Isomorphism of Rings True or False


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*Let $f:R\to S$ be a homomorphism of rings.  Then $f(1_R) = 1_S$

*Let $f:R\to S$ be a homomorphism of rings.  Then $f(0_R) = 0_S$

*Suppose $u$ is a unit in a ring $R.$  The $u$ cannot be a zero divisor. 

*Let $R$ be a ring.  Suppose $z$ is in the ring and is not a zero divisor. Then $z$ is a unit. 

*There is a bijection from $Z(6)$ onto $Z(2) \times Z(3)$.


If someone could help me with formatting a little I would appreciate it.  We went over isomorphisms in class.  Still not completely understanding of the concept yet.
My answers: F, T, T, F, T
 A: For (1), some books require a ring homomorphism to map the multiplicative identity to the multiplicative identity, so if that's the case for you, it's trivially true. If the definition doesn't require it, then the statement is false; consider $\varphi : R \to R \times R$ defined by $r \mapsto (r, 0)$.  In this case $\varphi(1_R) = (1, 0) \neq (1, 1) = 1_{R \times R}$. Moreover, if $\varphi(1_R) \neq 1_S$, then $\varphi(1_R)$ is a zero divisor.
(2) is true as $f(0_R) = f(0_R + 0_R) = f(0_R) + f(0_R)$, subtracting $f(0_R)$ from both sides yields $f(0_R) = 0_S$ as desired.
(3) is true: Suppose $u$ is a unit and suppose, by way of contradiction that $u$ is a zero divisor, then there exists $0 \neq v$ such that $uv = 0$, but notice that $u^{-1}(uv) = (u^{-1}u)v = v = 0$, contradicting the fact that $0 \neq v$. Conclude that $u$ is not a zero divisor.
(4) is false. Consider $\mathbb Z$ as the comments pointed out repeatedly. The only units are $\pm 1$ and the only zero divisor is $0$ (it has no nonzero zero divisors as it is an integral domain). For instance $2$ is not a unit and it is not a zero divisor.
(5) is true. Define $\varphi : \mathbb Z_6 \to \mathbb Z_2 \times \mathbb Z_3$ by $[x]_6 \mapsto \big([x]_2, [x]_3\big)$. The first question we need to ask is: is this map well-defined? Suppose $[x]_6 = [y]_6$, then $x \equiv y \pmod 6$ which means $x - y = 6s$ for some $s \in \mathbb Z$. In particular, notice that $x - y = 2(3 \cdot s)$ which means $x \equiv y \pmod 2$ and $x - y = 3(2 \cdot s)$ which means $x \equiv y \pmod 3$. Hence we have $$\varphi\big([x]_6\big) = \big([x]_2, [x]_3\big) = \big([y]_2, [y]_3\big) = \varphi\big([y]_6\big)$$ and we can conclude that $\varphi$ is well-defined.
That it's a ring homomorphism follows from the additive and multiplicative properties of equivalence classes: $[x + y] = [x] + [y]$, $[xy] = [x][y]$.
Is $\varphi$ injective? Suppose $\varphi\big([x]_6\big) = \varphi\big([y]_6\big)$, then $x \equiv y \pmod 2$ and $x \equiv y \pmod 3$. This means that $x - y = 2s = 3t$ where $s, t \in \mathbb Z$. Notice that since $2 \nmid 3$, it must follow that $2 \mid t$ which means $t = 2v$ for some $v \in \mathbb Z$. Hence $x - y = 3t = 3 \cdot (2v) = 6v$ which means $x \equiv y \pmod 6$ and thus $[x]_6 = [y]_6$. Conclude that $\varphi$ is injective.
Is $\varphi$ surjective? Yes, notice that
\begin{align*}
[0]_6 &\mapsto \big([0]_2, [0]_3\big) \\
[1]_6 &\mapsto \big([1]_2, [1]_3\big) \\
[2]_6 &\mapsto \big([0]_2, [2]_3\big) \\
[3]_6 &\mapsto \big([1]_2, [0]_3\big) \\
[4]_6 &\mapsto \big([0]_2, [1]_3\big) \\
[5]_6 &\mapsto \big([1]_2, [2]_3\big)
\end{align*}
Conclude that $\varphi$ is a bijection between $\mathbb Z_6$ and $\mathbb Z_2 \times \mathbb Z_3$.
