# The ring of homogeneous polynomials

I think I found an error in my textbook, but I am not completely sure. The book is Hulek, Elementary algebraic geometry, pag. 73. There is a theorem showing that $U_i$ and the affine space $\mathbb{A}^n$ are Zariski-homeomorphic, where $U_i$ is the complement in $\mathbb{P}^n$ of the hyperplane defined by $X_i=0$. In the proof, it considers the ring $S^h$ defined as follows: $$S^h:=\{f\in k[X_0,\ldots,X_n] :f \textrm{ is homogeneous}\}$$

How can $S^h$ be a ring?? The polynomials in $S^h$ are not of fixed degree! Do you think it is a typo? Then it says there are maps $$\alpha(f):=f(1,x_1,\ldots,n_n)$$ and $$\beta(g):=x_0^{\operatorname{deg}(g)}\cdot g\big(\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0}\big)$$ for every $f\in S^h$ and $g\in k[X_0,\ldots,X_n]$, and again $S^h$ is called a ring. Why?

• Indeed $S^h$ is a set. – Uri Brezner Mar 14 '14 at 12:47

$$S^{h}$$ is indeed not a ring because the sum of two homogeneous polynomials of unequal degree is not homogeneous in general.