Let $R,S$ be rings and $\varphi : R\to S$ be a ring homomorphism. Verify that

  1. $\varphi(na) = n\varphi(a)$ for all $n\in\mathbb Z$ and $a\in R$.
  2. $\varphi(a^n) = (\varphi(a))^n$ for all $n\in\mathbb Z^+$ and all $a\in R$.
  3. If $A$ is a subring of $R$, then $\varphi(A) = \{\varphi(a):a\in A\}$ is a subring of $S$.

For (1) I have the following: $$ \begin{split} \varphi(na) &= \varphi((n-1)a+a) \\ &= \varphi((n-1)a)+\varphi(a) \\ &= (n-1)\varphi(a)+\varphi(a)\\ &= n\varphi(a). \end{split} $$

For (2): For $n=1$, we have $\varphi(a) = \varphi(a)$. Suppose for some $n\in\mathbb Z^+$ that $\varphi(a^n)=(\varphi(a))^n$. Observe: $$ \begin{split} \varphi(a^n) & =(\varphi(a))^n \\ \varphi(a)\cdot \varphi(a^n) & =\varphi(a)\cdot (\varphi(a))^n \\ \varphi(a \cdot a^n) & =(\varphi(a))^{n+1} \\ \varphi( a^{n+1}) & =(\varphi(a))^{n+1} \\ \end{split} $$

For (3): I verify for $a,b\in\varphi(A)$ then $a-b\in\varphi(A)$ and $a\cdot b\in \varphi(A)$ (for brevity).

  • Well, $0\in A \implies 0 \in \varphi(A) $
  • For $a,b \in \varphi(A)$ we have $\varphi(a) - \varphi(b) = \varphi(a-b)$, thus, verified, because $a-b\in A$.
  • For $a,b \in \varphi(A)$, we have $\varphi(a) \cdot \varphi(b) = \varphi (a\cdot b)$, thus, verified, because $a\cdot b \in A$.

Did I do these right? Feedback appreciated!

  • 3
    $\begingroup$ Instead of writting "For $n=1$, we have $\varphi(a)=\varphi(a)$" I would write "For $n=1$, we have $\varphi(a^n) = \varphi(a) = (\varphi(a))^n$". $\endgroup$
    – azif00
    Nov 10, 2021 at 17:23

2 Answers 2


For the first property, use induction to show that $\phi(na) = n\phi(a)$ for all $a$ and $n\geq 0$. This is actually what you have done.

For negative $n$ one must be careful. For $n>0$ put $(-n)a = -(na)$, which is the additive inverse of $na$. This part is then true simply by definition.


I agree with Wuestenfux's answer, and have some feedback to give. There is some error in your answer to part $(1)$, which is similar to one in part $(2)$ as well. How do you know that $\phi((n-1)a)=(n-1)\phi(a)$? For $(2)$ you suppose that there exists some $n\in\mathbb{Z}^+$ satisfies $\phi(a^n) = (\phi(a))^n$, but it could be the case that $\{n\in\mathbb{Z}^+|\phi(a^n) = (\phi(a))^n\}$ is empty. For both $(1)$ and $(2)$, one can use induction after showing the properties hold with base case $n=2$.

For problem $(3)$, remember that if $b = \phi(a)$ for some $a\in A$ then $b\in \phi(A)$. It is only once we write an element $b$ in the form $\phi(a)$ that the containment $b\in \phi(A)$ becomes clear. For example, given $0\in A$, we know by properties of ring homomorphisms that $$\phi(0) =0.$$ Hence $0 = \phi(0) \in \phi(A)$.

"For $a,b\in \phi(A)$, we have $\phi(a) - \phi(b) = \phi(a-b)$." This does not show that $a-b \in \phi(A)$, because $a \neq \phi(a)$ and $b\neq \phi(a)$. Again, to say that $a\in \phi(A)$ means there exists some $x\in A$ for which $\phi(x) = a$.

  • $\begingroup$ Follow up question, apologies for tardiness. Is the question then supposed to have been for $n\in\mathbb Z^+$ for (1) as well? $\endgroup$
    – Seong
    Nov 11, 2021 at 1:55
  • $\begingroup$ In (1), you can show that the result holds for all $n\in\mathbb{Z}$. A method for doing so is showing that (1) holds for $n\in \mathbb{Z}^+$, then using this to prove that it holds for all $\mathbb{Z}$. See Wuestenfux's answer for more detail. $\endgroup$
    – Matt E.
    Nov 11, 2021 at 19:54

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