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What is the meaning of the phrase "Up to isomorphism " in Abstract Algebra in context of-

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  • FIELDS
Urysohn Kutrov
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    When someone says something like "$G$ is the only group (or ring or field) with property X, up to isomorphism," what they mean is that any other group (or ring or field) with property X must be isomorphic to $G$. E.g., $\mathbb{Z}$ is the only infinite cyclic group up to isomorphism. There are other infinite cyclic groups, like $\langle a\rangle$, but that group is isomorphic to $\mathbb{Z}$. – kccu May 10 '17 at 15:02
  • It means that two things are the same in the sense that they are isomorphic to each other, i.e., there exists an isomorphism between them. It is similar to saying two functions ($f(x), g(x)$) are identical up to a constant factor, i.e., $f(x) = g(x) + c, , c \in \mathbb{R}$. – David May 10 '17 at 15:02
  • Did you not find the google search results helpful? A couple are other questions here on MSE even. – anon May 10 '17 at 15:02
  • rejecting the isomorphism.. – Nitin Uniyal May 10 '17 at 15:02
  • In field theory, the "up to isomorphim" concept is different because even if $E \cong F$ it doesn't mean there is no field $K$ containing $E$ and $F$ as two different subfields. So I think we say the field $F$ is unique up to isomorphism iff any field $K$ containing $E \cong F$ contains no other subfield $\cong F$ (this is the case if $F$ is a finite field or if $F$ is $\mathbb{Q}$ or $\mathbb{C}$) – reuns May 10 '17 at 15:04
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    @user1952009 That's not particular to fields; two isomorphic groups may be distinct subgroups of a bigger group (or partially ordered sets, or topological spaces, or Riemannian manifolds, or whatever). Your notion of up to isomorphism for fields seems too stringent. – anon May 10 '17 at 15:04
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    Related: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker May 10 '17 at 15:06
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    @EthanBolker More related (and possible duplicate), obtained from google searching "up to isomorphism." – anon May 10 '17 at 15:06
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    @user1952009: Keep in mind that a lot of people tend to say "field" when they really mean "extension field of $F$ or "field embedded in $E$"... and in those contexts "homomorphism" is meant to be a homomorphism of extensions of $F$ or of embeddings in $E$. –  May 10 '17 at 15:16
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    @user1952009, subobjects are not classified only with their structure, but also how exactly they are embedded in ambient object. Thus, two fields can be isomorphic, yet embedded differently in some other field. Even easier example is $\mathbb Z\cong 2\mathbb Z$, yet they are different subgroups of $\mathbb Z$. – Ennar May 10 '17 at 15:20
  • I know there are other effectively duplicate questions here on Math.SE but I really appreciate the clarity of the answers here and the briefness of the question. "Up to isomorphism" is actually quite a subtle and beautiful use of language. It's deceptively elegant, and being able to explain and concretise the statement is a good gauge of how well you understand isomorphism. It's an important question. – Benjamin R May 11 '17 at 05:30

2 Answers2

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In mathematics it matters less what things are than how they behave.

Things that behave in exactly the same way can be thought as being the same.

Isomorphisms make "behave in exactly the same way" a precise notion.

With the appropriate definition of isomorphism, this reasoning can be extended to other mathematical objects, such as vector spaces, topological spaces, differentiable manifolds, etc.

lhf
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The other answer and the comments have already said much.

Here is an example to think about. Consider the set $Z = \{0,1\}$. Consider the operation of addition modulo $2$. That is $$\begin{align} 0 + 0 &= 0 \\ 0 + 1 & = 1 \\ 1 + 0 &= 1 \\ 1 + 1 &= 0 \end{align} $$ Under this operation $Z$ becomes a group. (You can easily check the axioms.) Now consider another group.

Let $Y = \{1, -1\}$. Then $Y$ and $X$ are different sets. Consider the operation of multiplication. That is $$\begin{align} 1\cdot 1 &= 1 \\ 1\cdot (-1) &= -1 \\ -1\cdot 1 &= -1 \\ (-1)\cdot(-1) &= 1 \end{align} $$ Under this operation $Y$ is a group. It is technically a different group because $X$ and $Y$ are different sets. One has addition as operation and the other multiplication. But, as you can see from the above, the groups behave exactly the same way. There is no essential difference between the two groups. Saying that is kinda vague, so we have the concept of groups being isomorphic. Two isomorphic groups are exactly "essentially the same".

So here it the point: Take any two symbols/elements. Now make the set of these two elements into a group by defining some operation on the set. The fact is that no matter how you do this, you will always get a group that is isomorphic to one of the two above. That way, there is essentially only one group of order $2$ (i.e. a group with two elements).

This is exactly what we mean when we say that there is, up to isomorphism, only one group of order $2$.

Thomas
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