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Theorem 3.7 in the chapter on ring extension on page 352 of the latest edition of Lang's "Algebra" appears redundant in its phrasing to me. Specifically, if $g_s$ is a polynomial of total degree $< d_s$, isn't any pure power of one of its variables $Z_s$ automatically occuring with a power strictly less than $d_s$?

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What could he have meant with this statement? (If it just plain redundant and no deeper meaning to it, it would also be good to be ensured.)

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Picture of cited theorem i53.tinypic.com/qzj9jd.jpg –  Myself May 13 '11 at 21:57

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The phrasing is indeed quite confusing - since $A$ is a domain, if the total degree of $g_s$ is $< d_s$, then any individual term of (in particular any pure power occurring in) $g_s$ has degree $< d_s$. I believe that the condition on pure powers is redundant.

To justify this, we can examine the proof - which, incidentally, has its own shortfalls. The $s$ in the $2^\text{nd}$ line ("for some $s$") is, a priori, in no way related to the $s$ in the $3^\text{rd}$ line ("pick an index $s$ such that") - another index $t$ should be used, i.e. $\varphi(z_j/z_t) \ne \infty$ for all $j$. However, $\varphi(z_s) = \infty$ and $\varphi(z_s/z_t) \ne \infty$ together imply $\varphi(z_t) = \infty$ (if not, $\varphi(z_s) = \varphi((z_s/z_t) \cdot z_t) = \varphi(z_s/z_t) \cdot \varphi(z_t) \ne \infty$), so replacing $s$ by $t$, we may assume $s = t$. Then, as long as $g_s$ has total degree $< d_s$, each term of $g_s$ has degree $\le d_s - 1$, so writing $\displaystyle g_s = \sum_{i=0}^{d_s-1} g^{(i)}_s$ where $g^{(i)}_s$ is homogeneous of degree $i$,

$$\frac{g_s(z_j)}{z_s^{d_s}} = \frac{1}{z_s}\frac{g_s(z_j)}{z_s^{d_s-1}} = \frac{1}{z_s} \sum_{i=0}^{d_s-1}\frac{g^{(i)}_s(z_j/z_s)}{z_s^{d_s-i-1}}$$

and $\varphi$ of each term in the sum is finite whereas $\varphi(1/z_s) = 0$, so $\varphi(g_s(z_j)/z_s^{d_s}) = 0$, as desired.

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