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Question 1: In A Singular Introduction to Commutative Algebra, page 222, there is written:

enter image description here

How can I check that this isomorphism actually holds? I would really prefer a computational proof (using a computer), because I just can't do this manually for every single case.

I've tried the surjective map $f\!:K[x,y]\!\rightarrow\!K[-t^3+6t,t^4-6t^2]$, defined by $x\mapsto -t^3+6t$ and $y\mapsto t^4-6t^2$, but the SINGULAR code

ring A=0,(x,y),dp;
ring B=0,(t),dp;
map F=A,-t3+6t,t4-6t2;
ideal I=0;
setring A;

tells me that its kernel is $\langle x^4-84x^2y-y^3-32400x^2-360y^2-32400y\rangle$ and not $\langle x^4+6x^2y-y^3\rangle$, so this isn't the right map. Am I supposed to guess what the right map is?

Question 2: If I have $R:=K[x,y,z]/\langle z^2-xy\rangle$, how can I find a suitable subring of $K[t_1,\ldots,t_n]$ (for some $n$), that is isomorphic to $R$?

EDIT: I would just like to add that after restarting the computer and running the code again, SINGULAR, for some mysterious reason, now returned the correct answer $\langle x^4+6x^2y-y^3\rangle$.

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Your two questions are essentially unrelated. Next time, please do ask unrelated questions in separate questions. – Mariano Suárez-Alvarez Jan 2 '12 at 2:14
up vote 2 down vote accepted

Q1: If I do the computation in Macaulay2, things work as expected...

Macaulay2, version 1.4
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra,

i1 : R = QQ[x,y];

i2 : S = QQ[t];

i3 : phi = map(S, R, matrix{{-t^3 + 6*t, t^4 - 6*t^2}});

o3 : RingMap S <--- R

i4 : ker phi

            4     2     3
o4 = ideal(x  + 6x y - y )

o4 : Ideal of R

Q2: The elements $a=x^2$, $b=y^2$ and $c=xy$ of $K[x,y]$ generate a subring of $K[x,y]$ which can be presented as $K[a,b,c]/(ab-c^2)$.

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This is very strange. I also tried the SINGULAR code ring A=0,(x,y),dp; LIB "algebra.lib"; ring B=0,(t),dp; ideal I=-t^3+6t,t^4-6t^2; map F=A,I; alg_kernel(F,A); but the result is still $\langle x^4-84x^2y-y^3-32400x^2-360y^2-32400y\rangle$. My and your kernel are not equal. What am I doing wrong? – Leon Jan 4 '12 at 6:17
@LeonLampret, I am much less familiar with SINGULAR than M2, but probably you should really be asking on a SINGULAR-related mailing list rather than here. – Mariano Suárez-Alvarez Jan 4 '12 at 6:19
Regarding Q2, I really had in mind a more general procedure. I know that I didn't formulate the question that way, sorry, I thought I would be able to recognize the procedure from your answer. So, is there some general procedure (preferably a command in some software) that, given an ideal $I\!\unlhd\!K[x_1,\ldots,x_n]$, finds some $m$ and some homomorphism $f\!:K[x_1,\ldots,x_n]\!\rightarrow\!K[y_1,\ldots,y_m]$ with $\mathrm{Ker}(f)=I$? Should I ask this in another post? – Leon Jan 5 '12 at 7:04
Every subring of a polynomial ring is a domain, so the kernel of every morphism into such a ring is a prime ideal: $(x_1x_2)$ is not a prime ideal. For the second example, suppose $f:k[x_1,x_2]/(x_1x_2-1)\to k[y_1,\dots,y_n]$ is an injective morphism of $k$-algebras; then the image of the class of $x_1$ is an invertible element; since the only invertible elements in $k[y_1,\dots,y_n]$ are the scalars, there exists a $\lambda\in k$ such that $x_1-\lambda$ is in $(x_1x_2-1)$; but this is absurd. – Mariano Suárez-Alvarez Jan 5 '12 at 7:21
No. In general, there are many ring morphisms $K[x_1,\ldots,x_n]\!\rightarrow\!K[y_1,\ldots,y_m]$ which are not $K$-algebra morphisms. Already (taking $n=m=0$) there are usually many ring morphisms $K\to K$ which are not $K$-algebra morphisms, that is, which are not $K$-linear: to find examples, it is sufficient to find a field $K$ which has non-trivial automorphisms. – Mariano Suárez-Alvarez Jan 10 '12 at 7:36

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