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Let $R$ be a Noetherian local ring with infinite residue field. Let $\text{gr}_IR$ denote the associated graded ring $\bigoplus_{n=0}^{\infty} I^n/I^{n+1}$. Let $P_1,...,P_n$ be $n$ homogeneous ideals of $\text{gr}_IR$, and each $P_i$ doesn't contain all the elements of $\text{gr}_IR$ of positive degree. Then there is a homogeneous element $h=x+I^2$ where $x\in I$ such that $h$ is not contained in any $P_i$ for $i=1,...,n$. The context can be found in the second paragraph of the proof on page 170 of proposition 8.5.7 in this book. It is claimed one can choose such $h$ according to prime avoidance for graded Noetherian rings with infinite residue field.

However, if $I_1=\bigoplus_{n=1}^{\infty} I^n/I^{n+1}$, $I_2=\bigoplus_{n=2}^{\infty} I^n/I^{n+1}$. Then $I_1\not\subset I_2\cup P_1\cup P_2\cup\dots\cup P_n$. By usual prime avoidance one can choose the desired $h$ (By theorem A.1.3 from the linked book). Here I didn't use the infinite assumption of residue field. Where did I go wrong?

Theorem A.1.3 (Prime Avoidance) Let $R$ be Noetherian $\mathbb{N}$-graded ring. Let $P_1, . . . , P_s$ be homogeneous ideals in $R$, at most two of which are not prime ideals. If $I$ is a homogeneous ideal generated by elements of positive degree and not contained in $P_1 \cup · · · \cup P_s$, then there exists a homogeneous element $x \in I$ that is not in any $P_i$.

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  • $\begingroup$ How do you know $I_1\not\subset I_2\cup P_1\cup P_2\cup...\cup P_n$? $\endgroup$ Commented Nov 12, 2023 at 3:21
  • $\begingroup$ @EricWofsey It follows because of prime avoidance. The number of non-prime ideals is one, so it works. $\endgroup$ Commented Nov 12, 2023 at 3:38

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You can find $h\in I_1$ that is not in $I_2$ or any of the $P_i$, but that $h$ is not necessarily homogeneous. It is helpful to consider an example (which is generally what you should do if you seem to have proved something false--run through the proof with an example and find the first statement that is wrong). If your graded ring is the polynomial ring $\mathbb{F}_2[x,y]$, then $I_1=(x,y)$, $I_2=(x,y)^2$, and you could have $P_1=(x)$, $P_2=(y)$, and $P_3=(x+y)$ and these will together cover all the homogeoneous elements of degree $1$. However, $I_2\cup P_1\cup P_2\cup P_3$ is still not all of $I_1$, since for instance there are elements like $x^2+y$ that have a nonzero degree $1$ part and are not divisible by $x,y,$ or $x+y$.

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  • $\begingroup$ That makes sense to me. So it seems that theorem A.1.3 I just quoted in my post is not true? $\endgroup$ Commented Nov 12, 2023 at 4:10

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