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In the exercise from chapter $2$ of the book, "Introduction to Commutative Algebra" by Atiyah & Macdonald, I understand that the highlighted $A$-algebra $B$ is the the direct limit but I want to ask the following questions:

  1. How can we say that the direct limit is the tensor product of the given family? Shouldn't one also prove the universal property of the tensor product here? If yes than how can we prove that universal property since here we have arbitrary family of $A$-algebras which may be infinite.

  2. How the arbitrary element of the $A$-algebra $B$ will look like? My intuition says: For each $\lambda$, fix an element $b_\lambda$ $\in$ $B_\lambda$, then, any arbitrary elements of $B$ is of the form $\otimes$$_\lambda$$x_\lambda$ where $x_\lambda$=$b_\lambda$ for all but finitely many values of $\lambda$. But the problem is that how the tensor map $\otimes$$_\lambda$$x_\lambda$ will be defined since $\lambda$ runs over an arbitrary family.

  3. Can one define the tensor products in the same way for directed systems of arbitrary families of, say $A$-modules or say vector spaces of a field? Will the direct limits exist in those cases?

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How can we say that the direct limit is the tensor product of the given family?

Well, Atiyah and Macdonald are defining the tensor product of an infinite family of $A$-algebras. We're free to make whatever definitions we want...

Shouldn't one also prove the universal property of the tensor product here?

Yes, that would provide some justification that the definition is a reasonable one. But be careful: in the category of $A$-algebras, the universal property satisfied by the tensor product of two algebras $B\otimes_A C$ is the universal property of the coproduct! It is the universal property of the infinite coproduct that is satisfied by Atiyah and Macdonald's definition, not anything to do with multilinear maps. That is, what Atiyah and Macdonald are doing in this exercise is providing an explicit construction of the infinite coproduct in the category of $A$-algebras.

The reason why the underlying $A$-module of the coproduct of $A$-algebras $B$ and $C$ agrees with the tensor product of the underlying $A$-modules of $B$ and $C$ is that bilinear maps $B\times C\to D$ are closely related to pairs of maps $B\to D$ and $C\to D$. For example, if we have a pair of $A$-algebra homomorphisms $f\colon B\to D$ and $g\colon C\to D$, then we can form a bilinear map $B\times C\to D$ by $(b,c)\mapsto f(b)g(c)$. This relationship breaks down in the infinite case. Given a family of $A$-algebra homomorphisms $f_\lambda\colon B_\lambda\to D$ for all $\lambda\in \Lambda$, we can't get a multilinear map in the same way: multiplication of infinitely many outputs of the $f_\lambda$ doesn't make sense in $D$.

How the arbitrary element of the $A$-algebra $B$ will look like?

In general, the direct limit of a directed system of algebraic structures can be described as the union of all the structures in the system, modulo the equivalence relation defined by $c\in C$ is equivalent to $d\in D$ if and only if $c$ and $d$ agree later in the system, i.e. there is some structure $E$ in the system with maps $f\colon C\to E$ and $g\colon D\to E$ such that $f(c) = g(d)$.

In this particular case, the canonical $A$-algebra homomorphism $B_J\to B_{J'}$ that Atiyah and Macdonal refer to is the one that extends a tensor by $1$s. E.g. if $B_J = B_1\otimes_A B_2$ and $B_{J'}$ is $B_1\otimes_A B_2\otimes_A B_3 \otimes_A B_4$, then the map $B_J\to B_{J'}$ is determined by $x_1\otimes x_2\mapsto x_1\otimes x_2\otimes 1\otimes 1$. So the elements of the direct limit are all elements of finite tensor products from the family, where we view two elements as equal if they are equal after we extend them both by $1s$ to put them in the same finite tensor product.

It turns out that this is the same as considering all finite linear combinations of infinite tensors $\bigotimes_{\lambda\in \Lambda} x_\lambda$, where all but finitely many of the $x_\lambda$ are equal to $1$, modulo the usual relations defining the tensor product. See Eric Wofsey's answer here for more details and a sketch of the proof that this construction satisfies the universal property of the coproduct.

