The first isomorphism theorem is probably the most important result one will learn in a first course on algebra. We will state and prove the theorem for groups. There is, of course, an analogous result for rings. We will state the theorem for rings (the proof is identical to that for groups) and then look at an example using the theorem.

**Theorem. **If is a homomorphism of groups, then is a normal subgroup of and .

The proof is not terribly complicated; in fact, it is rather “computational”, meaning, we just have to verify a few definitions. It is just a matter of checking that the kernel satisfies the definition of normal and then defining a function from to and checking that the function satisfies the definition of an isomorphism. The “hardest” part of the proof, I think, is actually defining a function from to . However, as we will see, the choice of such a function is quite natural.

*Proof.* First, we will show that the kernel of is a normal subgroup of *. *We check the following:

*Nonempty. *Since is a homomorphism, it must map the identity in to the identity in . Thus, which shows that the kernel is nonempty.* *

*Closure. *Let and be in the kernel of . We want to show that is in the kernel too. We compute that

Thus, is indeed in the kernel of , which shows that is a subgroup of .

*Normal.* Let be an element of and let be an element in the kernel of . It suffices to show that is also in the kernel. We have that

Thus, is in the kernel of as desired. This shows that is a normal subgroup of .

Next, we will show that is isomorphic to . Define by . I claim that is a well-defined isomorphism. Indeed, we check the following:* *

*Well-Defined. *Suppose that . Then for some element in the kernel of . We wish to show that Evaluating we see,

Thus, is well defined.* *

*Homomorphism. *Let and be cosets in the factor group . By multiplying cosets we observe that Consequentially,

Thus, is a homomorphism.* *

*Injective. *Suppose that . We wish to show that . Consider the following,

Since is an element of and is a group, exists. Therefore,

which shows that is in the kernel of . It follows that and so is injective.

*Surjective. *Let be an element in the image of . Then there exists an element in such that . Thus, which shows that is surjective.

We conclude that is a bijective homomorphism, hence an isomorphism.

We now state the analogous result for rings.

**Theorem. **If is a homomorphism of rings, then is an ideal of and .

**Example.** Let be a commutative ring with unity. Prove that

Define the function to be the evaluation homomorphism for any in . By the first isomorphism theorem, it suffices to check that is a surjective homomorphism with . We check the following:

*Homomorphism. *Let and be in . We compute that

.

Thus, is indeed a ring homomorphism.

*Surjective. *Let be an element in . Then is in and which shows that is surjective.

*Kernel.* Let be in the kernel of . Then which means that is a factor of and so is in the ideal . Therefore, . Now, let be in the ideal . Then for some in . Thus, which shows that is in the kernel of . Therefore, . We conclude that .

By the first isomorphism theorem we have that

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