How to answer 'why the following ring homomorphisms cannot exist'? 
Question 2.3 ($8$ marks)
  Explain why the following ring homomorphisms cannot exist.
  a) $\varphi_1 \colon \mathbb{Z}_{20} \to \mathbb{Z}_{15}$ surjective.
  b) $\varphi_2 \colon M_2(\mathbb{Z}_2) \to \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ injective.
  c) $\varphi_3 \colon \mathbb{R} \to \mathbb{Z}$ surjective.
  d) $\varphi_4 \colon \mathbb{Z}_{24} \to \mathbb{Z}_2 \times \mathbb{Z}_{12}$ surjective.

In a question like the one I have attached, would it be sufficient to come up with my own possible homomorphism and show that it doesn't work? Or does that only show that MY specific homomorphism doesn't work, but others may be? If it's the latter, how should I go about doing this? Thanks very much. 
 A: The problem is asking you to show that (e.g.) there is no homomorphism from $\mathbb{R}$ to $\mathbb{Z}$ which is surjective. To show this, it's not enough to come up with some surjective map and show that it's not a homomorphism, or to come up with some homomorphism and show it's not surjective; you need to prove that no possible map is both surjective and a homomorphism.
HINT for this part: suppose $f:\mathbb{R}\rightarrow\mathbb{Z}$ is surjective. Let $f(\alpha)=1$ (why does such an $\alpha$ exist?). Assuming that $f$ is a homomorphism, what is $f({\alpha\over 2})$?
A: (a) Suppose such $f$ exists. By first isomorphism theorem, $\mathbb{Z}_{20}/\ker(f) \cong \mathbb{Z}_{15}$. Now, $\ker(f)$ is a subgroup of the additive abelian group $\mathbb{Z}_{20}$. By Lagrange's Theorem, $|\mathbb{Z}_{20}/\ker(f)|$ divides $20$, but $|\mathbb{Z}_{20}/\ker(f)|=15$, a contradiction.
(b)Note $M_2(\mathbb{Z_2})$ has nontrivial nilpotent elements, whereas $(\mathbb{Z_2})^4$ does not. Let $A \in M_2(\mathbb{Z_2})$ be nilpotent and nontrivial. If $f:M_2(\mathbb{Z_2}) \rightarrow (\mathbb{Z_2})^4 $ is an injective ring homomorphism, $f(A) \neq 0$ but $f(A)^n=0$ for some $n$, a contradiction.
(c) $\mathbb{R}$ is a field, a surjective ring homomorphism $f$ would produce an ideal in $\mathbb{R}$, namely $\ker(f)$. Then either $\ker(f)=0$ or $\mathbb{R}$. The former would imply an isomorphism and hence $\mathfrak{c}= |\mathbb{Z}|$ a contradiction. The latter would imply $f$ is not surjective, another contradiction.
(d) By first isomorphism theorem, such a map would give $\mathbb{Z_{24}}/\ker(f) \cong \mathbb{Z}_2 \times \mathbb{Z}_{12}$, so $|\mathbb{Z_{24}}/\ker(f)|=24$, hence $\ker(f)$ is trivial and $f$ is actually an isomorphism. But $\mathbb{Z}_2\times \mathbb{Z}_{12}$ has a two element subfield, while $\mathbb{Z_{24}}$ has no such thing.
A: Hint for part (a): Consider $\mathbb Z_{20}$ and $\mathbb Z_{15}$ as additive groups; then the order of $\varphi_1\left(\mathbb Z_{20}\right)$ is $\dfrac{\left|\mathbb Z_{20}\right|}{\left|\ker\varphi_1\right|}$ where $\ker\varphi_1$ is the kernel of $\varphi_1$ considered as a group homomorphism.
Hint for part (b): Let $I$ be the identity matrix and $M=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$; then $\varphi_2(M)^2=(1,1,1,1)=\varphi_2(I)$. However the only element $\zeta\in\mathbb Z_2^4$ such that $\zeta^2=(1,1,1,1)$ is $\zeta=(1,1,1,1)$.
Noah Schweber has provided a hint for part (c).
Hint for part (d): Since $\mathbb Z_{24}$ and $\mathbb Z_2\times\mathbb Z_{12}$ have the same number of elements, if $\varphi_4$ is surjective, then it must be injective as well; hence it would be an isomorphism between the additive groups of $\mathbb Z_{24}$ and $\mathbb Z_2\times\mathbb Z_{12}$. Are these two isomorphic as groups?
A: For $(a)$, by the first isomorphism theorem $15$ would be a divisor of $20$ which clearly is not the case. 
For $(b)$, there would have to be a noncommutative subring of the target ring, since $M_2(\mathbb{Z}_2)$ clearly contains elements that do not commute. However, the target ring is commutative so its subrings are commutative.
For $(c)$, we know $2 \mapsto 2$, but $\frac{1}{2} \times 2 \mapsto 1$, so that $\frac{1}{2} \mapsto \frac{1}{2} \notin \mathbb{Z}$, a contradiction. 
For $(d)$, note that because the domain and range have the same finite number of elements, the homomorphism would be injective also, so that $\mathbb{Z}_{24} \cong \mathbb{Z}_2 \times \mathbb{Z}_{12}$. This is not true however, these aren't even isomorphic as additive groups, since a generator would be of the form $(1,a)$, but this element has order at most $12$. 
