I Meaning of ##\coprod_{x\in\mathrm{Spec}R}R/\mathfrak{p}_x##

[elias001]

TL;DR Summary

I would like to know how to interpret the notation: $\coprod_{x\in\mathrm{Spec}R}R/\mathfrak{p}_x$ in the quoted passage

The following is taken from a chapter from Algebraic geometry and commutative algebra by S Bosch

Background

We want to sketch the basics of this approach in brief terms. For $f\in R$ and $x\in\mathrm{Spec}R$ define $f(x)$ as the residue class of $f$ in $R/\mathfrak{p}_x$. Here $\mathfrak{p}_x$ is a second notation instead of $X$, just to remember that the $\textit{point x}$ is, in reality, a prime ideal in $R$. This way, elements $f\in R$ can be interpreted as functions $$f:\mathrm{Spec}R\to \coprod_{x\in\mathrm{Spec}R}R/\mathfrak{p}_x.\quad (*)$$ In particular, assertion like $f(x)=0$ or $f(x)\neq 0$ make sense and just means $f\in \mathfrak{p}_x$ or $f\not\in \mathfrak{p}_x$.

and $\mathrm{Spec}R=\{\mathfrak{p}\subset R:\mathfrak{p}\text{ prime ideal in }R\}$.

For reference, I included a screenshot of the typed out quoted passage above below:

Questions

I would like to know the notation: $\coprod_{x\in\mathrm{Spec}R}$ in $\coprod_{x\in\mathrm{Spec}R}R/\mathfrak{p}_x$, does it mean coproduct or product. Also, is the function $f$ in $(*)$ defined as: $f(x)=(f+(\mathfrak{p}_x), x)$ if the codomain is a coproduct, or a product: $f(x_i)=(f+(\mathfrak{p}_{x_1}),f+(\mathfrak{p}_{x_2}),\ldots)$?




Thank you in advance.

 

[PeroK]

How many textbooks are you studying from concurrently?

 

[elias001]

@PeroK I am going through Hungerford's Abstract Algebra, along with Dummit and Foote's abstract algebra. Those two are my main ones. Along the way, when I don't understand something on a topic, I consult others. I went through a text in category theory, which I made post about, a few exercises I can't do on here as well an older textbook on module theory. This post I made, I was trying to understand something in the topic of localization.

 

[martinbn]

Disjoint union.

 

[elias001]

@martinbn thank you so much for clearing that up. In the text, I can't tell if the notation meant product or disjoint union. So I am to interpret the notation as coproduct/disjoint union.

 

[elias001]

@fresh_42 For the above passage:

"We want to sketch the basics of this approach in brief terms. For $f\in R$ and $x\in\text{Spec }R$ define $f(x)$ as the residue class of $f$ in $R/\mathfrak{p}_x$. Here $\mathfrak{p}_x$ is a second notation instead of $X$, just to remember that the point $x$ is, in reality, a prime ideal in $R$. This way, elements $f\in R$ can be interpreted as functions
$$f:\text{Spec }R\to \coprod_{x\in\text{Spec }}R/\mathfrak{p}_x.\quad (*)$$
In particular, assertion like $f(x)=0$ or $f(x)\neq 0$ make sense and just means $f\in \mathfrak{p}_x$ or $f\not\in \mathfrak{p}_x$. "

I have the following listed of definitions, observations, examples, etc from Adamek's text Theory of Mathematical Structures.(Is a really old text.)

'
$\textbf{Definition 1:}$ A coproduct of a collection of objects $A_i,i\in I,$ is an object $A$ together morphism (called injections)$\\\\$

$$\epsilon_i:A_i\to A \text{ for all } i\in I$$

which have the following universal property:$\\\\$

$\quad$ For each collection of morphisms $f_i:A_i\to B,i\in I,$ there exists a unique morphism $f:A\to B$ with $f_i=f\cdot \epsilon_i$ for each $i\in I.\\\\$

Notation: The above object $A$ is denoted by $\coprod_{i\in I}A_i$ (in case $I=\{1,2\}$ also by $A_1+A_2,$ analogously $A_1+A_2+A_3,$ etc.). A category is said to have coproducts if each collection of its objects has a coproduct.$\\\\$

$\textbf{Example:}$ Coproducts of sets. $\\\\$

(a) Let $X_i,i\in I,$ be pairwise disjoint sets. Their coproduct in $\textbf{Set}$ is their union
$$X=\cup_{i\in I}X_i$$

together with the inclusion maps$\\\\$

$$v_i:X_i\to X,i\in I.$$

Indeed, for each collection of maps$\\\\$

$$f_i:X_i\to Y.i\in I,$$

we have a unique map$\\\\$

$$f:X\to Y,$$

which coincides with $f_i$ on the subset $X_i\subset X$ for all $i\in I,$ in other words, a unique map with$\\\\$

$$f_i=f\circ v_i\text{ for each }i\in I.$$

(b) In case the sets $X_i$ fail to be pairwise disjoint, we first "separate" them. This is done by writing their elements as pairs $(x,i)$ where $x$ is an (arbitrary) element of $X_i$ and $i$ states explicitly which set is being considered. Thus, instead of the set $X_i$ we work with$\\\\$

$$X_i\times\{i\}=\{(x,i):x\in X_i\}:$$'

