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I am currently studying for an abstract algebra final exam. I am trying to disprove the statement "Consider the homomorphism $f: R \rightarrow S$ where R and S are rings. Prove/Disprove: If $a \in R$ is not a unit in $R$, then $s = f(a)$ is NOT a unit in S.

I know this is false because taking $f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_3$,

then f looks like $f([a]_6) = [a]_3$. Then if we let $a = 4$, a is not a unit in $\mathbb{Z}_6$ because it's a zero divisor. However, $a$ IS a unit in $\mathbb{Z}_3$, because $[4]_3 = [1]$, which is a unit.

My trouble is coming in where proving $f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_3$ is indeed a homomorphism.

Here is my attempt:

Let $a, b \in R$. So $a = 6n + r_1$ and $b = 6m + r_2$ for $n, m, r_1, r_2 \in \mathbb{Z}$. So
$f([a]_6 + [b]_6) = f(a) + f(b)$
$f([6n + r_1 + 6m + r_2]_6) = [6n + r_1]_3 + [6m + r_2]_3$

After this line, I'm not exactly sure how to convert the elements in $\mathbb{Z}_6$ to look like elements in $\mathbb{Z}_3$.

If anyone could help me out that would be greatly appreciated.

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  • $\begingroup$ Here is an easier counter example: consider the quotient map $\pi : \mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$. Then, $5 \in \mathbb{Z}$ is not a unit but $\pi(5) = 1 \in \mathbb{Z}/2\mathbb{Z}$ is. $\endgroup$ – Aniruddh Agarwal Dec 11 '18 at 2:59
  • $\begingroup$ To prove that the map you constructed is a homomorphism, consider using the universal property of quotients. $\endgroup$ – Aniruddh Agarwal Dec 11 '18 at 3:00
  • $\begingroup$ The only way we've learned to prove homomorphisms is show that the function preserves addition and multiplication $\endgroup$ – Mari Dec 11 '18 at 3:01
  • $\begingroup$ You are probably looking at $\mathbb Z_6$ and $\mathbb Z_3$ sets of equivalence classes. (My answer user different viewpoint where $\mathbb Z_6=\{0,1,2,3,4,5\}$. I appologize if that lead to confusion.) So you can simply write $f([a]_6+[b]_6) = f([a+b]_6) = [a+b]_3=[a]_3+[b]_3=f([a]_6)+f([b]_6)$. As pointed out in Aniruddh Agarwal's, the fact that $f$ preserves operation is the easy part when using this viewpoint. What is more relevant here is to show that $f$ is well-defined. $\endgroup$ – Martin Sleziak Dec 11 '18 at 14:08
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Your function is given by $f(a)=a\bmod 3$. And you are asking whether $$f(a+_6b)=f(a)+_3f(b).$$ The two operations are given by $a+_6b=(a+b)\bmod 6$ (on $\mathbb Z_6=\{0,1,2,3,4,5\}$ and $a+_3b=(a+b)\bmod3$ (on $\mathbb Z_3=\{0,1,2\}$. So this is equivalent to the question whether $$((a+b)\bmod 6)\bmod 3 = (a\bmod 3+b\bmod 3)\bmod 3.$$ You should be able to show that \begin{align*} ((a+b)\bmod 6)\bmod 3 &= (a+b)\bmod 3\\ (a\bmod 3+b\bmod 3)\bmod 3 &= (a+b)\bmod 3 \end{align*}

Just notice that the two numbers give the same remainder modulo $3$ if their difference is a multiple of $3$, i.e., $$x\bmod 3=y\bmod 3 \qquad\Leftrightarrow\qquad 3\mid x-y.$$ Which is the same as saying that $x$ and $y$ are congruent modulo $3$, if you are familiar with this notion.

So all you can to do is check whether $(a+b)-((a+b)\mod 6)$ and $(a+b)-(a\bmod 3+b\bmod 3)$ are multiples of $3$.

You can use similar arguments to show that $f$ preserves also multiplication, i.e., that $f(a\cdot_6 b)=f(a)\cdot_6 f(b)$. (And, as I mentioned, the whole problem becomes much simpler if you are familiar with basic properties of congruences and modular arithmetic.)

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  • $\begingroup$ Thank you very much. Maybe now I have done it correctly: $f([a]_6 + [b]_6) = f(a) + f(b)$ $f([6n + r_1 + 6m + r_2]_6) = [6n + r_1]_3 + [6m + r_2]_3$ $(6(n + m) + r_1 + r_2)mod3 = (6n + r_1)mod3 + (6m + r_2) mod3 $\endgroup$ – Mari Dec 11 '18 at 3:22
  • $\begingroup$ Thank you very much. Maybe now I have done it correctly? $f([a]_6 + [b]_6) = f(a) + f(b)$ $f([6n + r_1 + 6m + r_2]_6) = [6n + r_1]_3 + [6m + r_2]_3$ $(6(n + m) + r_1 + r_2)mod3 = (6n + r_1)mod3 + (6m + r_2) mod3 $ = (6(n + m) + r_1 + r_2)mod3$ This still does not seem right? Have I done something wrong? Does this suffice? $\endgroup$ – Mari Dec 11 '18 at 3:32
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Remember to also check that $f$ is well-defined upto choice of representative. Here is how I would write the proof:

Fix $a + 6\mathbb{Z}, b + 6\mathbb{Z} \in \mathbb{Z}/6\mathbb{Z}$. Then,

$$ \begin{align} f((a + 6\mathbb{Z}) + (b + 6\mathbb{Z})) &= f((a + b) + 6\mathbb{Z}) \\ &= (a + b) + 3\mathbb{Z} \\ &= (a + 3\mathbb{Z}) + (b + 3\mathbb{Z}) \\ &= f(a + 6\mathbb{Z}) + f(b + 6\mathbb{Z}) \end{align} $$

You can similarly show that $1$ goes to $1$ and that multiplication is preserved (although the latter follows from the fact that addition is preserved).

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