ring isomorphism from $\mathbb Z_3\times \mathbb Z_2$ to $\mathbb Z_6$

Problem is ring isomorphism from $\mathbb Z_3\times \mathbb Z_2$ to $\mathbb Z_6$.

My solution:

Let $f=\mathbb Z_3\times \mathbb Z_2 \to \mathbb Z_6$ defined as $f(a_3,b_2)=ab_6$.

I prove it is isomorphism:

It is obviously that $|\mathbb Z_6|$=$|\mathbb Z_3\times\mathbb Z_2 |$ (the same cardinality). Surjectivity is obviously because of the Chinese Remainder Theorem.

Injectivity : Let $f(a_3,b_2)=f(c_3,d_2)$. Then $f(a_3-c_3,b_2-d_2)=f(a_3,b_2)-f(c_3,d_2)=0$. If $f(a_3,b_2)=ab_6=0_6$ then because a_2 is divisible by 2 and b_3 by 3, then ab_6 is divisible by 6. Therefore ab_6=0.

It is correct ? Thanks

• This is not a homomorphism for addition. – Bernard Mar 20 '16 at 0:01
• It does not define a function either. $1_2=3_2=b_2$, $0_3=3_3=a_3$, but $(0\cdot1)_6\ne(3\cdot3)_6$. – user228113 Mar 20 '16 at 0:09

The correct homomorphism is $$f(a_3,b_2)=(4a+3b)_6$$ See if you can prove this is a ring isomorphism.
• @joe for example, the only way for $(4a+3b)_6$ to be 0 is if $2\mid b$ because we already have $2\mid 4a$. Thus in that case $b_2=0$. You can also show that $a_3$ must be 0, so it is injective. – Matt Samuel Mar 20 '16 at 15:05
Note that any ring homomorphism $\Bbb Z_6 \to \Bbb Z_2\times \Bbb Z_3$ is determined by the image of $1$. Consider mapping $1\to (1,1)$. This is an isomorphism. Indeed, it is injective since if $n\to (n,n)=(0,0)$ then $2,3$ divide $n$, and then so does $6$, hence $n=0$. Since both rings have six elements, this is a bijection. One readily checks this is a ring homomorphism, so the claim follows. The inverse mapping sends $(1,0) = 3(1,1)$ to $3$ and $(0,1) = 4(1,1)$ to $4$.