# Abstract Algebra Homework Solution - Check Ring Homomorphism

• NanoMath
In summary: The first argument is more general, and might be more convincing to someone who doesn't know about countability.
NanoMath

## Homework Statement

Hello guys
So I have the following problem, given the mapping above I have to check weather it's ring homomorphism, and
maybe monomorphism or epimorphism.

## The Attempt at a Solution

So the mapping is obviously well defined, and I have proven it's homomorphism, and it's obviously not monomorphism because a polynomial P(x)= 5 - x2 is in the kernel so kernel is not trivial.
I am not sure how to prove if the function is surjective or not, obviously if the codomain were integers for every integers C , I could just use constant function p(x) = C and function would be surjective.

NanoMath said:
I am not sure how to prove if the function is surjective or not

It is just an intuition, but I doubt there are polynomials of ##\mathbb{Z}[X]## such that ##p(\sqrt{5}) \in \mathbb{Q}-\mathbb{Z} ##.

Showing the map is a homomorphism shouldn't be too difficult. Simply show the map satisfies the properties of a homomorphism.

The rest of the question is asking if the map is also a bijection, or something more specific. Try applying the first isomorphism theorem if you know it, and use the fact that ##x^2 - 5## is the minimal polynomial.

NanoMath said:
I am not sure how to prove if the function is surjective or not, obviously if the codomain were integers for every integers C , I could just use constant function p(x) = C and function would be surjective.

Isn't it pretty obvious that the range is contained in the set ##a+b \sqrt{5}## where ##a## and ##b## are integers? Why isn't that all of ##R##?

I managed to show that function is not surjective with the hint that every element in the range is of the form ## a+b \sqrt{5} ## because for example ##\sqrt{2}## doesn't get hit by any element in domain. Is it also valid argument that function can't be surjective because ##\mathbb{R}## is uncountable whilst ##\mathbb{Z}[X]## is countable?

NanoMath said:
I managed to show that function is not surjective with the hint that every element in the range is of the form ## a+b \sqrt{5} ## because for example ##\sqrt{2}## doesn't get hit by any element in domain. Is it also valid argument that function can't be surjective because ##\mathbb{R}## is uncountable whilst ##\mathbb{Z}[X]## is countable?

Both of those arguments are good. The second makes it obvious if you know cardinality.

## 1. What is a ring homomorphism?

A ring homomorphism is a function between two rings that preserves the algebraic structure of the rings. This means that the function preserves the operations of addition and multiplication, and also preserves the multiplicative identity and the distributive property.

## 2. How do you check if a function is a ring homomorphism?

To check if a function is a ring homomorphism, you need to verify that the function preserves the algebraic structure of the rings. This means that the function should satisfy the following conditions:

• Preserves addition: f(a+b) = f(a) + f(b)
• Preserves multiplication: f(ab) = f(a)f(b)
• Preserves multiplicative identity: f(1) = 1
• Preserves distributive property: f(a(b+c)) = f(a)f(b) + f(a)f(c)
If the function satisfies all these conditions, then it is a ring homomorphism.

## 3. What is the purpose of checking a ring homomorphism?

Checking a ring homomorphism is important because it ensures that the function preserves the algebraic structure of the rings. This is necessary for the function to be useful in abstract algebra, as it allows us to extend our understanding of one ring to another ring through the homomorphism.

## 4. Can a function be a ring homomorphism between two different types of rings?

Yes, a function can be a ring homomorphism between two different types of rings. The only requirement is that the function must satisfy the conditions for being a ring homomorphism, as mentioned in the answer to question 2.

## 5. How is a ring homomorphism different from a ring isomorphism?

A ring homomorphism preserves the algebraic structure of the rings, while a ring isomorphism not only preserves the structure, but also has a one-to-one correspondence between the elements of the two rings. This means that a ring homomorphism may map different elements of one ring to the same element in another ring, while a ring isomorphism must preserve distinct elements.

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