Implicit isomorphism involved in extension/sub fields/structures?

  • #1
Implicit isomorphism involved in extension/sub fields/structures???

This has been bugging me for a while. I'm pretty sure I'm correct but I'd just like to verify to put my mind at ease. I'd like to know if there is an implicit isomorphism involved when we say, for example, F is a substructure of E, or for something more specific, F is a subfield of E or E is an extension field of of F.

For an obvious example I can say that the complex numbers, C, are an extension field of the real numbers R. Now what I'm thinking when I see that statement is that there is an isomorphism, f, between a subfield of C and R, given by f(a + 0i) = a. That is, R is isomorphic to a subfield of C. If you said R IS a subfield of C, then the difference would be that the elements of R are elements of C.

In every place I look it's simply stated that real numbers are complex numbers and that's that. However, when you get into more abstract stuff like using factor groups to create extension fields then it's not clear at all that the elements of a field ARE elements IN the extension field. For example regarding Kronecker's Theorem, it is said that given a field F then, F[x]/<p(x)> is an extension field of F, where F[x]/<p(x)> is the factor ring of F[x] by <p(x)>, F[x] are the set of polynomials with coefficients from the field F and <p(x)> = {g(x)p(x) : g(x) element of F[x]} is the maximal ideal generated by p(x), an element of F[x]. Thus F[x]/<p(x)> = { f(x) + <p(x)> : f(x) element of F[x] }. Now with all of that said it is clear that a single element of F is not the same thing as an entire set of polynomials. Since the elements of F[x]/<p(x)> are entire sets, it is clear that something does not add up here. Thus it seems like an isomorphism is necessary for the statement to mean anything at all... Say m(a) = a + <p(x)> for all elements a of F. However no source ever explicitly states any sort of isomorphism is going on and refers to F as a SUBfield of F[x]/<p(x)>, when the elements of F by themselves are not entire sets of polynomials. It's like saying that 2 = { f(x) + p(x)g(x) : f(x), g(x) elements of R[x] the set of polynomials with real coefficients }, it does not make any sense, yet I keep seeing it.

Why is there no source that will simply explicitly state that there is an isomorphism going on? Is there a reason to try to ninja stuff to purposely make it more confusing?

Any clarification is extremely appreciated. Thank you for your time and have a good day.
 

Answers and Replies

  • #2
606
1


This has been bugging me for a while. I'm pretty sure I'm correct but I'd just like to verify to put my mind at ease. I'd like to know if there is an implicit isomorphism involved when we say, for example, F is a substructure of E, or for something more specific, F is a subfield of E or E is an extension field of of F.

For an obvious example I can say that the complex numbers, C, are an extension field of the real numbers R. Now what I'm thinking when I see that statement is that there is an isomorphism, f, between a subfield of C and R, given by f(a + 0i) = a. That is, R is isomorphic to a subfield of C. If you said R IS a subfield of C, then the difference would be that the elements of R are elements of C.


*** Correct and nicely put ***


In every place I look it's simply stated that real numbers are complex numbers and that's that. However, when you get into more abstract stuff like using factor groups to create extension fields then it's not clear at all that the elements of a field ARE elements IN the extension field. For example regarding Kronecker's Theorem, it is said that given a field F then, F[x]/<p(x)> is an extension field of F, where F[x]/<p(x)> is the factor ring of F[x] by <p(x)>, F[x] are the set of polynomials with coefficients from the field F and <p(x)> = {g(x)p(x) : g(x) element of F[x]} is the maximal ideal generated by p(x), an element of F[x].


**** Be careful in all the above: there's no such thing as "<p(x)> is the maximal ideal generated by p(x)", but ONLY "<p(x)> is the (principal) ideal generated by p(x)". Its being maximal follows iff p(x) is an irreducible polynomial ****


Thus F[x]/<p(x)> = { f(x) + <p(x)> : f(x) element of F[x] }. Now with all of that said it is clear that a single element of F is not the same thing as an entire set of polynomials. Since the elements of F[x]/<p(x)> are entire sets,


**** This might be accurate but pretty misleading if it isn't completely understood. It'd be better to say, imo, that the quotient ring is the set of equivalence classes determined by the ideal <p(x)> WHEN we choose one single representative of each equiv. class to work with under the standard definition of operations in the quotient.
Of course, basically you're right but it can make things pretty confusing, I think, to actually consider those elements merely "as sets". ****



it is clear that something does not add up here. Thus it seems like an isomorphism is necessary for the statement to mean anything at all... Say m(a) = a + <p(x)> for all elements a of F. However no source ever explicitly states any sort of isomorphism is going on and refers to F as a SUBfield of F[x]/<p(x)>


**** You're right, but this is due, I believe, to the fact that when one reaches this level one already understands how to embed algebraic structures in other ones, and in this case the embedding is a rather trivial one: map any [itex]a\in F[/itex] to [itex]a+<p(x)>[/itex] , which is correct since any element in the field can be seen as a a polynomial of degree zero (or -oo or undefinded, for the zero element) and it's easily seen to be a monomorphism.****


, when the elements of F by themselves are not entire sets of polynomials. It's like saying that 2 = { f(x) + p(x)g(x) : f(x), g(x) elements of R[x] the set of polynomials with real coefficients }, it does not make any sense, yet I keep seeing it.


**** Then it should be almost obvious that it is you who isn't completely understanding something and not that "it makes no sense", wouldn't you say?
Remember that as elements in a quotient ring R/I, we have that [itex]a+I=b+I\Longleftrightarrow a-b\in I[/itex] , and this is precisely what makes the monomorphism mentioned above to work so nicely. You must be sue you understand this in general rings (or, at least, commutative unitary rings) before you attack the matter of polynomial rings and stuff.

DonAntonio ****


Why is there no source that will simply explicitly state that there is an isomorphism going on? Is there a reason to try to ninja stuff to purposely make it more confusing?

Any clarification is extremely appreciated. Thank you for your time and have a good day.
....
 
  • #3
795
7


This has been bugging me for a while. I'm pretty sure I'm correct but I'd just like to verify to put my mind at ease. I'd like to know if there is an implicit isomorphism involved when we say, for example, F is a substructure of E, or for something more specific, F is a subfield of E or E is an extension field of of F.

For an obvious example I can say that the complex numbers, C, are an extension field of the real numbers R. Now what I'm thinking when I see that statement is that there is an isomorphism, f, between a subfield of C and R, given by f(a + 0i) = a. That is, R is isomorphic to a subfield of C. If you said R IS a subfield of C, then the difference would be that the elements of R are elements of C.
When you're doing normal math you say that R is a subset of C. When you're explicitly constructing the complex numbers as a set of pairs of reals, for example, you then note that C has a subfield isomorphic to R, and you agree to simply say from now on that R is a subset of C. Because as sets, the reals as defined in real analysis are a different set than the copy of the reals inside C.

This is something that nobody cares about except mathematicians, and even then only when they're learning or teaching the construction of C.

By the way we can think about this with respect to the various ways we can construct C. We can call C the set of ordered pairs of reals with addition and multiplication defined appropriately; or we can call C the ring quotient of R[X] by the ideal generated by x^2+1; then we have two isomorphic models of the complex numbers, but the models are different sets.

If you ask, well, what are the complex numbers really, then perhaps they are sort of the Platonic ideal of all the different ways we could use set theory to build the complex numbers. That's a question for the philosophers.

(edit) As far as why nobody ever explicitly mentions the isomorphic embeddings that are implicitly all over the place in math, I did have one grad level abstract algebra professor who hammered this point home all the time. I think that's how I got it clear in my mind.
 
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