Is $\phi:\mathbb{Z}_5\to\mathbb{Z}_{30},$ $\phi(x)=6x$ a homomorphism?

Question

Is $\phi:\mathbb{Z}_5\to\mathbb{Z}_{30},$ $\phi(x)=6x$ a homomorphism?

They can help me solve this problem. I know the answer is no. And I think it's a trick that I still don't see

Answer 1

There's some ambiguity here.

First of all, the map is a group homomorphism, which is what is suggested by your tags. If, however, you're asking whether it is a ring homomorphism (per the comments), it depends on what definition of ring and ring homomorphism you're using. Specifically, many texts - most, in my experience - include the existence of a multiplicative unit as part of the definition of ring, and correspondingly require ring homomorphisms to preserve that unit. Other texts don't make these requirements. When there's a danger of ambiguity the terms "unital ring (homomorphism)" and "non-unital ring (homomorphism)" are used.

For example, consider the set $2\mathbb{Z}$ of even integers with the usual notions of addition and multiplication. This is a non-unital ring; whether or not it's considered a ring outright depends on what text you're using. Relatedly, consider the function $$f: \mathbb{Z}\rightarrow\mathbb{Z}: x\mapsto 2x.$$ Both domain and codomain are unital rings, but $f$ itself is only a homomorphism in the non-unital sense. Basically, what we have here is a non-unital ring homomorphism between two non-unital rings which happen to have units. :P More jargonily, the category of unital rings is a non-full subcategory of the category of non-unital rings.

But back to your question. The map $$\phi:\mathbb{Z}_5\rightarrow\mathbb{Z}_{30}: x\mapsto 6x$$ is only a ring homomorphism in the non-unital sense, and so - again, in my experience (I'm not an algebraist) - most texts would not consider it to be a ring homomorphism.

Answer 2

Hints: As, $6$ is a idempotent in $\mathbb{Z}_{30}$

$\phi(xy)=6xy=(6x)(6y)=\phi(x) \phi(y) $