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The following is a lemma in a note of graded ring, however, I do not know how to prove it. Please help me. Thanks.

Let $R$ be a commutative reduced graded ring where $R_{0}$ is a field and let $u\in R_n\setminus \lbrace0\rbrace$. Then $u$ is transcendental over $R_{0}$.

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Suppose there exists an integer $m$ and $a_{m - 1}, \ldots, a_0 \in R_0$ such that $$ u^m + a_{m - 1}u^{m - 1} + \cdots + a_0 = 0. $$ Now, $u^m$ is a non-zero [$R$ is reduced] element of degree $nm$. The above equation gives the additive inverse for $u^m$, which must also be a non-zero element of $R_{nm}$. But given that $R = \bigoplus R_d$, is this possible? What is the degree of $a_iu^i$?

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I am so sorry for my stupid question but how do we know that $u^m$ has degree $mn$? There does not exist such an element of the form $a_{i}u^{i}$ in the ring R, does not it ? –  variete May 23 '12 at 2:24
    
@variete No need to be sorry. My reasoning is that in general we require $R_dR_e \subset R_{d+e}$, so in particular $u^2 \in R_{2n}$ and so on. I'm afraid that I can't understand the second comment—could you elaborate? –  Dylan Moreland May 23 '12 at 2:55
    
Thank you very much Dylan Moreland. I have a mistake in the second comment, please forget that comment. Can you tell me what is the degree of $a_{i}u_{i}$ ? Is it $n+i$? –  variete May 23 '12 at 3:19
    
@variete All the $a_i$ are degree zero, so the degree of $a_iu^i$ is $in$. –  Dylan Moreland May 23 '12 at 3:24
    
Ok, I got the light of the proof. Thank you so much Mr Moreland. –  variete May 23 '12 at 3:28

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