# Prime avoidance for graded Noetherian ring with infinite residue field.

This question is related to this question.

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 $$P_i$$.

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

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 $$x+y$$.