Question

Fun Exercise (Abstract Algebra): Let \(F\) be a field, let \(R\) be a ring, and let \(\varphi:F\to R\) be a ring homomorphism. Show that \(\varphi\) is either trivial or injective.

Answer 1

Am I missing something here. Take R be the field F and the identity ring homomorphism. A field is by definition a ring and clearly the identity map is bijective and hence surjective. Is this the trivial map?

Answer 2

We really cannot put restrictions on \(F\) and \(R\) since they are arbitrary by hypothesis. The trivial homomorphism is the "zero homomorphism," i.e., it maps any \(x\in F\) to the zero element \(0_R\in R\).

Answer 3

How am I suppose to interpret the problem? In the case of R=F \(\phi\) can be bijective and hence surjective. This seems to be a counterexample of what we sought to prove.

Answer 4

When I searched for what a trivial homomorphism is, I saw a search engine result in planetmath.org for the exact same thing. I didn't click on the link since that would spoil the fun. Are you missing some sort of premise here?

Answer 5

A counter-example would be a homomorphism that is neither injective nor trivial. The identity homomorphism in your case is injective.

Answer 6

I should have said "zero homomorphism" to remove any kind of ambiguity.

Answer 7

I can see why this is true. Fields are rings with more constraints and if there exists a surjective ring homomorphism then the field is in some sense "larger" than the ring despite fields are more constrained...

Answer 8

Don't expect answers anytime soon. I am just using the textbox as a scratch paper.

Answer 9

kernels are ideals, every field has 2 ideals, the entire thing or trivial. if kernel is trivial then map is 1-1, if kernel is whole thing map sends everything to 0.

Answer 10

So is there a bijection between ideals and kernals? If so then I think your proof correct?

Answer 11

what?

Answer 12

I know nothing about ring theory so what I am saying may not even make sense.

Answer 13

F is a idal and {e} is an ideal. Surely there is no bijection between them for all fields F ({e} is not a field). But my proof holds fine....

Answer 14

@zzr0ck3r is right. What he is saying is that a field can only have two ideals: the trivial ring and itself. This is because every element of a field other than \(0\) is a unit.

Answer 15

this would be really hard if you have no ring theory :(

Answer 16

Not really.

Answer 17

well you need to understand what a ring is... lol

Answer 18

and what a ring homomorphism is. which is VERY hard to know if you dont know ring theory. By definition

Answer 19

\[
x\in F\\
xx^{-1}=1_F\\
\varphi(xx^{-1})=\varphi(x)\varphi(x^{-1})=\varphi(1_F)=1_R
\]
\(\varphi\) must be injective because the invertible elements of \(R\) are a subset of \(R\)? I couldn't point out what bad things will happen if \(\varphi\) is surjective though.

Answer 20

see^

Answer 21

you just said ident maps to ident

Answer 22

which is always true for any homomorphism in any monoid

Answer 23

actually I assumed (stupidely) that you meant 0. You are assuming there is a 1 in R

Answer 24

also if there is a 1, it is not necessary that mult ident maps to mult idents.

Answer 25

I thought the definition of ring homomorphism requires multiplicative identity maps to multiplicative identity...

Answer 26

only additive idents have that property

Answer 27

nope :(

Answer 28

subsets of rings can have different mult ident than there parent

Answer 29

subrings*

Answer 30

I was reading of Wikipedia and that's what Wikipedia said lol.

More explicitly, if R and S are rings, then a ring homomorphism is a function f : R → S such that[1][2][3][4][5][6]

f(a + b) = f(a) + f(b) for all a and b in R
f(ab) = f(a) f(b) for all a and b in R
f(1R) = 1S.

Answer 31

consider any homomorphism R to {0}

Answer 32

rings do not always have unity, so how can we say every ring homo sends unity to unity?

maybe they mean additive ident

Answer 33

keep reading...

Answer 34

Depending on definition of rings. Some say rings have multiplicative identity and some say they don't. Those who say rings must have multiplicative identity call rings without multiplicative identity rng.

Answer 35

Silly.

Answer 36

plenty of nice rings without 1

Answer 37

It is just names lol. Names are sometimes silly.

Answer 38

Yeah, this is like the natural number thing....

Answer 39

Apparently statisticians think that heteroscedasticity is a better term than unequal variance. Go figure.

Answer 40

hehe

Answer 41

I love graph theory, they just dont use stupid pellet and we all agree.

Answer 42

like there is no empty graph. We mean vertices with no edges when we say empty graph, not {}

Answer 43

I think that is called the singleton graph or something like that...

Answer 44

that is v(X) = {a}

Answer 45

empty graph is v(X) = {a,b,c,d,s,....} e(X) = {}

Answer 46

empty graph on n vertices....

Answer 47

in Godsil anyway ...

Answer 48

Buttttt empty graph is not empty!!!!! Lol.

I wonder what brilliant answer is across going to put out...

Answer 49

Graph theory is one of the most badass subjects I've looked into. And thanks for bringing up that great point, @zzr0ck3r: I had forgotten that if \(R\) and \(S\) are rings with multiplicative identities and \(\varphi\) is a homomorphism between them, then it need not be the case that \(1_R\mapsto1_S\) (only if \(S\) is an integral domain).

Answer 50

It actually doesn't get better than saying that \(\ker\varphi\) is either trivial or \(F\), immediately proving the conclusion.

Answer 51

That lemma proves allot of things :)

Answer 52

Suppose \(F\) is field with ideal \(I\) and suppose \(a\in I\) s.t. \(a\ne0\) then \(a*a^{-1}=1\in I\) (by def of ideal) so \(b*1=b\in I\) for all \(b\) in F, so \(I=F\)

Answer 53

I'm running out of nice questions that are not impregnable, xD.

Answer 54

I t.a.'d for this class at my university. The professor calls the Ideal property "black hole property" . It really helps people remember :)