L be algebric  extension of K.Show any subring R of L  s.t  K subset R subset L  is  a  field?
446810
Let r be an element if R, and hence in L.  Since L is an algebraic extension of K, all elements of L, there is an irreducible polynomial:\n\n    P(x) = a0 + a1x + a2x^2 + ... anx^n\n\nwith the ai's in K such P(r) = 0.\n\nSubtracting a0 from both sides, we see:\n\n    -a0 = r ( a1 + a2r + a3r^2 .. + anr^(n-1))\n\nDividing both sides by -a0 we get (a0 is not zero because P is irreducible):\n\n      1 = r (-a1/a0 - a2/a0 r - a3/a0 r^2 ...)\n\nThe right side is in R since the the coefficients -ai/a0 are in K and r is in r.  So we have an inverse of r in R.
460866
Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
204094
ok, i think i have an idea for this one. \nSketch of a proof:\nSuppose that a is an element of order k. Also suppose that f is an isomorphism. Then we know that f(a*a*a...k times)=f(a)*f(a)*f(a)*...k times (homomorphism definition). We know that f(a)^k is the identity because isomorphisms take the identity to the identity. Now see that k must be the first number with the quality f(a)^k=e. This comes as a result of the isomorphism having a kernel with one element. If we suppose for the sake of contradiction that f(a)^m=e with m<k, then we would have f(a*a*a*a...m times)=e, so that means that f(a^m)=e at the same time as f(a^k)=e with a^m not equal a^k, this is a contradiction.\n\nQED...ish
764991
Let [a, b, c] denote the vector <4a+3b, 0, a+b+c, c-2a>. Vectors of this form are a subset of R^4. In order to show that they form a subspace, we must show that they are closed under linear combinations. It suffices to show that they are closed under vector addition and scalar multiplication. [a, b, c] + [a', b', c'] = <4a+3b, 0, a+b+c, c-2a> + <4a'+3b', 0, a'+b'+c', c'-2a'> = <4(a+a') + 3(b+b'), 0, (a+a') + (b+b') + (c+c'), (c+c') - 2(a+a')> = [a+a', b+b', c+c'], so the set is closed under addition. Similarly, k[a, b, c] = k<4a+3b, 0, a+b+c, c-2a> = <4ka+3kb, 0, ka+kb+kc, kc-2ka> = [ka, kb, kc], so the set is closed under scalar multiplication.\n\nThe spanning set is {[1, 0, 0], [0, 1, 0], [0, 0, 1]}. Writing this out in full: {<4, 0,1, -2>, <3, 0, 1, 0>, <0, 0, 1, 1>}
1159041
If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.
79530
http://mathforum.org/library/drmath/sets/high_calculus.html
830686
A subset of a set contains any or all of the elements of the set.\n\nA proper subset contains any of the elements of the set but not all of them.
394530
1) Let G be a group and let H be a subgroup of G with index 2 in G.  The two right cosets of H in G must be H and G - H = { a in G | a is not in H }.  The two left cosets of H in G must be H and ... G - H.  Once you notice that it is easy to show that H is a normal subgroup of G.\n\n2) Let G be a cyclic group with generator a, so that G = < a > = { a^n | n in Z }.  Then\n\na^n a^m = a^(n+m) = a^(m+n) = a^m a^n
1093109
Yes in both cases! (although for vector spaces, it depends whether you're allowed to pick which field you're working over.  It's not true for vector spaces over the real numbers, for example)\n\nHere's a hint: think about the smallest group there is that isn't cyclic.\n\nOnce you figure it out for groups, you'll realize the same answer works as a vector space over an appropriate (small!) field.
457089
G=uys\nF=***\nE=verthing
761029
So you just need to show that for any x,y in A, f(x)=f(y) implies that x=y.  So let x,y be elements of A so that f(x)=f(y).  Since f(x)=f(y), and g is a function, then g(f(x))=g(f(y)).  Since gof is injective by assumption, this implies that x=y.  So we start with f(x)=f(y), and this implies x=y.  Therefore f is injective.  Cheers!
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
757825
There has been no development in this field, but you can check out the following result pertaining to set theory and group theory.\n\nThe power set P(x) of a set X, is a not a group under (U) union operation.\n\nProof : \nTo prove this we try to prove three properties \n1. Associativity \nFor A, B, C are in P(x) then \n(AUB)UC = A U (BUC)  [This is more or less obvious or simple to prove]\n\n2. Identity element\nThere is an an element(e) in P(x) such that for all A in P(x)\ne U A = A U e = A\nAnd, this element for power set is 'empty set' . [Sorry, but I dont have 'phi' symbol here.]\n\n3. Inverse element\nFor each A  in P(x) there exist another element B in P(x) such that\n(AUB) = (BUA) = e(empty set)\n\nThis property is not possible because you can not produce an empty set with union operation with any other set(irrespective wether its empty or not)\n\nSo, its not a group under Union.\n\n\nPlease noe that : The power set P(x) of a set X, is a not a group under intersection operation.
738536
Take any non-Noetherian ring R (with 1) and consider a non-finitely generated ideal I of R. R is a finitely generated R-module (generated by 1), but I is not a finitely generated R-module. For a concrete example, take R=Z[x1,x2,x3,x4,...] where Z is the ring of integers and x1,x2,x3,... are independent transcendental elements (i.e. take the polynomial ring over the integers in infinitely many variables)  and I=(x1,x2,x3,...).
773496
I suppose by Z you mean the set of integers.\n the torsion subgroup T of the group Z is the subgroup of Z consisting of all elements that have finite order. That is, the subgroup of element a such that a^n = 1 for some n.\n\nTherefore T = {-1, +1}
818651
(a) Suppose f is NOT injective, that is, f( x') = f(x'') for some x' and x'' in X. Then also g(f( x')) = g(f( x'')), which means that g o f is NOT injective. (This is proof by transposition: instead of showing directly that A implies B, we show that Not B implies Not A.)\n\n(b) Take any t in T. Then t = g(f(x)) for some x (since g o f is surjective). This means that t = g(y) for some y, namely, for y = f(x). So g is surjective.\n\n(c) By Part (a), f is injective. By Part (b), applied to f o g in place of g o f, f is surjective. Beig both injective and surjective, f is bijective.
682200
go online, click on search and type in MATH 012 : Intermediate algerbra course. something should pop up. and then write down the evidence