Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\phi\colon R \to R'$ be a ring homomorphism. Prove that if $R$ is a field then either $\phi$ 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.

share|cite|improve this question
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
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

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)

share|cite|improve this answer
+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

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.

share|cite|improve this answer

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!)

share|cite|improve this answer

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 $a$?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.