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I'm reading the book "a first course in abstract algebra" by Faleigh on page 290. The theorem is: Every field F has an algebraic closure F‾. In the middle of the writer's proof, he start to build a towel of algebraic extension fields.

statement: All algebraic extension fields E_j of F, form a set S={E_j|j in J} that is partially ordered under our usual subfield inclusion.

Let T={E_j_k} be a chain in S, and let W=∪_k(E_j_k).Then we make W into a field.

Here is my problem: If ∪_k(E_j_k) is the finite union, then it has no problem, we can pick the largest extension field and define the sum and product operation on it. Now, if ∪_k(E_j_k) is infinite union, how do we define the field operation (sum and product) in consistent way, the book I refer just say that the operation is well defined

,but in my thinking it is not so natural. I think in infinite case, we can't define the operation to be one of the extention field's operation. From one field to its subfield, the subfield can inherit the field operation of the larger field, but from the field to its extension field, I 'm not sure whether the operation of the extention field is the same with the original field.

Thanks for clarify my concept~~!

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    $\begingroup$ Can someone please come up with an acceptable concept that can be called an algebraic towel? I need more material for my dry sense of humor. $\endgroup$ – RghtHndSd Oct 10 '14 at 2:08
  • $\begingroup$ Hello David Holden, thanks for your note. But if the other elements aren't in n^th object, then can we use the same n^th operation? $\endgroup$ – 扁頭科學麵Kevin Lee Oct 10 '14 at 2:14
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this set-up is encountered in many situations where we have such towers of algebraic objects. the point is that any element in the union must lie in all the objects after a certain $n$. if there are two objects, then take $n = \max(n_1,n_2)$. now both elements coexist in the same ($n^{th }$) object and operations can be defined there.

added

suppose $a,b$ are elements of a group $G$. the operation $ab$ has a well-defined result (by definition of a group - closed under 'multiplication'). this result does not change if i regard $G$ as a subgroup of another group $H$.

from the point of view of an inhabitant of $G$ there might be a question: "i know how to relate to my fellow-members of $G$ but how should i combine with the 'new' members of $H$?"

but this is already taken care of by the fact that $G$ is a sub group of $H$. the details need not detain us - again, this is a merely a matter of definition.

look at $\mathbb{Q}$ - the rationals - as an additive group. let us number the primes in order, so $p_1$ is $2$. we shall define a tower $T_n$ for $n=,0,1,2,...$, beginning with $T_0=\mathbb{Z}$

now define the sets $G_n$ so that the elements of $G_n$ are all possible square-free products of the primes $p_1,p_2,...,p_n$, so that, for example: $$ G_4 = \{1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210\} $$ let $T_n$ be the subgroup of $\mathbb{Q}$ of elements whose denominator is one of the members of $G_n$. you can see that each $T_n$ is closed under addition, because by construction $G_n$ contains the least common multiple of any subset of its elements. also: $$ \mathbb{Z} \subset T_1 \subset T_2 \subset \cdots $$ now consider the claim that: $$ \mathbb{Q^*} = \cup_{n=0}^{\infty}T_n $$ is a well-defined group.

suppose now we take sets $H_n$, constructed like $G_n$ except that we only employ primes of the form $4k+1$, leaving out those of type $4n+3$. the tower constructed for the $H_n$ will give rise to a subgroup $\mathbb{Q^{**}}$ of $\mathbb{Q^*}$

i shan't labour the point, but ask you to think about this example in the light of your question, and my initial explanation (first paragraph of this answer). if anything is still unclear, perhaps you can express it in terms of this simple case-study, and someone will no doubt be able to set you straight

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  • $\begingroup$ thx! i'll try to smooth the path... $\endgroup$ – David Holden Oct 10 '14 at 2:44

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