Is it true if $R = m\Bbb{Z}/md\Bbb{Z}$ is isomorphic to $\Bbb{Z}/d\Bbb{Z}$, then it must have a unit element? This is a question I ask myself, but I'm not certain of this answer. Is anyone could explain to why this is (or is'nt) true?

That question comes from the result :

$m\mathbb{Z}/md\mathbb{Z} \cong \mathbb{Z}/d\mathbb{Z}$ as rings without identity if and only if $(d,m)=1$.

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    $\begingroup$ Isn't the element $1\in\mathbb{Z}/d\mathbb{Z}$ always a unit element in this ring? And therefore it's image in $R$ would also be a unit? $\endgroup$ – Alex Mathers Feb 12 '16 at 18:02
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    $\begingroup$ @taj Do you realize you need to distinguish ring or group homomorphisms? Sometimes students forget to do this and only were really talking about a group homomorphism. If it's a ring homomorphism, then yes, since the image of the identity is the identity of the image. $\endgroup$ – rschwieb Feb 12 '16 at 18:03

Yes, this is true, since $A = \Bbb Z/d\Bbb Z$ has a unit element, i.e. an element $1$ such that $1 \cdot x =x=x\cdot 1$ for all $x \in A$.

If $f : A \to B$ is a surjective ring homomorphism, then $f(1)$ is a unit element of $B$ (in particular for $B= m\Bbb Z/md \Bbb Z$) : if $b \in B$, then $b=f(a)$ for some $a \in A$, so that $$f(1)\cdot y = f(1 \cdot x)=f(x)=y=y \cdot f(1)$$

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    $\begingroup$ To avoid confusion it should be noted that $A\to B$ refers to the direction $\Bbb Z/d\Bbb Z\to m\Bbb Z/md\Bbb Z$ here $\endgroup$ – Hagen von Eitzen Feb 12 '16 at 18:03
  • $\begingroup$ Also, $f$ must be surjective. $\endgroup$ – Slade Feb 12 '16 at 18:30
  • $\begingroup$ Thank you for your comments. I typed too quickly. I corrected my answer. $\endgroup$ – Watson Feb 12 '16 at 18:54

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