$\phi : R \to R’$ is an onto ring homomorphism. $1 \in R$. Then $\phi(1) \in R’$ is the unit element of $R’$.

Question

Let $\phi : R \to R’$ be an onto ring homomorphism from a ring $R$ with unit element to a ring $R’$. Then $\phi(1) \in R’$ is the unit element of $R’$.

I am a little uneasy about not explicitly using the fact that $\phi$ is surjective. Can someone please explain to me at what point do we use this fact in this proof?

My attempt:

Let $a \in R$ then $\phi(1a) = \phi(1) \phi(a)$. Since $1a=a$, then $\phi(1)\phi(a)=\phi(a)$. Similarly $\phi(a)\phi(1)=\phi(a)$. Therefore, $\phi(1)$ is, by definition of a unit element, the unit element of $R’$. $\Box$

Answer 1

It can be shown that netrual elements are unique. Thus, $\varphi(1)$ is the neutral of $R'$ if you can manage to show that for every $x'$ in $R'$ we have $x' = \varphi(1) x' = x' \varphi(1).$ This makes surjectivity obvious, every $x' = \varphi(x)$ for some $x$ and what you wrote applies.

Answer 2

(I'm assuming from the context of the question that you don't require ring homomorphisms to preserve $1$.)

All you've shown so far is that $\phi (1)$ acts as a multiplicative identity for elements of the form $\phi (a)$: that is, elements in the image of $\phi$. Since $\phi$ is surjective, that includes all elements of $R'$ and so $\phi (1)$ is the identity. However, if $\phi$ wasn't surjective, it would be possible for there to be elements not in the image where $\phi(1) x \neq x$.

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As a slightly contrived example, let $R$ be the ring whose elements are $\mathbb{Z}$-linear combinations of the variables $\{X_i | i \in \mathbb{N} \}$, with multiplication defined by $X_i X_j = X_{max(i,j)}$, extended linearly. The $1$ of this ring is $X_0$.

Now define $\phi: R \rightarrow R$ by $\phi(X_i) = X_{i+1}$, extended linearly to make it a ring homomorphism. Looking at the image of the $1$ of the ring $R$ under $\phi$, we see that $\phi(X_0) = X_1$ is a multiplicative identity for every element of the ring in the image since everything in the image doesn't have an $X_0$ term. However, $\phi(X_0)$ is not an identity for the whole ring since $\phi(X_0) X_0 = X_1 X_0 = X_1 \neq X_0$.