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So basically this question is from Serge Lang "Undergraduate Algebra" Third Edition, III, $\S$3, exercise 2.

Show that a ring-homomorphism of a field $K$ into a ring $R \neq {0}$ is an isomorphism of $K$ onto its image

My attempt:

let $f : K \rightarrow R$ is a ring-homomorphism.

Need to show that $K \cong f(K)$.

It is enough to show that $f$ is bijective with respect to sets $K$ and $f(K)$.

(injectivity) $$ \begin{align*} x &\neq y \\ x - y &\neq 0 \\ f(x-y) &\neq 0 \\ f(x) - f(y) &\neq 0 && (*) \\ f(x) &\neq f(y) \end{align*} $$ (surjectivity)

From the definition $f(K)$ is an image of $K$ under $f$, so it must be surjective.

From this follows that $f: K \rightarrow f(K)$ is a bijection, and thus two sets are isomorphic.

(note) In $(*)$ we assume that $R \neq 0$, otherwise injectivity fails

Question:

I am not sure whether this solution is correct or not, because I did not use the fact that $K$ is a field. Could you please verify it?

Nuraly
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  • The fact that $K$ is a field is used in $x-y\neq0 !\implies! f(x-y)\neq0$. That's because the kernel of $f$ is an ideal of $K$, and the only ideals of a field are the trivial ones. – azif00 Dec 21 '23 at 01:56
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    You didn't explain why $x-y\ne 0$ implies $f(x-y)\ne 0$, which is the only nontrivial step. – Mark Dec 21 '23 at 01:56
  • @azif00 This is an answer to the question, not just a comment. Comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore, please post your answer as an answer. – Martin Brandenburg Dec 21 '23 at 02:31

2 Answers2

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Proof for injectivity:

You have showed that $\ker f=\{0\}$. We will continue from this observation.

Select arbitrary $x,y\in K$ s.t. $x\ne y$. Since $x-y\ne 0$, we yield $x-y\notin\ker f$. By definition of kernel, \begin{equation} a\in \ker f\iff f(a)=0\cdots(\star). \end{equation} Since ($A\iff B$) $\implies$ ($\neg A\iff\neg B$), we infer from ($\star$) that $$ a\notin \ker f\iff f(a)\ne 0 \cdots(\star\star).$$

Since $x-y\notin \ker f$, from ($\star\star$) we infer $$ f(x-y)\ne 0. $$ Therefore $f(x)-f(y)=f(x-y)$ is non-zero, i.e. $f(x)\ne f(y)$.

Asigan
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Thanks for azif00 and Mark for useful comments. Now I think my solution is complete:

(injectivity) $$ \begin{align*} x &\neq y \\ x - y &\neq 0 \\ f(x-y) &\neq f(0) && (*)\\ f(x-y) &\neq 0 \\ f(x) - f(y) &\neq 0 && (**) \\ f(x) &\neq f(y) \end{align*} $$ (surjectivity)

From the definition $f(K)$ is an image of $K$ under $f$, so it must be surjective.

From this follows that $f: K \rightarrow f(K)$ is a bijection, and thus two sets are isomorphic.

(note)

In $(*)$ we know that $\text{ker} f$ is either 0 or unit ideal. However it cannot be unit ideal, since this would imply that $R={0}$, which is not the case by definition. Hence, $f(0)=0$

In $(**)$ we assume that $R \neq 0$, otherwise injectivity fails

UPD (thanks to Allen Bell for pointing out where I was wrong):

Actually the correct way to explain (*) $$ f(x-y) \neq f(0) \Rightarrow f(x-y) \neq 0 $$ Would be this:

Since K is a field, we have only two ideals: $0$ and $K$ (see this question). $\text{ker} f$ is an ideal, since $f$ is ring-homomorphism (see this). So $\text{ker} f$ is either $\{0\}$ or $K$ itself. $\text{ker} f \neq K$, otherwise $f$ would map every element in K to 0. Assume $\text{ker} f = K$, then $f(1_K)=0$, which is a contradiction, since $f$ is a ring-homomorphism and identity of $K$ should be mapped into identity of $R$.

So $\text{ker} f = \{0\}$, and $f(0) = 0$, since $0 \in \text{ker} f$.

Nuraly
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  • (1) What is your definition of ring homomorphism? Is $f(x)=0$ $\forall x\in K$ a homomorphism in your definition? If it is, then it is not injective. (2) I guess by "Hence, $f(0)=0$" you actually want to say "$f(x)=0$ admits the only solution $x=0$". These two are totally different statements. Please think about the logical difference between them. – Asigan Dec 21 '23 at 16:00
  • @Asigan my definition of ring homomorphism (assuming that rings are unital) is a function $f: R \rightarrow R'$ such that for $x,y \in R$, $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ and $f(e)=e'$, where $e'$ is a multiplicative identity of $R'$. – Nuraly Dec 21 '23 at 16:47
  • @Asigan thank you for pointing out logical flaws. I am not considering case when $f(x) = 0 \forall x \in K$, since this would imply that $R={0}$, which is not true, by assumptions – Nuraly Dec 21 '23 at 16:50
  • Would it be correct to say that $f(0)=0$, since $0$ is in the kernel of our homomorphism, and in fact, it is the only element in the kernel. – Nuraly Dec 21 '23 at 16:52
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    Your justification for (*) is not quite complete. If $\ker f=K$, this only implies that $f(K)=0$ not that $R=0$. Since $f(1_K)=1_R\ne 0_R$ (this is where you use $R\ne 0$), we cannot have $f(K)=0$. – Allen Bell Dec 21 '23 at 18:04
  • @AllenBell thank you! I updated the solution – Nuraly Dec 21 '23 at 18:44
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    It is perhaps more clear to write the last sentence as "So $\ker f={0}$. Therefore $f(x-y)\ne 0$ since $x-y\notin \ker f$." Please carefully think about the differences between this sentence and the last sentence of yours. – Asigan Dec 21 '23 at 23:58
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    I am sorry to say your proof still seems incorrect, The error is not at the implication "$f(x-y)\ne f(0)\implies f(x-y)\ne 0$". This implication is correct since $f(0)=0$ is true for any ring homomorphism. It is the previous implication that is wrong, i.e. "$x-y\ne 0\implies f(x-y)\ne f(0)$ (wrong)". Recall that you intend to prove injectivity. How does this implication hold without assuming $f$ is injective? – Asigan Dec 22 '23 at 04:10
  • @Asigan yeah I guess that doesn't make sense, otherwise the proof becomes circular. Thank you – Nuraly Dec 22 '23 at 10:39
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    @Nuraly It is good that you have overcome the logical issue. I have written an answer, which might be a reference of the correct argument. – Asigan Dec 22 '23 at 12:47