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Let $R,S$ be a rings and let $\varphi:R\to S$ be a rings homomorphism on $S$. Prove or disprove with counter example:

A. if $R$ is a commutative ring then $S$ is commutative

B. if $R$ has a unit then $S$ has a unit.

Attempt:

A. Take $r_1,r_2\in R$ then $\varphi(r_1\cdot r_2)=\color{blue}{\varphi(r_1)\varphi(r_2)}=\varphi(r_2\cdot r_1)=\color{blue}{\varphi(r_2)\varphi(r_1)}\implies S$ commutative.

B. No, let $\varphi:\mathbb Z\to 2\mathbb Z$ there is a unit at $\mathbb Z$ but not in $2\mathbb Z$

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  • $\begingroup$ Is there an error in your title "comotativicy"? (If you mean commutativity, you have the correct spelling in your question box). $\endgroup$ – Michael Burr Apr 16 '17 at 21:19
  • $\begingroup$ Can you edit please? my english is not the best $\endgroup$ – Error 404 Apr 16 '17 at 21:22
  • $\begingroup$ What is the definition of a ring homomorphism in your world? In mine, any ring homomorphism must send 1 in R to 1 in S. $\endgroup$ – Chickenmancer Apr 16 '17 at 21:28
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    $\begingroup$ For B you should use $\varphi\colon \mathbb{Z}\to\mathbb{Z}\times 2\mathbb{Z}$, $\varphi(x)=(x,0)$. $\endgroup$ – egreg Apr 16 '17 at 21:30
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    $\begingroup$ @Michael Burr Yes, I am aware of this. Consider that S has no identity, then 1 must be sent to something 2-idepmotent or zero, else the map is not well defined. $\endgroup$ – Chickenmancer Apr 16 '17 at 21:33
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Hints:

  • For $A$, this only works for elements in the image of $\varphi$. What if $S$ has more elements than are images of $\varphi$? As an example, consider the map $$ \mathbb{Z}\rightarrow M_{2,2} $$ the map from the integers to $2\times 2$ matrices where $a\mapsto\begin{bmatrix}a&0\\0&a\end{bmatrix}$.

  • For $B$, what is your map from $\mathbb{Z}\rightarrow 2\mathbb{Z}$? Is it the multiplication by $2$ map? If so $2=\varphi(1)=\varphi(1\cdot 1)=\varphi(1)\cdot\varphi(1)=2\cdot 2=4$. So, the map is not well-defined. What if you consider the zero map $\mathbb{Z}\rightarrow2\mathbb{Z}$? (Or the example of @egreg in the comments above).

  • As a side note to $B$, consider the map $$ \mathbb{Z}\rightarrow\operatorname{Diag}_{2,2} $$ the map from the integers to $2\times 2$ diagonal matrices where $a\mapsto\begin{bmatrix}a&0\\0&0\end{bmatrix}$. In this case, both $\mathbb{Z}$ and $\operatorname{Diag}_{2,2}$ have identities, but the map does not take the identity of $\mathbb{Z}$ to the identity of $\operatorname{Diag}_{2,2}$.

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  • $\begingroup$ For A, this type of matrices is also commutative and also $\mathbb Z$ $\endgroup$ – Error 404 Apr 16 '17 at 21:41
  • $\begingroup$ @Error404 Just the image is commutative. The set of all $2\times 2$ matrices is not commutative. The codomain is much larger than the image. $\endgroup$ – Michael Burr Apr 16 '17 at 21:42
  • $\begingroup$ I got it for A. for B I meant if there is a unit necessary (not must be the same unit as in S) @Michael Burr $\endgroup$ – Error 404 Apr 16 '17 at 21:59
  • $\begingroup$ No, a unit is not necessary, either use the zero map or the example that @egreg gives in the comments. $\endgroup$ – Michael Burr Apr 16 '17 at 22:07
  • $\begingroup$ if the homomorphism was onto?, for B $\endgroup$ – Error 404 Apr 16 '17 at 23:02

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