Can one define the tensor products in the same way for directed systems of arbitrary families of, say $A$-modules or say vector spaces of a field?

No, the ring structure is crucial here, since we use $1$ to define the canonical maps $B_J\to B_{J'}$. For infinite tensor products of modules or vector spaces, one has to consider multilinear maps. See the discussion here.

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    $\begingroup$ @PrinceKhan That's indeed relevant. Note that the linked document defines an infinite tensor product $\otimes_{i\in I}^t E_i$ of vector spaces $(E_i)_{i\in I}$ relative to a choice $t = (t_i\in E_i)_{i\in I}$ of an element $t_i$ in each space $E_i$. And the universal property proven on p. 5 is with respect to multilinear mappings from the restricted product $\Pi^t_{i\in I} E_i$ defined on p. 1. $\endgroup$ – Alex Kruckman Feb 18 '20 at 17:52
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    $\begingroup$ So I maintain that the answer to "can one define the tensor product in the same way for arbitrary families of vector spaces?" is No. But the answer to "can one define the tensor product in the same way for arbitrary families of pointed vector spaces?" (vector spaces with a chosen non-zero element) is Yes, as long as you're ok with the universal property represented by the tensor product depending on the choice of points. $\endgroup$ – Alex Kruckman Feb 18 '20 at 17:57
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    $\begingroup$ Yes you were right but I have some problems in understanding with their construction thatshwy I shared the link. What I understood is that the family $Π^t_{i∈I}E_i$={${(x_i)\in Π_{i∈I}Ei|x_i=t_i}$ for all but finitely many i}. If the product $Πt^t_{i∈I} E_i$ is really what I wrote above than it's not a vector subspace of the product $Π_{i∈I}E_i$. Can you please explain it to me. Thanks! $\endgroup$ – Prince Khan Feb 18 '20 at 20:21
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    $\begingroup$ I think the restricted product should be like this $Π^t_{i∈I}E_i$={${(x_i)\in Π_{i∈I}E_i|x_i\in <t_i>}$ for all but finitely many i}, Where $<t_i>$ is the vector subspce of $E_i$ generated by the element $t_i$. But if I assume this than I don't know how to prove the universal property in the next page. $\endgroup$ – Prince Khan Feb 18 '20 at 20:28
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    $\begingroup$ @PrinceKhan I'm not familiar with this construction (this conversation is the first time I've heard about it), and I don't have time to sort it out right now. I suggest you ask a new question which is explicitly about the construction in the notes you linked to. $\endgroup$ – Alex Kruckman Feb 18 '20 at 21:36
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Note that the tensor product of algebras is in fact the coproduct in the category of $A$-algebras. What AM defines there should therefore satisfy the universal property of an infinite coproduct. I am not sure about the universal property with respect to multilinear maps of the underlying modules though. I believe one can derive something like this, yet one has to ask oneself, if it would be worth the effort.

I recall doing this construction in some exercise quite a while ago and would prefer not to reconsider how the elements look like. So I am afraid I wont be of much use regarding your second question. Maybe someone else can help out there. Personally I strive to do as much as possible using universal properties only...

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  • $\begingroup$ can you please give an independent proof for the universal property of tensor product without using the argument that the tensor product of algebras over a commutative ring is the coproduct? $\endgroup$ – Prince Khan Feb 1 '20 at 2:12
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    $\begingroup$ Well, as I said and as is backed up by Alex Kruckmans answer, this construction is a coproduct construction. I didn’t mean it as an argument to derive anything about tensor products and in fact I don’t find the word tensor product appropriate in the context of algebras. Maybe it helps to take the perspective that it is a mere coincident that the underlying module of a coproduct of algebras is a tensor product of the underlying modules. Imo the universal property of an infinite tensor product is simply the wrong universal property to consider here! $\endgroup$ – PrudiiArca Feb 1 '20 at 8:51

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