Below is a clear, simple, and exam-oriented explanation of Direct Products and Morphisms in Abstract Algebra (Rings/Groups).
This is exactly what you need to write in BCA/MCA/University exams.

---

⭐ DIRECT PRODUCT & MORPHISMS

Direct product is a way to combine two algebraic structures into a new one.
Morphisms describe structure-preserving maps between these products.

Direct products are defined for:

• Groups
• Rings
• Modules
• Fields (with restrictions)

Let’s focus on rings and ring morphisms, since this follows your earlier topics.

---

⭐ 1. DIRECT PRODUCT OF RINGS

Let (R_1) and (R_2) be two rings.

Their direct product is defined as:

[
R_1 \times R_2 = {(a,b) \mid a \in R_1,\ b \in R_2}
]

with the operations defined component-wise:

Addition

[
(a_1, b_1) + (a_2, b_2) = (a_1 + a_2,\ b_1 + b_2)
]

Multiplication

[
(a_1, b_1)\cdot (a_2, b_2) = (a_1 a_2,\ b_1 b_2)
]

Zero element

[
(0,0)
]

Identity (if exists)

[
(1,1)
]

👉 The direct product of two rings is again a ring.

---

⭐ 2. Projection Morphisms

Given a direct product ring (R_1 \times R_2), there are two natural ring morphisms called projections.

First projection:

[
\pi_1 : R_1 \times R_2 \to R_1
]
[
\pi_1(a,b) = a
]

Second projection:

[
\pi_2 : R_1 \times R_2 \to R_2
]
[
\pi_2(a,b) = b
]

These are ring homomorphisms because they preserve addition and multiplication:

[
\pi_1((a_1,b_1)+(a_2,b_2)) = a_1 + a_2 = \pi_1(a_1,b_1)+\pi_1(a_2,b_2)
]

[
\pi_1((a_1,b_1)(a_2,b_2)) = a_1 a_2 = \pi_1(a_1,b_1)\pi_1(a_2,b_2)
]

Same for (\pi_2).

---

⭐ 3. Inclusion Morphisms (Diagonal Morphism)

Another important morphism is the diagonal map:

[
\Delta : R \to R \times R
]

[
\Delta(a) = (a,a)
]

This is also a ring homomorphism.

---

⭐ 4. Universal Property of Direct Product

Direct product is characterized by the following universal property:

Given rings (A, R_1, R_2) and ring homomorphisms:

[
f_1 : A \to R_1
]
[
f_2 : A \to R_2
]

There exists a unique ring homomorphism:

[
f : A \to R_1 \times R_2
]

such that:

[
\pi_1 \circ f = f_1
]
[
\pi_2 \circ f = f_2
]

And explicitly:

[
f(a) = (f_1(a),\ f_2(a))
]

This shows direct products are the product object in category of rings.

---

⭐ 5. Kernel of Projection Morphisms

For (\pi_1: R_1 \times R_2 \to R_1):

[
\ker(\pi_1) = {(0,b) \mid b \in R_2}
]

For (\pi_2):

[
\ker(\pi_2) = {(a,0) \mid a \in R_1}
]

➡ These kernels are ideals in the product ring.

---

⭐ 6. Example (Very Common in Exams)

Let (R_1 = \mathbb{Z}), (R_2 = \mathbb{Z}_6).

Direct product ring:

[
\mathbb{Z} \times \mathbb{Z}_6
]

Projection morphism:

[
\pi_1(n,[m]) = n
]
[
\pi_2(n,[m]) = [m]
]

Kernel:

[
\ker(\pi_1) = {0} \times \mathbb{Z}_6
]
[
\ker(\pi_2) = \mathbb{Z} \times {[0]}
]

---

⭐ 7. Relationship with Ideals & Quotient Rings

The direct product ring has two big ideals:

[
I_1 = \ker(\pi_2) = R_1 \times {0}
]
[
I_2 = \ker(\pi_1) = {0} \times R_2
]

And we have:

[
(R_1 \times R_2)/I_1 \cong R_2
]
[
(R_1 \times R_2)/I_2 \cong R_1
]

This is a direct application of the First Isomorphism Theorem.

---

⭐ Quick Exam-Oriented Summary

✔ Direct product of rings:
[
R_1 \times R_2 = {(a,b)\mid a\in R_1\ , b\in R_2}
]

✔ Operations component-wise:

• ((a_1,b_1)+(a_2,b_2)=(a_1+a_2,,b_1+b_2))
• ((a_1,b_1)(a_2,b_2)=(a_1 a_2,,b_1 b_2))

✔ Projection morphisms:

• (\pi_1(a,b)=a)
• (\pi_2(a,b)=b)

✔ Diagonal morphism:

• (\Delta(a)=(a,a))

✔ Kernels become ideals:

• (\ker(\pi_1)={0}\times R_2)
• (\ker(\pi_2)=R_1\times{0})

✔ Direct product satisfies universal property.

---