Proving that Euler's Totient Function is multiplicative by Abstract Algebra

Question

I do know that similar question was already answered, but i didn't understand how does the proof work.

If there exist rings $R$ and $S$ (meaning that its set of elements have two binary operations: multiplication and addition, under addition, ring is an abelian group, under multiplication it is associative, or otherwise known as monoid). Two new rings can be created by these rings $R × S$ and $R + S$. If $R$ and $S$ are finite rings, amount of elements in $R$ multiplied by amount of elements in $S$ is equal to amount of elements in $R × S$.

Statement above seems to be very simply understandable, then comes the "equation", If there exists A and B that are coprime, then:

$ℤ/⟨AB⟩≅ℤ/⟨A⟩×ℤ/⟨B⟩$

As i understand, the "equation" above states that under positive integers group generated by multiplication of $A$ and $B$, is isomorphic (has the same properties) to group generated by $A$ multiplied by group generated by $B$.

This seems to be the abstract algebraic definition of multiplicative function, but i'm unable to find the proof. Also author of the answer mentions that this is the Chinese Remainder Theorem, but as a computer scientist who researches asymmetric cryptography, i thought the purpose of CRT in asymmetric cryptography was to find a new method other than $c^d$ (where d is private key, multiplicative inverse) to increase speed of decryption. But now i found out that there are different purposes as well.

Is the isomorphism above the proof that Euler's totient function is multiplicative? If so, how? If not, how can abstract algebra be utilized to prove this?

Thank you!

Answer 1

Yes, that's the core of the proof.

We can see $\varphi(a)$ as the number of units in the ring $\mathbb{Z}/a\mathbb{Z}$.

It is obvious that, if $R$ and $S$ are finite rings, then the number of units in $R\times S$ is the product of the number of units in the two rings, because an element $(r,s)$ is a unit in $R\times S$ if and only if $r$ is a unit in $R$ and $s$ is a unit in $S$.

Now the Chinese remainder theorem asserts that, when $a$ and $b$ are coprime, $$ \mathbb{Z}/a\mathbb{Z}\times\mathbb{Z}/b\mathbb{Z}\cong\mathbb{Z}/ab\mathbb{Z} $$ as rings, so $\varphi(ab)=\varphi(a)\varphi(b)$ follows.

How can we prove the theorem? Consider the map $$ f\colon\mathbb{Z}\to\mathbb{Z}/a\mathbb{Z}\times\mathbb{Z}/b\mathbb{Z} \qquad f(x)=(x+a\mathbb{Z},x+b\mathbb{Z}) $$ which is readily seen to be a ring homomorphism. Its kernel is $$ \ker f=a\mathbb{Z}\cap b\mathbb{Z} $$ and, in general, $a\mathbb{Z}\cap b\mathbb{Z}=m\mathbb{Z}$, where $m$ is the lowest common multiple of $a$ and $b$. In the particular case when $a$ and $b$ are coprime, we have $m=ab$. By the general homomorphism theorem, $f$ induces an injective homomorphism $$ \mathbb{Z}/\ker f=\mathbb{Z}/ab\mathbb{Z}\to \mathbb{Z}/a\mathbb{Z}\times \mathbb{Z}/b\mathbb{Z} $$ Since both sets have the same cardinality $ab$, this homomorphism is an isomorphism.

Answer 2

The CRT theorem started out as a simple number theory result. For example, if a number is even and leaves a remainder of one when divided by three, then it leaves a remainder of four when divided by six. There is nothing special about two and three except they have to be relatively prime. This simple general result is the CRT. After the development of group theory this exact same result was recast in the form of a theorem about cyclic groups. This is the isomorphism that you are asking about. Namely, the product of two cyclic groups is cyclic iff the orders are coprime.

This simple, but fundamental, theorem becomes more interesting when it is applied to the multiplicative group of units of a finite ring. This is where the application to cryptography comes from.