# the variety of an irreducible polynomial sum of two homogeneous polynomials of degree m, m+1 is rational

This problem has two parts, the first I did it. But I have problems with the second.

Given an irreducible polynomial F over $k[T]$. Prove that the ring $k\left[ {T_1 ,...,T_n } \right]/\left( {I\left( {V\left( F \right)} \right)} \right)$, is an integral domain. Where given a variety W, we denote $I(W)$ the ideal of all the polynomials over $k[T_1,...,T_n]$ that vanish in W, and $V\left( F \right) = \left\{ {\left( {x_1 ,...,x_n } \right) = x \in K^n ;F\left( x \right) = 0} \right\}$ , where K it´s probably algebraically closed.

But now... The second part :

Since it's an integral domain, it field of fractions exist. How can we prove the following:

Let $F(T_1,...,T_n) = G(T_1,...,T_n)+H(T_1,...,T_n)$ where G is a homogeneous polynomial of degree $m-1$ and H is a homogeneuous polynomial of degree $m$. Assume that F is irreducible in $k[T_1,...,T_n]$ where k is a field.

$$Frac\left( {k\left[ {T_1 ,...,T_n } \right]/\left( {I\left( {V\left( F \right)} \right)} \right)} \right)$$ it´s isomorphic to a field of rational functions $k\left( {T_1 ,...,T_m } \right)$ for some m.

I have no idea how to use the hypothesis of being sum of two irreducible homogeneous polynomials.

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To simply your question, you can already notice that $I(V(F))$ is generated by $F$ by Nullstellensatz because $F$ is irreducible. So in the field you are looking at, if you denote by $t_i$ the image of $T_i$, then $F(t_1,...,t_n)=0$. Now divide this equality by $t_1$ (can suppose $F(T_1,\dots, T_n)$ is not linear), and try to prove that your fraction field is generated by $t_2/t_1, \dots, t_n/t_1$. – user18119 Apr 25 '12 at 13:01