Checking whether a map satisfies being homomorphism

Question

My question might seem silly ,but excuse me for it as I've just started my hand on Ring theory.

My question is : Like we've in case of group homomorphisms that we can check that a map $\phi :G \rightarrow H$ is in actual a homomorphism by simply checking that it satisfies the relations in the presentation of group $H$ after homomorphism.

Example:$\,\,\,\,\,$ consider group homomorphism $\phi :G \rightarrow G$ given by $\phi :g \rightarrow g^2$ If G was the group $\mathbb Z_6\times \mathbb Z_2$=$\langle a,b;a^6=1,b^2=1\rangle$ then you need to verify that $a^{\text12}=1$ and that $b^4=1$. These relations all clearly hold, so the map $a\rightarrow a^2,b\rightarrow b^2$ is a homomorphism.

But in case of Ring homomorphism what do we check to see whether a map satisfies being a ring homomorphism? Can anyone explain it to me with an example .Please help....

Answer 1

To check that a map $\phi: G \to H$ between groups is a (group) homomorphism, it's enough to verify that $\phi$ respects the product: $$\phi(g_1 g_2) = \phi(g_1) \phi(g_2) \text{ for all } g_1, g_2 \in G.$$

(Note that your example $\phi : g \mapsto g^2$ is in general not a group homomorphism $G \to G$, but it will be for certain groups.)

Now, rings have $2$ operations, which I'll denote by $+$ and juxtaposition, and ring homomorphisms $R \to S$ must respect both, and (for unital rings) it must map the multiplicative identity of $R$ to that of $S$. (If it satisfies these conditions, it automatically sends $0_R$ to $0_S$, and the same happens for the group homomorphism condition above.)

1. $\phi(1_R) = 1_S$,
2. $\phi(r_1 + r_2) = \phi(r_1) + \phi(r_2)$, and
3. $\phi(r_1 r_2) = \phi(r_1) \phi(r_2)$.

A basic but important example of a ring homomorphism is the map $$\phi: \mathbb{Z} \to \mathbb{Z} / n \mathbb{Z}$$ that reduces modulo $n$, i.e., the map that sends $a \in \mathbb{Z}$ to the equivalence class $[a] \in \mathbb{Z} / n \mathbb{Z}$. It's perhaps easy but instructive to check the above conditions (1)-(3) to show that this is a (ring) homomorphism. This generalizes to an important class of ring homomorphisms: Given any ring $R$ and ideal $I$, we can form the quotient ring $R / I$, and the map $R \to R / I$ that sends an element to its equivalence class in $R / I$ is always a ring homomorphism. In fact, the kernel of any ring homomorphism is an ideal, so all ideals arise this way.

Answer 2

A presentation of a group $G$ can (and should be!) thought of as the choice of a free group $F$ (where the generators are our list of symbols), a group homomorphism $\pi:F\to G$, and a list of generators of $\ker{\pi}$. In this case, we identify $G$ as $F/\ker{\pi}$.

For example, the group $G=\langle a,b;a^6=1,b^2=1\rangle$ is formed by taking $F$ to be the free group on two generators $x$ and $y$, and the generators for $\ker{\pi}$ are $x^6$ and $y^2$ (the point is that these combinations of symbols must both be sent to the identity when we replace $x$ with $a$ and $y$ with $b$).

For rings, the situation is basically identical, but the way free objects look has changed. A presentation of a ring $R$ is a choice of a polynomial ring $F$ with coefficients in $\mathbb{Z}$. And now, when we look at the projection $\pi: F\to R$, $\ker{\pi}$ is an ideal of $F$.

For example, the ring $\mathbb{Z}[i]$, which is the ring "generated" by $i$ with $i^2+1=0$, can be described as the quotient of $\mathbb{Z}[X]$ by the ideal $(X^2+1)$.

And your observation holds true for rings as well as groups. If we try to define a function based on the presentations, this is like defining a map between polynomial rings. This is a morphism of rings precisely when you can, using this map, verify the relations of the target ring using the relations of the source ring.