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I want to write $k[t^6,t^7,t^{15}]$ in the form $k[x,y,z]/(...)$; but I even don't know how to start.

Is there in general a way that one can write $k[t^{a_1},t^{a_2},t^{a_3}]$ in the form $k[x,y,z]/(\cdots)\cdots(\cdots)$?

thanks.

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  • $\begingroup$ One place to start would be the ideals $(x^7-y^6), (x^5 - z^2), (y^{15} - z^7)$. $\endgroup$
    – Arthur
    Commented Dec 6, 2014 at 15:20
  • $\begingroup$ If you are interested in reading a paper, then Herzog wrote one which covers your case I believe, Generators and Relations of Abelian Semigroups and Semigroup rings. Of course, you will find more refined versions and modern treatment as @JyrkiLahtonen mentioned above. $\endgroup$
    – Youngsu
    Commented Dec 7, 2014 at 7:43

3 Answers 3

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Sketch mostly: I think that the essential parts are in here, but the argument probably isn't entirely convincing yet.

Mapping $\phi: x\to t^6$, $y\to z^7$ and $z\to t^{15}$ when the question becomes one of finding generators for the kernel of $\phi:k[x,y,z]\to R\subset k[t]$, where $R$ is your ring. Denote that kernel by $I$.

There are $(6-1)(7-1)/2=15$ powers $t^i$ missing from $k[t^6,t^7]$ namely those with $i$ outside the numerical semigroup $S(6,7)$ generated by $6$ and $7$ $$ \Bbb{N}\setminus S(6,7)=\{1,2,3,4,5,8,9,10,11,15,16,17,22,23,29\}. $$ Of those numbers $15,22,29$ are included in $S(6,7,15)$. Therefore we can deduce that $$ \dim_k\phi(k[x,y,z])/\phi(k[x,y])=3. $$

Let us first find $I':=I\cap k[x,y]$. By looking at $S(6,7)$ we see that it is disjoint union of the sets $6\Bbb{N}+7\ell$ with $\ell=0,1,2,3,4,5$. Therefore there cannot be relations with $y$ appearing with an exponent $<6$. It follows that $I'$ is generated by $x^7-y^6$.

The extra powers of $t$ missing from $\phi(k[x,y])$ are $\phi(z)$, $\phi(zy)$ and $\phi(zy^2)$. Thus we need to be able get rid of anything involving $z^2$, $zx$ or $zy^3$. To that end we can use the polynomials $$ x^5-z^2, zx-y^3, zy^3-x^6\in I. $$ If we denote be $J$ the ideal $$ J=\langle z^2-x^5, zy^3-x^6, zx-y^3, y^6-x^7\rangle $$ we see that $J\cap k[x,y]=I'$ and that the $k$-space $(k[x,y,z]/J)/(k[x,y]/I')$ is spanned by $z,zy,zy^2$. The dimensions match, so $J=I$.

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    $\begingroup$ Cannot put more time to this now. Leaving this here in the hope that this may lead somebody to see the light. My approach does not generalize very well. It depended on the very special fact that the difference between the semigroups $S(6,7)$ and $S(6,7,15)$ is a finite set owing to the fact $\gcd(6,7)=1$. For a more general $a_1,a_2,a_3$ with $\gcd(a_1,a_2,a_3)=1$ but all pairwise gcds $>1$ the difference will be infinite and another argument will be needed. $\endgroup$ Commented Dec 6, 2014 at 17:29
  • $\begingroup$ Comparing this with the Macaulay output provided by Mariano I noticed that $zy^3-x^6$ is unnecessary: $$zy^3-x^6=x(z^2-x^5)-z(zx-y^3).$$ My thinking is a bit Gröbnerish, but I'm not at all conversant with that either. $\endgroup$ Commented Dec 7, 2014 at 7:48
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    $\begingroup$ Now that you opened the discussion let me add: $y^6-x^7=x^2(z^2-x^5)-(xz-y^3)(xz+y^3)$. $\endgroup$
    – user26857
    Commented Dec 7, 2014 at 9:36
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When thinking about an algebra of the form $k[x,y,z]/(...)$, you should be thinking about generators and relations. What are the generators of $k[t^6, t^7, t^{15}]$? Hint: They're staring you in the face. What are the relations? This is usually harder. There are usually the "obvious" ones - such as $6+6+6+6+6 = 15+15$ here, but then one needs to prove that the obvious ones are the only ones or find ones which aren't obvious. This can be much more difficult.

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One way to do the computation using Macaulay2:

i1 : R1 = QQ[t];

i2 : R2 = QQ[x,y,z];

i3 : f = map(R1, R2, {t^6, t^7, t^(15)});

o3 : RingMap R1 <--- R2

i4 : I = ker f

             3         5    2
o4 = ideal (y  - x*z, x  - z )

o4 : Ideal of R2
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