# A question about ring homomorphisms

• Artusartos
In summary: Yes, that is correct. In Frac[R(x)], addition is defined as (f, g) + (h, k) = (fk+gh, hk). This means that \phi(f + g, h + k) = \frac{fk+gh}{hk}, which is equal to \phi(f,h) + \phi(g,k). Therefore, \phi is a ring homomorphism and Frac[R(x)] is isomorphic to F(x).
Artusartos

## Homework Statement

If R is a domain with F=Frac(R), prove that Frac(R[x]) is isomorphic to F(x).

## The Attempt at a Solution

Let $$\phi : Frac(R[x]) \rightarrow F(x)$$ be a map sending (f(x),g(x)) to f(x)/g(x). We need to show that $$\phi$$ is a ring homomorphism. Let f,g,h,k be in R[x] such that f/h and g/k is in Frac(R[x]).

We know that

$$\phi(1,1) = 1/1= 1$$

$$\phi (fg, hk) = \frac{fg}{hk} = \frac{f}{h}\frac{g}{k} = \phi(f,h)\phi(g,k)$$

But I'm confused with the addition part...

$$\phi(f+g,h+k) = \frac{f+g}{h+k}$$

$$\phi(f,h)+\phi(g,k) = \frac{f}{h}+\frac{g}{k} = \frac{kf+gh}{h+k}$$

But now $$\phi(f+g,h+k) \not= \phi(f,h) + \phi(g,k)$$

Can anybody help with this?

You appear to be interpreting "Frac[R(x)]" as pairs, (f(x), g(x)). How are you defining addition in Frac[R(x)]?

HallsofIvy said:
You appear to be interpreting "Frac[R(x)]" as pairs, (f(x), g(x)). How are you defining addition in Frac[R(x)]?

Oh...ok I think I get what you mean...because $$\phi(f,h + g,k) = \phi(fk+gh, hk)$$, right?

Last edited:

## 1. What is a ring homomorphism?

A ring homomorphism is a function between two rings that preserves the algebraic structure of the rings. In other words, it maps elements of one ring to elements of another ring in a way that preserves addition, multiplication, and the identity element.

## 2. How is a ring homomorphism different from a group homomorphism?

While both a ring homomorphism and a group homomorphism are functions that preserve the algebraic structure of their respective objects, a ring homomorphism also preserves multiplication, while a group homomorphism only preserves group operations.

## 3. What are some examples of ring homomorphisms?

Some examples of ring homomorphisms include the identity homomorphism, the zero homomorphism, and the evaluation homomorphism. The identity homomorphism maps every element of a ring to itself, the zero homomorphism maps every element to the additive identity, and the evaluation homomorphism maps polynomials to their values at a specific point.

## 4. Can a ring homomorphism be both injective and surjective?

Yes, a ring homomorphism can be both injective (one-to-one) and surjective (onto). This means that the function is both preserving and bijective, so every element in the target ring is mapped to by exactly one element in the source ring.

## 5. How are ring isomorphisms related to ring homomorphisms?

Ring isomorphisms are a special case of ring homomorphisms, where the function is also bijective. This means that not only does the function preserve the algebraic structure of the rings, but it also has an inverse function that maps elements of the target ring back to elements of the source ring.

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