$\textbf{Definition 2:}$ The disjoint union of sets $X_i,i\in I,$ is the set$\\\\$

$$X=\cup_{i\in I}X_i\times \{i\}.$$

$\text{Observations:}$ (i) The disjoint union $X$ is the coproduct of the sets $X-i,i\in I.$ with respect too the injections$\\\\$

$$\epsilon_i:X_i\to X\quad (i\in I)$$

defined by$\\\\$

$$\epsilon_i(x)=(x,i)\quad (i\in I;x\in X_i).$$

I want to give rigorous definitions for the following in the above quoted passage for the following notations:$\\\\$

$$1)\;f:\text{Spec }R\to \coprod_{x\in\text{Spec }}R/\mathfrak{p}_x, x\in \text{Spec }R, \mathfrak{p}_x, \coprod_{x\in\text{Spec }}R/\mathfrak{p}_x.$$

and the phrase:$\\\\$

$2)\;f(x)$ as the residue class of $f$ in $R/\mathfrak{p}_x.\\\\$

$\textbf{(1)}\\\\$

For $\text{Spec } R, \text{Spec } R = \{ \mathfrak{p} \subset R : \mathfrak{p} \text{ is a prime ideal in } R \}\\\\$.

For $\mathfrak{p}_x$. It means the prime ideal in $R$ corresponding to $x$, or equivalently, the point $x$ corresponds to a prime ideal $\mathfrak{p}_x \subset R.\\\\$

As for $R / \mathfrak{p}_x,$ it is the set of cosets $R / \mathfrak{p}_x = \{ a + \mathfrak{p}_x : a \in R \}= \{ a + b : b \in \mathfrak{p}_x \}.\\\\$

For $x \in \text{Spec } R\\\\$

Since $x \in \text{Spec } R$ corresponds to a prime ideal $\mathfrak{p}_x$, we can write:$\\\\$

$$\{ x : x \in \text{Spec } R \} = \{ \mathfrak{p}: \mathfrak{p} \subset R, \mathfrak{p} \text{ is a prime ideal in }R\}$$

and each $x$ is identified with a prime ideal $\mathfrak{p}_x\\\\$

For $\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x\\\\$

From the quoted materials in Adamek from above,$\\\\$

The coproduct of sets $X_i, i \in I$ is the disjoint union$\\\\$

$$\coprod_{i \in I} X_i = \bigcup_{i \in I} X_i \times \{ i \},$$

with injections $\epsilon_i : X_i \to \coprod_{i \in I} X_i, \epsilon_i(x) = (x, i).\\\\$

Applying to the case for the family of quotient rings $R / \mathfrak{p}_x$ indexed by $x \in \text{Spec } R\\\\$

is defined as:

$$\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \bigcup_{x \in \text{Spec } R} (R / \mathfrak{p}_x) \times \{ x \}$$

Expanding the definition even further using set builder notations, we have:

$$\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \{ (a + \mathfrak{p}_x, x) : x \in \text{Spec } R, a + \mathfrak{p}_x \in R / \mathfrak{p}_x\}$$

where $R / \mathfrak{p}_x$ has already been defined, and the pair $(a + \mathfrak{p}_x, x)$ means that the element $a + \mathfrak{p}_x$ belongs to the copy of $R / \mathfrak{p}_x$ indexed by $x,\\\\$

The injections for the coproduct are:

$$\epsilon_x : R / \mathfrak{p}_x \to \coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x, \quad \epsilon_x(a + \mathfrak{p}_x) = (a + \mathfrak{p}_x, x)$$

and expanding the definition for the map $f : \text{Spec } R \to \coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x$ even further in terms of set builder notation, for each $x \in \text{Spec } R$

$$f(x) = (f + \mathfrak{p}_x, x) \in (R / \mathfrak{p}_x) \times \{ x \} \subset \coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x$$

Then universal property for the coproduct is defined, as for any set $Y$ and maps $g_x : R / \mathfrak{p}_x \to Y$, there exists a unique map $h : \coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x \to Y$ such that $g_x = h \circ \epsilon_x$, defined by $h((a + \mathfrak{p}_x, x)) = g_x(a + \mathfrak{p}_x).\\\\$

