Easy related theorem: given any function f: X -> Y, we can uniquely factor f as a composite X -> I -> Y where the map X -> I is onto, and I -> Y is an inclusion. The subset I is exactly the image of f.
That separates f into a part involving X and (part of) Y, and a part wholly about Y. Can we have a part wholly about X?
Define a quotient p : X -> Q to be a function that's 1:1, where each p(x) is a subset of X and contains x, and if y is in p(x) then p(y) = p(x). So the collection Q = {p(x)} is a bunch of subsets exactly covering X. You can look up a version of this definition here. They are hard to count (check out the asymptotics!).
Now, given any function f:X -> Y, we can factor it uniquely as X ->> Q -> I -> Y, a quotient followed by a perfect correspondence Q -> I followed by the inclusion of I as a subset of Y.
What do quotients of vector spaces look like? Now p: V -> Q, and we could ask that Q be a vector space, and p be linear. Then (theorem) the elements of Q are the translates of ker(p).
Definition: if W is a subspace of V, let V/W be the set of all translates of W in V, and p : V -> V/W be the quotient map taking v to v+W := {v+w : w in W}, an element of V/W.
It's not hard to prove that dim(V/W) = dim V - dim W, if dim V is finite.
A ring is a set with a +,-,x,0,1 satisfying some fairly obvious conditions, like 0+r = r = 1r, and multiplication distributes over addition. Examples: Z, R, Z[x], {evens,odds}, Z/nZ. We only care about "commutative" rings, where multiplication should be commutative. (So not, e.g., NxN matrices which is a perfectly good noncommutative ring. Nor the quaternions.)
If R is a ring, and p : R -> Q is a quotient map, we could ask that Q be a ring and p take +,-,1,times on R to the corresponding operation on Q, i.e. p(rs) = p(r) p(s). Theorem: the elements of Q must be the R-translates of an ideal I in R.
(Next time: given an ideal, we can form the quotient R/I, and we should!)
That separates f into a part involving X and (part of) Y, and a part wholly about Y. Can we have a part wholly about X?
Define a quotient p : X -> Q to be a function that's 1:1, where each p(x) is a subset of X and contains x, and if y is in p(x) then p(y) = p(x). So the collection Q = {p(x)} is a bunch of subsets exactly covering X. You can look up a version of this definition here. They are hard to count (check out the asymptotics!).
Now, given any function f:X -> Y, we can factor it uniquely as X ->> Q -> I -> Y, a quotient followed by a perfect correspondence Q -> I followed by the inclusion of I as a subset of Y.
What do quotients of vector spaces look like? Now p: V -> Q, and we could ask that Q be a vector space, and p be linear. Then (theorem) the elements of Q are the translates of ker(p).
Definition: if W is a subspace of V, let V/W be the set of all translates of W in V, and p : V -> V/W be the quotient map taking v to v+W := {v+w : w in W}, an element of V/W.
It's not hard to prove that dim(V/W) = dim V - dim W, if dim V is finite.
A ring is a set with a +,-,x,0,1 satisfying some fairly obvious conditions, like 0+r = r = 1r, and multiplication distributes over addition. Examples: Z, R, Z[x], {evens,odds}, Z/nZ. We only care about "commutative" rings, where multiplication should be commutative. (So not, e.g., NxN matrices which is a perfectly good noncommutative ring. Nor the quaternions.)
If R is a ring, and p : R -> Q is a quotient map, we could ask that Q be a ring and p take +,-,1,times on R to the corresponding operation on Q, i.e. p(rs) = p(r) p(s). Theorem: the elements of Q must be the R-translates of an ideal I in R.
(Next time: given an ideal, we can form the quotient R/I, and we should!)
