# Exercise 2.17(d) of Eisenbud's Commutative Algebra

First some notation: Let $P$ be a homogeneous prime ideal of a $\Bbb{Z}$ - graded ring $R$, $U$ the multiplicative subset of all homogeneous elements not in $P$. Suppose that there exists a homogeneous element $f$ of degree $1$ that is not in $P$.

The problem in Eisenbud is to show that the image of $P$ (which we call $Q$) in the ring $R/(f-1)$ is a prime ideal. In fact what I am trying to prove (something stronger actually) is that the complement of $Q$ in $R/(f-1)$ is the multiplicative set $\bar{U}$, that is the image of $U$ under the canonical projection $\pi : R \longrightarrow R/(f-1)$. Recall because $\pi$ is surjective $Q$ is already an ideal and hence

$$\text{Q is a prime ideal in R/(f-1) \iff R/(f-1) - Q is multiplicatively closed}$$

Now we want to show that $Q^{c} = \bar{U}$. To show these containments we can assume wlog that we show containment for just homogeneous elements. Now one direction I have already shown, namely that $Q^{c} \subseteq \bar{U}$. The other direction is tantamount to showing that $\bar{U} \subseteq Q^{c}$.

Now suppose this does not hold. Then there is $x \in \bar{U}$ such that $x$ is also in $Q$. I.e. there exist $u \in U$ and $p \in P$ such that $x = \pi(u) = \pi(p)$ so that

$$\pi(u - p ) = 0 \implies u- p = (f-1)r$$

for some $r \in R$. Now if $r \in p$ we have our desired contradiction. If $r$ is not in $P$ then we can write

$$r = x_{n_1} + \ldots x_{n_k}$$ where each $x_{n_i}$ for $1 \leq i \leq k$ is homogeneous of degree $x_{n_i}$ and there is at least one $x_{n_i}$ (which we can take to be $x_{n_1}$ ) that is not in $P$. Then rearranging the equation above we have that

$$p = u - fx_{n_1} - \ldots f_{n_k} + x_{n_1} + \ldots x_{n_k}.$$

Recall we assumed that $x_{n_1} \notin P$. Now where I am stuck is I want to do some argument in comparing degrees to derive a contradiction. However what if say only $u$ and $x_{n_1}$ have the same degree as $p$ so that $u + x_{n_1} = p$? How can I get a contradiction out of something like that?

This is where I am stuck because the rest of the analysis seems to be bashing out cases like this and I do not get the desired contradictions. Any hints on how to proceed on the problem would be greatly appreciated.

Thanks.

Edit 1: I believe I have my desired contradiction at least in the case that $r$ is homogeneous. The proof is as follows. Suppose that $u-p = fr - r$. Then rearranging this equation we have $p = u - fr + r$. Now if $u$ has degree different than the other two terms we are stuffed because recall $P$ is a homogeneous prime ideal so then $u \in P$, a contradiction. Furthermore $u$ can only have the same degree as at most 1 of $fr$ or $r$. This is because $\deg fr \neq \deg r$. We divide this out into two cases:

Case 1: $u$ has the same degree as $r$.

Now because $p = u - fr + r$ and $u$ has the same degree as $r$, we conclude by comparing degrees that $p = u + r$ or $p = -fr$. Now in the former case we must have $-fr = 0 \in P$ contradicting $fr \notin P$ because $f \notin P$ and $r \notin P$. The latter case also gives the same contradiction.

Case 2: $u$ has the same degree as $-fr$.

Now $p$ being a homogeneous element means by comparing degrees that $$p = u-fr, \hspace{2mm} r = 0 \hspace{3mm} \text{or} \hspace{3mm} p = r, \hspace{3mm} u-fr = 0.$$ In the former case we have a contradiction because $r = 0 \implies u = p$ and in the latter we have a contradiction too because $r \notin P$.

Edit 2: The one direction that I claimed to have shown, namely that $Q^{c} \subseteq \overline{U}$ is false. Namely because I can't just assume to show containment for homogeneous elements because $U$ and $\overline{U}$ don't have any additive structure on them.

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Your claim is not what the exercise asks you to do. In fact, it is not true in general that the image of $\frak p$ in $R/(f-1)$ is a prime ideal. For example, let $R$ be the polynomial ring $\mathbb{Q}[x]$ over the field of rational numbers, $\frak p$ be the principal ideal $(x+1)$ generated by $x+1$, and $f=x$. So the image of $\frak p$ is the whole ring. If you assume $\frak p$ is homogeneous, then the claim is correct. For the direction you hope to show, please read my comment below. In fact, it may be helpful to realize that the ring $R/(f-1)$ is canonically isomorphic to the ring $R_{(f)}$.
Your claim does not work because $(x+1)$ is not a homogeneous prime ideal. For example if we put the usual grading on $\Bbb{Q}[x]$ we must have that because $x$ and $1$ have different degrees and are homogeneous that $x \in \mathfrak{p}$ and $1 \in \mathfrak{p}$ which is ridiculous. – user38268 May 31 '12 at 0:46
Have you ever required that $\frak p$ is homogeneous? – Joy-Joy May 31 '12 at 0:49