$\textbf{(2)}\\\\$

2) To define $f(x)$ as the residue class of $f \in R / \mathfrak{p}_x,\\\\$

- Let $R$ be a commutative ring with identity.$\\\\$
We know that the spectrum $\text{Spec } R$ is the set of all prime ideals of $R:\\\\$

$$\text{Spec } R = \{ \mathfrak{p} \subset R : \mathfrak{p} \text{ is a prime ideal in } R\}.$$

Then for any element $f \in R$, the residue class of $f \in R / \mathfrak{p}_x$ is the coset$\\\\$

$$f + \mathfrak{p}_x = \{ f + b : b \in \mathfrak{p}_x \}.$$

The map $f : \text{Spec } R \to \coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x$ is defined such that for each $x \in \text{Spec } R ,\\\\$

$$f(x) = f + \mathfrak{p}_x \in R / \mathfrak{p}_x.$$

where $f(x)$ assigns to each prime ideal $\mathfrak{p}_x$ the equivalence class of $f$ in the quotient ring $R / \mathfrak{p}_x$. The value $f(x) = 0 \in R / \mathfrak{p}_x$ (i.e., the zero coset $\mathfrak{p}_x$) occurs if and only if $f \in \mathfrak{p}_x$, and $f(x) \neq 0$ means $f \not\in \mathfrak{p}_x.\\\\$

This means the elements $f \in R$ are consider as ``functions" on $\text{Spec } R$, with values in the disjoint union of the quotient rings $R / \mathfrak{p}_x.\\\\$

I have avoid mentioning anything having to do with category theory.

 

[elias001]

@fresh_42 For the entire post 6 above, I just want to know if the following two notations are defined correctly:$\\\\$

$x \in \text{Spec } R:=\{ x : x \in \text{Spec } R \} = \{ \mathfrak{p}: \mathfrak{p} \subset R, \mathfrak{p} \text{ is a prime ideal in }R\}\\\\$

$\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \{ (a + \mathfrak{p}_x, x) : x \in \text{Spec } R, a + \mathfrak{p}_x \in R / \mathfrak{p}_x\}\\\\$

I mean we can inferred from the second notation that $f$ is defined as $f(x) = f + \mathfrak{p}_x \in R / \mathfrak{p}_x.\\\\$

 

[fresh_42]

@fresh_42 For the entire post 6 above, I just want to know if the following two notations are defined correctly:$\\\\$

$x \in \text{Spec } R:=\{ x : x \in \text{Spec } R \} = \{ \mathfrak{p}: \mathfrak{p} \subset R, \mathfrak{p} \text{ is a prime ideal in }R\}\\\\$

$\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \{ (a + \mathfrak{p}_x, x) : x \in \text{Spec } R, a + \mathfrak{p}_x \in R / \mathfrak{p}_x\}\\\\$

I mean we can inferred from the second notation that $f$ is defined as $f(x) = f + \mathfrak{p}_x \in R / \mathfrak{p}_x.\\\\$

It is a bit confusing since it looks as if $x=\mathfrak{p}_x ,$ which is a mess in my opinion. So the first line is correct. We have the Zariski-topological space $X=\operatorname{Spec}(R)$ where every point in it is a prime ideal. That's why the author identifies its elements $x$ with prime ideals $\mathfrak{p}_x ,$ depending on whether he wants to do topology where he uses $x,$ or algebra, where he uses $\mathfrak{p}_x.$

The second line is problematic as you missed defining the role of $a,$ which appears to be $a=f.$

Let us instead look at what is going on. I introduce different letters in order to avoid further confusion. Let's take a ring element $r\in R$ and a prime ideal $\mathfrak{q}\subseteq R.$ Then we can consider its image under the canonical projection $\pi_\mathfrak{q}\, : \,R \twoheadrightarrow R/\mathfrak{q}.$ Yes, you can write this as $\pi_\mathfrak{q}(r)=r+\mathfrak{q}$ but this isn't really helpful in this case.

Now, if we vary the prime ideals $\mathfrak{q}$ then we get different epimorhisms $\pi_\mathfrak{q}$ and different images $\pi_\mathfrak{q}(r).$ What we do next is think of them written in a sequence
$$
(\pi_{\mathfrak{q}_1}(r),\pi_{\mathfrak{q}_2}(r),\pi_{\mathfrak{q}_3}(r),\ldots)
$$
However, we do not know whether we can enumerate all prime ideals, nor do we know how many prime ideals exist in the ring $R.$ This is why we write
$$
\left(\coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} \pi_\mathfrak{q}\right)(r)=\coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} \pi_\mathfrak{q}(r)
$$
instead of the list. It is basically a direct product with possibly uncountably many members.
At this point, we have a function
$$
\rho=\coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} \pi_\mathfrak{q}\, : \,R\longrightarrow \coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} \pi_\mathfrak{q}(R)=\coprod_{\mathfrak{q}\in \operatorname{Spec}(R)}R/\mathfrak{q}
$$
with $\displaystyle{\rho(r)=\coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} \pi_\mathfrak{q}(r).}$

