# Prove some properties of ring homomorphism

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!

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

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 $$\phi(x) = a$$.

• Follow up question, apologies for tardiness. Is the question then supposed to have been for $n\in\mathbb Z^+$ for (1) as well? Nov 11, 2021 at 1:55
• 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. Nov 11, 2021 at 19:54