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Let $\phi\colon R \to R'$ be a ring homomorphism. Prove that if $R$ is a field then either $R$ is an isomorphism or $\phi(r) = 0$ for all $r \in R$.

I am stuck on this problem and don't know where to begin. I feel like I'm very weak in writing proofs.

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Hint $\ $ There are only a couple possibilities for the ideal ker $\phi$ given that R is a field. –  Bill Dubuque May 3 '11 at 20:44
    
I'm sorry I don't know what your hint refers to. My professor skips around and sometimes goes back whenever a student reminds him that he hasn't proven something that he thinks is obvious –  Person May 3 '11 at 20:56
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You must be using "isomorphism" to mean "one-to-one homorphism"; otherwise, $\mathbb{R}\hookrightarrow\mathbb{C}$ would be a counterexample (not to mention $\mathbb{R}\hookrightarrow \mathbb{R}\times\mathbb{Z}$). –  Arturo Magidin May 3 '11 at 21:01
    
As stated, this is not true: Consider for instance the inclusion of $\mathbb Q \to \mathbb R$. So, you need some other assumption, like $\phi$ is surjective. Also, I assume you mean $\phi$ is an isomorphism, not $R$. –  lhf May 3 '11 at 21:04
    
@lhf: Some books use "isomorphism" to mean "one-to-one", and "isomorphism onto" to mean bijective morphism. E.g., Herstein's Topics in Algebra. –  Arturo Magidin May 3 '11 at 21:21
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the kernel of $\phi$ (referred to in the comments) is the set $\{r\in R : \phi(r)=0\}$. exercise 1: this is an ideal of $R$ (an additive subgroup $I$ of $R$ with the property that $rI\in I$ for every $r\in R$, need to be a little more specific if $R$ is noncommutative etc.). exercise 2: a field $F$ only has two ideals, $0$ and $F$. so if $R$ is a field then the kernel of $\phi$ is either $0$ (in which case $\phi$ is injective) or all of $R$ (in which case $\phi$ is the zero map)

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+1; I prefer this approach but did not want to assume an exercise. –  Hans Parshall May 3 '11 at 21:19
    
What will these 2 excercises tell me? I already know how to show that the kernel is an ideal in R and that F contains only 2 ideals. I don't get why you said that if R is a field then the kernel of $\phi$ is either 0 or all of R –  Person May 3 '11 at 21:57
    
@Person: The kernel of $\phi$ is an ideal of $R$. If $R$ is a field, then what are the possible ideals of $R$ that $\operatorname{ker}\phi$ could be? –  Michael Chen May 3 '11 at 23:41
    
The ring itself and {0} –  Person May 4 '11 at 6:39
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Here's a proof sketch from first principles. This should work even if your professor "skips around".

THEOREM $\ $ TFAE for a field $\rm\:R\:$ and a ring hom $\rm\ f\:: R\to R'$

$\rm (1)\ \ \ f\:$ is not one-one

$\rm (2) \ \ \ f(r) = 0\ $ for some $\rm\ r\ne 0,\ \ r\in R$

$\rm (3) \ \ \ f(1) = 0$

$\rm (4) \ \ \ f(R) = 0$

Proof $\rm\ (1\Rightarrow 2)\ \ \ a\ne b,\ f(a) = f(b)\ \Rightarrow\ f(a-b) = f(a)- f(b) = 0$

$\rm\ (2\Rightarrow 3)\ \ \ r\ne 0\ \Rightarrow 1/r\in R\ \Rightarrow\ f(1) = f(r\cdot 1/r) = f(r)\ f(1/r) = 0$

$\rm\ (3\Rightarrow 4)\ \ \ f(r) = f(1\cdot r) = f(1)\ f(r) = 0$

$\rm\ (4\Rightarrow 1)\ \ \ R$ a field $\rm\Rightarrow 1\ne 0\:,\:$ so $\rm\ f(1) = f(0) = 0\ \Rightarrow\ f\:$ is not one-one.

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I hope I'm not giving away too much, but I did this problem only recently! Note if $R$ is a field, then it's simple i.e. it's ideals are only the ${0}$ ideal or $R$, Kernel is an ideal also... I hope you get my drift... if you need any further help let me know! (though I doubt you do!)

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Suppose $R$ is a field. If $\phi(r) = 0$ for all $r \in R$, we're done. So suppose this is not the case. If we can now show that $\phi$ is an isomorphism, we'd be done.

Let $r \in R$ with $\phi(r) \neq 0$. Let $a$ be nonzero. Then it has an inverse $a^{-1}$ since $R$ is a field. Further, $$\phi(r) = \phi(aa^{-1}r) = \phi(a)\phi(a^{-1})\phi(r).$$ Can you conclude from this that $\phi$ is injective? That is, can you show that $\phi(a) \neq 0$ for our arbitrary nonzero $a$?

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For injectivity aren't you supposed to start off by saying that suppose a,b belongs to R and phi(a) = phi(b) and then conclude that a = b? To prove that phi is isomorphic I think I would have to show that Kernel of phi is 0. I don't know I'm not completely sure. –  Person May 3 '11 at 21:01
    
@Person: Recall (I hope you've learned this!) that $\phi$ is injective iff the kernel of $\phi$ is exactly 0. I claim you should be able to do that from the above. –  Hans Parshall May 3 '11 at 21:06
    
So is it alright if I say " phi(r) = phi(y) " Since phi(r) = phi(a)phi(a^-1)phi(r), phi(a)phi(a^-1)phi(r) = phi(y). Multiply both sides by phi(y)^-1 and then get phi(r)phi(y^-1) = e. By hypthesis, ry^-1 belongs to the kernel so ry^-1 = e. Multiply both sides by y to get r = y –  Person May 3 '11 at 21:12
    
It's true that you want to show this is injective, and your idea is a standard way to start proving that a function is injective. However, with a homomorphism, it's frequently (much) easier to prove instead that it has a trivial kernel. Any ring homomorphism is injective if and only if it has a trivial kernel. This is not to say that your idea won't work, but I don't think it does as written. –  Hans Parshall May 3 '11 at 21:16
    
In what you've written above, it appears you're assuming $R'$ is a field. How do you know $\phi(y)$ is invertible? –  Hans Parshall May 3 '11 at 21:17
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