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Just as in Group theory we look at maps which "preserve the operation", in Ring theory we look at maps which preserve both operations.

**Definition**

A map *f* : *R*→ *S* between rings is called a **ring homomorphism** if

*f*(*x* + *y*) = *f*(*x*) + *f*(*y*) and *f*(*xy*) + *f*(*x*)*f*(*y*) for all *x*, *y* ∈ *R*.

**Remarks**

- The operations on the left are in the ring
*R*; those on the right are in*S*. - Since such a homomorphism is a group homomorphism from (
*R*, +) to (*S*, +) it maps 0_{R}to 0_{S}.

Even if the rings*R*and*S*have multiplicative identities a ring homomorphism will not necessarily map 1_{R}to 1_{S}. - It is easy to check that the composition of ring homomorphisms is a ring homomorphism.

A ring homomorphism which is a bijection (one-one and onto) is called a **ring isomorphism**.

If *f* : *R* → *S* is such an isomorphism, we call the rings *R* and *S* isomorphic and write *R* *S*.

**Remarks**

- Isomorphic rings have all their ring-theoretic properties identical. One such ring can be regarded as "the same" as the other.

- The inverse map of the bijection f is also a ring homomorphism.

- The map from
**Z**to**Z**_{n}given by*x*↦*x*mod*n*is a ring homomorphism. It is not (of course) a ring isomorphism. - The map from
**Z**to**Z**given by*x*↦ 2*x*is a group homomorphism on the additive groups but is not a ring homomorphism. - The map from
**Z**to the ring of 2 × 2 real matrices given by x ↦ is a ring homomorphism which does not map the multiplicative identity to the multiplicative identity.Of course the map

*x*↦*x*mod*n*is a homomorphism which does map the identity to the identity.

- The "evaluation at
^{1}/_{2}map" from the ring of continuous real-valued functions on the interval [0, 1] to**R**given by (*e*_{½})*f*=*f*(^{1}/_{2}) for*f*belongs*C*[0, 1] is a ring homomorphism. - The "evaluation at 1 map" from the ring
**Q**[*x*] to**Q**given by (*e*_{1})*p*=*p*(1) for*p*∈**Q**[x] is a ring homomorphism.

That is:*e*_{1}(*a*_{0}+*a*_{1}*x*+*a*_{2}*x*^{2}+ ... +*a*_{n}*x*^{n}) =*a*_{0}+*a*_{1}+*a*_{2}+ ... +*a*_{n}. - The "evaluation at √2 map" from
**Z**[*x*] to**R**given by*p*↦*p*(√2) is a ring homomorphism.

This will turn out to be a surprisingly important example. - The map from
**C**to ring of 2 × 2 real matrices given by*a*+*bi*↦ is a ring isomorphism.**Proof**: Exercise. - Let
*R*be the field with 9 elements {*a*+*bx*|*a*,*b*∈**Z**_{3}} and the multiplication rule*x*^{2}= -1.

Let*S*be the field with 9 elements {*a*+*by*|*a*,*b*∈**Z**_{3}} and the multiplication rule*y*^{2}=*y*+ 1.

Then the map defined by 1 ↦ 1 and*x*↦*y*+ 1 defines a ring isomorphism.**Proof**: We'll see a neat way of proving this later.

The connection of this with the last section is given by:

**Definition**

The **kernel** of a (ring) homomorphism is the set of elements mapped to 0.

That is, if *f*: *R*→ *S* is a ring homomorphism, *ker*(*f*) = *f*^{-1}(0) = {*r* ∈ *R* | *f*(*r*) = 0_{S} }.

**Theorem**

The kernel of a ring homomorphism is an ideal.

Proof

An easy verification.

**Remarks**

- Note the similarity with the corresponding result for groups:
*the kernel of a group homomorphism is a normal subgroup.* - If the ring
*R*is not commutative, the kernel is a two-sided ideal.

- The kernel of the above map from
**Z**to**Z**_{n}is the ideal*n***Z**. - Just as in the group-theory case, the kernel of a homomorphism is {0} if and only if the homomorphism is one-one.

**Proof**

(=>) If*f*(*r*) =*f*(*s*) then*f*(*r*-*s*) = 0 and so*r*-*s*belongs*ker*(*f*) and we have*r*-*s*= 0.

(<=) If*a*∈*ker*(*f*) and*a*≠ 0 then*a*, 0 ↦ 0 and so the map is not one-one. - The kernel of the "evaluation at 0" map from
**R**[*x*] to**R**is the ideal <*x*> of polynomials with zero constant terms. - The kernel of the "evaluation at 0 taken modulo 2" map from
**Z**[*x*] to**Z**_{2}is the ideal < 2,*x*> of polynomials with even constant terms.

We will see later that every ideal is the kernel of a ring homomorphism. This is similar to the group theory result that every normal subgroup is the kernel of a group homomorphism.

The last result in this section also parallels the corresponding example in Group theory,

**Theorem**

The image of a ring homomorphism *f*: *R*→ *S* is a subring of *S*.

**Proof**

The image is the set *im*(*f*) = {*s* ∈ *S* | *s* = *f*(*r*) for some *r* ∈ *R* }. It is easy to do the verification.

**Example**

The image of *e*_{√2} from **Z**[*x*] to **R** is the subring {*a* + *b* √2 | *a*, *b* ∈ **Z** } we met earlier.

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