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There seems to be an error in the proof of Proposition I-18. The first inset equation (line 7) on page 20 only holds in $X_{f_a f_b}$. But it is used on line 15, where it needs to hold in $X_{f_b}$. This seems to invalidate the proof for rings with zero divisors.

As an example, use the scheme in Exercise I-20(b). Take $f_a$ to be the image (coset) of $x$ in $\mathbb{C}[x]/(x^2-x)$ and $f_b$ the image of $x-1$. Then $f_a f_b = 0$, $X = X_{f_a} \cup X_{f_b}$ and $X_{f_a} \cap X_{f_b} = X_{f_a f_b} = \varnothing$. So for any $g_a \in R_{f_a}$ and $g_b \in R_{f_b}$ there must be a $g \in R$ that becomes $g_a$ in $X_{f_a}$ and $g_b$ in $X_{f_b}$. To be specific take $g_a = 1/f_a$ and $g_b = 1/f_b$. Then $g$ is the image of $2x-1$ in $\mathbb{C}[x]/(x^2-x)$ but the proof of Proposition I-18 doesn't produce it. Am I confused, or is the proof wrong?

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2 Answers 2

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There is no error: the ends of the equation in line 7 (i.e. $f_b^Nh_a = f_a^Nh_b$) are meant to hold in $R$. The point is that $h_a, h_b$, and $N$ should be chosen for all the intermediate equalities to hold as well. So in your example $R = \mathbb{C}[x]/(x^2-x)$ with $f_a = x$, $f_b = x - 1$, $g_a = 1/f_a$, $g_b = 1/f_b$, taking $h_a = 1 = h_b$ will not work, as this does not satisfy $f_b^N h_a = (f_af_b)^N g_a$ in $R$.

In fact, we see that the simplest way to satisfy the equation above is by taking $h_a = f_a$, $h_b = f_b$. Thus $N = 2$ (note that $f_a$ is idempotent, but $f_b$ is not, as $f_b^2 = -f_b$), which yields $e_a = 1$, $e_b = 1$, so $g = e_ah_a + e_bh_b = f_a + f_b = 2x-1$, and indeed $f_b^2g = f_b^3 = f_b = f_b^2g_b$ in $R_{f_b}$ (and similarly in $R_{f_a}$).

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  • $\begingroup$ Thanks, I see it now. I'm nitpicking, but the proof as written was confusing (at least to me) because it says: $f_a^Ng_a \in R_{f_a}$ is the image of an element $h_a \in R$. This seems to imply that you could pick any such $h_a$. I think it would have been better to say: pick $N$ so that $(f_af_b)^Ng_a = (f_af_b)^Ng_b$ and set $h_a = f_a^Ng_a$, increasing $N$ if necessary so that $h_a \in R$. $\endgroup$ Sep 29, 2014 at 2:46
  • $\begingroup$ I agree that the proof was slightly confusing as written - I also agree with your wording. My answer previously had a sign error, which produced the wrong result, but has now been fixed (and now produces the desired section $g = 2x - 1$ you mentioned in your original post) $\endgroup$
    – zcn
    Oct 3, 2014 at 4:04
  • $\begingroup$ I'm finally understanding the proof of Prop 1-18. I think there is a step left out. If $X_{f_a} \cap X_{f_b} \neq \varnothing$ then from the requirement that $g_a$ and $g_b$ be equal in $X_{f_a} \cap X_{f_b}$ it follows that $(f_af_b)^{N'}g_a = (f_af_b)^{N'} g_b$. If $X_{f_a} \cap X_{f_b} = \varnothing$ (the missing step) then since $X_{f_a} \cap X_{f_b} = X_{f_a f_b}$ it follows that every prime ideal contains $f_a f_b$, in other words $f_a f_b$ is in the nilradical so $(f_af_b)^{N'} = 0$. Again $(f_af_b)^{N'}g_a = (f_af_b)^{N'} g_b$. Now the proof proceeds as before. $\endgroup$ Oct 5, 2014 at 18:03
  • $\begingroup$ Also, here are more details on the book proof. Set $h'_a = f_a^{N'} g_a$ and $h'_b = f_b^{N'} g_b$. By associativity in $R_{f_a f_b}$, $f_b^{N'}h'_a = (f_af_b)^{N'}g_a = (f_af_b)^{N'}g_b = f_a^{N'}h'_b$ so $f_b^{N'}h'_a = f_a^{N'}h'_b$ in $R_{f_a f_b}$. But since both terms are actually in $R$, it follows that $(f_af_b)^s f_b^{N'}h'_a = (f_af_b)^sf_a^{N'}h'_b$. Setting $N=N'+s$, $h_a = f_a^N g_a = af_a^sh'_a$ and $h_b = f_b^sh'_b$ gives $f_b^Nh_a = f_a^Nh_b$. As the proof in the book shows, this is all that is needed to check that $g = \sum_a e_ah_a$ has the desired properties. $\endgroup$ Oct 5, 2014 at 18:10
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The equality on line 7 holds in the ring $R$, that is, on the whole space $X$. The point is that by definition of a localization, if $x$ and $y$ are elements in $R$, then their images in $R_f$ are equal if and only if $f^Na = f^Nb$ in $R$. So that equality is true because the images of $g_a$ and $g_b$ are equal in $R_{f_af_b}$ but the equality itself is in $R$. So once the equality holds in $R$ it also holds in $R_{f_a}$ and $R_{f_b}$.

There's no issue in your example. Let $\mathbb{C}[x]/(x^2 - x) = R$. Then by Chinese remainder theorem, $R = \mathbb{C}[x]/(x) \times \mathbb{C}[x]/(x - 1) = R_1 \times R_2$. We have $R_x = R_2$ and $R_{x-1} = R_1$ with the natural maps being the same as the projections. Then the statement of the proposition is just that any $g_1 \in R_1$ and $g_2 \in R_2$ are the images of $(g_1,g_2) \in R_1 \times R_2$ which by Chinese remainder theorem we can take to be $g \in R$.

Geometrically, what is happening here is that $R$ corresponds to two distinct points in $\mathbb{A}^1$ and the statement is just that a function on $R$ is uniquely determined by it's value on each of the two points and that conversely if we pick values on each point we can find a function on $R$ that takes those values at the points because the points are two separate connected components. In this case the partition of unity in the proof of the proposition is $x - (x-1)$.

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  • $\begingroup$ Sorry, I'm not understanding your answer. Here's what I get when working through the proof using the scheme of Exercise I-20 with $f_a = \overline{x}$, $f_b = \overline{x-1}$, $g_a = 1/f_a$, $g_b = 1 /f_b$, $N=1$, $h_a = h_b = 1$. Here I use $\overline{x}$ to mean the image of $x$ in $\mathbb{C}[x]/(x^2-x)$. Then $e_a$ and $e_b$ are defined via $1 = e_af_a + e_bf_b$ so $e_a = 1$ and $e_b = -1$. Then $g = e_ah_a + e_b h_b = 0$. The proof claims to produce a $g$ with $g = g_a$ in $R_a$, but $ 0 \neq 1/f_a$ in $R_a$. $\endgroup$ Sep 28, 2014 at 18:43
  • $\begingroup$ Also you seem to be saying the first inset equation is true in $R$, where $R = \mathbb{C}[x]/(x^2-x)$. But that equation (with $N=1$) is $f_b h_a = (f_a f_b) g_a = (f_a f_b) g_b = f_a h_b$. Plugging in the values above gives $\overline{x-1} = 0 = 0 = \overline{x}$. This does not seem to be true in $R$. $\endgroup$ Sep 28, 2014 at 18:44

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