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

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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; preimage(B,F,I); 

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.2012-01-02

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

 Macaulay2, version 1.4 with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra,                TangentCone  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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    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.2012-01-10