$k[t^{a_1},t^{a_2},t^{a_3}]$ in the form $k[x,y,z]/(...)$

Question

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.

Answer 1

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.

Answer 2

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 $J=I$.

Answer 3

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