# Checking whether a map satisfies being homomorphism

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....

• Every commutative ring with $1$ is a $\mathbb{Z}$-algebra, i.e. has a presentation $R \cong \mathbb{Z}[x_i]/(f_j)$. Then for any ring $S$, giving a ring homomorphism $R \to S$ is equivalent to giving elements $y_i \in S$, one for each $x_i$, such that $f_j(y_i) = 0$ for each $j$. E.g. $\mathbb{Z}_6 \times \mathbb{Z}_2 \cong \mathbb{Z}[x]/(12,2(x-1),6x,x^2-x)$, so giving a ring map $\mathbb{Z}_6 \times \mathbb{Z}_2 \to S$ is equivalent to giving an element $y \in S$ such that $2(y - 1) = 6y = y^2 - y = 12 = 0$ in $S$ – zcn Sep 28 '14 at 6:41

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.

• @spectraa Here's another example: take the ring $\mathbb{Z}[\sqrt{2},\sqrt{3}]$. This is generated by two elements, $a$ and $b$, satisfying $a^2 = 2$ and $b^2 = 3$. If I want to know if I can make a homomorphism by sending $a$ and $b$ somewhere in some other ring, it's the same situation as what you mention: I just need to check that the images of $a$ and $b$ obey the correct relations in the target. – Slade Sep 28 '14 at 6:35
• @spectraa Oh, and keep in mind that rings can have "relations" that involve no generators. For example, $1+1=0$ is a relation that may hold. Only constants are involved, no symbols. But it's still an important relation that you have to pay attention to when dealing with homomorphisms—you can't have a ring with $1+1=0$ map homomorphically to another ring with $1+1\neq 0$. (all my rings have $1$ here, by the way—the situation is similar for rings without $1$, but a little more annoying to describe) – Slade Sep 28 '14 at 6:37
• @spectraa: It is in general a good idea to wait a little while before accepting an answer (also, you should not accept an answer if you do not feel that it answers your question), as accepting too early discourages people from adding an answer – zcn Sep 28 '14 at 6:43
• @zcn Of course, I can't resist the urge to describe $[X]$ as a $\$-algebra. And maybe we should deal with zero-divisors in the rng $A$ by passing to the quotient feld, and working with the $Q(\ )$-algebra $A\otimes_{}\ Q(\ )$. – Slade Sep 28 '14 at 7:24
• @Slade it'll be so kind of you if you can clarify these two things to me: $1.)$As you said rings can have "relations" that involve no generators.Can you help clarifying it to me with help of example in which there is such a case.. $2.$ I'm preparing for my ring theory exam which consists of ring homomorphism portion and just don't have any time to study what free groups are(I'll do this later on...) can you in a very simple language explain it to me the facts to be considered while writing a presentation of ring homomorphism,so that I can proceed with my preparation.... – spectraa Sep 28 '14 at 11:51

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.

• I understand that for being a ring homomorphism all these 3 conditions need to be satisfied.But my doubt is that to check these conditions are satisfied by all elements of ring will become tiresome.so we need to switch to relations as in case of groups in my example .What relations need to be satisfied in rings... – spectraa Sep 28 '14 at 5:56
• In practice, you don't have to check the conditions for all elements of a ring manually, you simply use the abstract properties of the ring elements. – Travis Willse Sep 28 '14 at 6:17
• Note that depending on your choice of definition of rings, not all rings have unit elements. – Alan Sep 28 '14 at 6:36
• Yes, that's true, I'll modify the answer to emphasize that this is for unital rings. – Travis Willse Sep 28 '14 at 6:40