Yep! Wedderburn's theorem on Division rings. It's a wee bit more complicated. A more elementary generalization would be that if an integral domain R contains a field K, and is a finite dimensional vector space over it, then it too is a field (this includes MathWonk's examples as well). It's...
Outlined only partially answers the question. Of course the rings Z/pZ are finite fields, but they are not the only fields. Perhaps the easiest (and perhaps most intuitive way) to explain this is to use elementary concepts from linear algebra.
Let K be any field and consider the map from...
Perhaps a little more intriguing is that upon expanding sin(x) and cos(x) into their respective Taylor series, we can regard sin(x) and cos(x) as elements of the formal power series ring over the rationals, which is also a vector space (over the rationals)!
I contemplated this, too and it would certainly make since in light of the fact that algebraic integer rings are Dedekind domains, but I feel it would have been specified and I looked in the textbook and I don't even think the concept of a Dedekind domain had been introduced.
Hm...I looked at the book and I didn't think that was the exercise verbatim...how odd, you're right, that is definitely not correct. I think what they mean to say is show that there is an isomorphism between the fields Rp/pRp and quotient field of R/P (this is certainly possible since R/P is an...
Well, in all honesty, it's a bit silly to talk about "an ideal of an ideal" since we define (particularly in commutative algebra) ideals to be subsets of rings and proper ideals are not rings (ie. they are not subrings).
Okay, so I'm trying to finish of a problem on integral closure and I am rather unsure if the following fact is true:
If L embeds into an algebraically closed field K and F is an algebraic extension of L, then it is possible to extend the embedding of L to F into K.
Now the case where F...
Why not check out past REUs and see what they've done? I know there's been an REU at Cornell that has done stuff along the lines of what you want to do. Why not email someone well-known in the field and ask them what's of interest right now in those fields? There's a geometric group theorist...
It's absolutely necessary. Consider the permutation group on three letters (ie. S3), then this is a group of order 6 = 2 *3 and is clearly not cyclic (and definitely not abelian either). However we do have this result:
If G is a group order pq, pq distinct primes say P < q and p does not...
What more is true, is that given any square free intgers m and n, Q(sqrt(m)) and Q(sqrt(n)) are nonisomorphic. Intution serves right when you say that it "does not respect the product" but being more rigorous, show that no ismorphism can possibly exist between the two fields by first showing...