One of the ways I find more useful to check if a given ideal $I$ of $K[X_1,\ldots,X_n]$ is prime, is to look at the quotient ring $K[X_1,\ldots,X_n]/I$.

If I'm able to show it is isomorphic to $K[f_1,\ldots,f_n]$ where $f_i$ are polynomials in several variables then I'm done.

For example, to check that $I=(x_1x_3-x_2^2,x_0x_2-x_1^2,x_0x_3-x_1x_2)$ is prime it is enough to show that the homomorphism $\varphi:D[x_0,x_1,x_2,x_3]\rightarrow K[u,v]$ that takes $x_0$ to $v^3$, $x_1$ to $uv^2$, $x_2$ to $u^2v$ and $x_3$ to $u^3$ has as kernel exactly $I$. And this can be easily done with Groebner Basis. So we have shown that the quotient ring is isomorphic to $K[v^3,v^2u,vu^2,u^3]$ which is clearly an integral domain.

So my question is if this can always be done, ie given a prime ideal $I$ of $K[x_1,\ldots,x_n]$ is it always possible to find an isomorfism between the quotient ring and $K[f_1,\ldots,f_n]$ where $f_i$ are polynomials in several variables.

If that is not possible, what do we have to ask to $I$ or to $V(I)$ so this is possible?

In case it can be done, is there any way to find these polynomials?


1 Answer 1


It can not always be done. In fact I guess this property of $I$ or $V(I)$ is quite restrictive.

If $K[x_1,\ldots ,x_n]/I\cong K[f_1,...,f_m]$ where the $f_k$ are polynomials in $r$ variables, then the fraction field $F$ of $K[x_1,\ldots ,x_n]/I$ is isomorphic over $K$ to a subfield of the field $K(y_1,\ldots ,y_r)$ of rational functions in $r$ variables. Thus for the transcendence degree $s$ of $F/K$ we have $s\leq r$. We can now write $K(y_1,\ldots ,y_r)$ as $L(y_{s+1},\ldots ,y_r)$, where $L=K(y_1,\ldots ,y_s)$ and apply a theorem of Roquette and Ohm (Archiv der Mathematik, Vol. 42, No. 2, On subfields of rational function fields, J. Ohm) and conclude that $F$ is isomorphic over $K$ to a subfield of $L$ already.

Now suppose that $V(I)$ is an algebraic curve. Then this argument shows that the function field $F$ of the curve is contained in $K(y)$, hence by Lüroth's theorem the curve has genus $0$.


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