Consider the homomorphism $ \phi :k[[x,y,z]] \to k[[t]]$ which sends $x \to t^3,y \to t^4, z \to t^5$. Find $\ker(\phi)$.

It is clear that $ (x^3-yz,y^2-xz,z^2-x^2y) \subseteq \ker(\phi)$. Any hints/ideas to prove the converse?

user26857
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Arpit Kansal
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  • Send $t\mapsto t^{1/3}, t^{1/4}, t^{1/5}$, using the power series of such expressions, then show that the composition with $\phi$, modulo the ideal you describe, is the identity. – Pax Oct 19 '15 at 16:09
  • How about noting that $\dim k[[x,y,z]]=3$. Then, show that the ideal you've written down is prime, and has height at least 2. But, the kernel is also prime of height at most $2$ (why?) which implies that they are equal. – Alex Youcis Oct 20 '15 at 14:35
  • Thanx @AlexYoucis this works,but we need some non trivial results for this argument! – Arpit Kansal Oct 20 '15 at 16:00

2 Answers2

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$\require{AMScd}$

With $A=k[[x,y,z]]$, $B=k[[t]]$ and $I = (x^3-yz, y^2-xz, z^2 -x^2 y)$ and $\bar{A} = A/I$ consider the diagram

$$ \begin{CD} \bar{A}_y @>{f}>> B_{t^4}\\ @AAiA @AAA \\ \bar{A} @>{}>> B \end{CD} $$

The map $f$ has a left-inverse $g$ given by $g(t) = z/y$. One can calculate

$x \to t^3 \to z^3/y^3 = x$ because $z^3 - x y^3 \in I$

$y \to t^4 \to z^4/y^4 = y$ because $z^4 - y^5 \in I$

$z \to t^5 \to z^5/y^5 = z$ because $z^5 - y^5 z \in I$

(I checked the $\in I$ with the ideal $J \subseteq R$ having the same generators as $I$, but in $R=k[x,y,z]$)

Now it remains to prove that $i$ is injective which follows from $(I:y^s) = I$. So we need only prove $(I:y) = I$. Now give $x,y,z$ the weights $3,4,5$, so $I$ is homogenous. Then $y h \in I$ implies $y h_d \in I$ for all weighted homogeneous parts $h_d$ of $h$. But this, I think, is equivalent to $y h_d \in J$ with the ideal $J$ introduced above. As $J$ is prime in $R$, we have $h_d \in J$, therefore $h_d \in I$. Going from the $h_d$ with lowest $d$ upward proves $h \in I$.

Jürgen Böhm
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This question is more or less a duplicate of Show the set of points $(t^3, t^4, t^5)$ is closed in $\mathbb A^{3}$.

If we have a surjective ring homomorphism $A\stackrel{f}\to B$, where $A,B$ are noetherian rings, and $I\subset A$ an ideal, then this gives rise to an exact sequence of $A$-modules $$0\to\ker f\to A\to B\to0$$ and by tensorizing with $\hat A$ (the completion of $A$ in the $I$-adic topology) we get $$0\to\ker f\otimes_A\hat A\to A\otimes_A\hat A\to B\otimes_A\hat A\to0.$$ This leads to the exact sequence $$0\to (\ker f)\hat A\to\hat A\stackrel{\hat f}\to\hat B\to0,$$ so $\ker\hat f=(\ker f)\hat A$. (The completion of $B$ is taken in the $f(I)$-adic topology.)

Now use the result for polynomial rings which I've mentioned above and extend it to the corresponding power series rings (which are the completions of the polynomials rings involved in our question).

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user26857
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