Next comes the transition from algebra to topology. This means, $X=\operatorname{Spec}(R)$ becomes our space of consideration. But what will we do with the elements of the ring? They have no place in $X$ since they aren't prime ideals and possibly not even contained in one. E.g., $1\in R$ isn't contained in any proper ideal. The solution to this problem is considering the ring elements as functions on $X.$ Hence, we need to define $r(\mathfrak{q}).$ We do so by setting $r(\mathfrak{q)}:=\pi_\mathfrak{q}(r).$ Over all prime ideals, i.e., all points of $X,$ this results in a function
$$
r\, : \,\operatorname{Spec}(R)\longrightarrow \rho(R)
$$
with $r(\mathfrak{q})=\pi_\mathfrak{q}(r) \in R/\mathfrak{q} \subseteq \rho(R).$

To complete the transition from algebra to topology, we write $f=r$ and $x=\mathfrak{p}_x=\mathfrak{q}.$

We now have a function $f$ on $X,$ and the image of the prime ideal $x$ under $f=r$ is $f(x)=\pi_x(f)=\pi_{\mathfrak{p}_x}(f)=f+\mathfrak{p}_x.$

The rest of post #6 is more or less notational ballast.

 

[elias001]

@fresh_42 I just got up from a nap. I am coming down from a cold due to change of seasonal weather again and hence a change of temperature.
the
Anyways, what does notational ballast mean

The book I got the passage from is Algebraic geometry and commutative algebra by S Bosch.

In trying to unpacked the notations, I got star trek's android Data's help and specific to it that the notations should be written out in accordance to the bits that are written out from Adamek's text.

This post is important because in trying to dealing with the correctness of the statement: $x \in \text{Spec } R:=\{ x : x \in \text{Spec } R \} = \{ \mathfrak{p}: \mathfrak{p} \subset R, \mathfrak{p} \text{ is a prime ideal in }R\}\\\\$, it is relevant to trying to show $\text{Spec }(A_\mathfrak{p})=\left\{\mathfrak{q}A_\mathfrak{p}\,|\,\mathfrak{q}\subseteq \mathfrak{p} \text{ and }\mathfrak{q}\subseteq A\text{ is prime }\right\}$. Actually, let's just finish dealing with the other post first before we continue with this one.

 

[fresh_42]

@fresh_42 I just got up from a nap. I am coming down from a cold due to change of seasonal weather again and hence a change of temperature.

Get well soon!

Anyways, what does notational ballast mean

It means that you shouldn't concentrate on the details in the definition of a coproduct. It is sufficient here to think of a "list". The reason why the author uses a coproduct and not a direct product is that it comes with an insertion
$$
R/\mathfrak{q} \hookrightarrow \coprod_{\mathfrak{q}\in \operatorname{Spec}(R)} R/\mathfrak{q},
$$
so we can "fill" the list member by member, even if uncountable.

 

[elias001]

@fresh_42 for $\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \{ (a + \mathfrak{p}_x, x) : x \in \text{Spec } R, a + \mathfrak{p}_x \in R / \mathfrak{p}_x\}\\\\$, without making reference to it as being the codomain of the function $f$. Is it correct, that is what I want to know. Sorry, I should have been a bit more specific. I know if i put that back as a piece for the function $f$, then I would need to let $a=f$ as you suggested it. Also, i want to unpack that upside down product looking sign as a disjoint union and unpack the notation. If I insist on that which is what Bosch meant I believe, what would I get. Anyways, lets continue with the other post.

 

[fresh_42]

@fresh_42 for $\coprod_{x \in \text{Spec } R} R / \mathfrak{p}_x = \{ (a + \mathfrak{p}_x, x) : x \in \text{Spec } R, a + \mathfrak{p}_x \in R / \mathfrak{p}_x\}\\\\$, without making reference to it as being the codomain of the function $f$. Is it correct, that is what I want to know. Sorry, I should have been a bit more specific. I know if i put that back as a piece for the function $f$, then I would need to let $a=f$ as you suggested it. Also, i want to unpack that upside down product looking sign as a disjoint union and unpack the notation. If I insist on that which is what Bosch meant I believe, what would I get. Anyways, lets continue with the other post.

I wouldn't write it that way, but technically yes, assuming $f=a=r$ (see my post #8). Carrying the ideal $x$ along the notation is a bit inconvenient, so it is usually dropped. That's why he wrote $x=\mathfrak{p}_x,$ which kind of carries the information $x$ in one expression rather than a pair.