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So I have a test on abstract algebra tomorrow. It is basically on rings and polynomials in a ring/field.

I just have a couple of questions about terms. The definition of an isomorphic ring states that

there is a function $f: R \rightarrow S$ such that:

i) $f$ is injective;

ii) $f$ is surjective;

iii) $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for all $a, b \in R$

What does it mean to be injective and surjective? The book is not clear enough. If anyone can put it in simple english, it would be appreciated. This is my first abstract algebra course.

Examples (the prof said to have a lot of examples in mind) would be helpful. He said examples of homomorphisms would be good. For example:

The function $f: Z \rightarrow Z_6$ is given by $f(a) = [a]$.

I can verify axiom three myself, but the example states that the homomorphism is surjective but not injective.. Why is that? Is it because the function maps an integer to its integer congruence class? And hence $ a \neq b$ like PEV puts it ?

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Why did you not ask your prof for the examples?! – Mariano Suárez-Alvarez Jan 25 '11 at 2:03
The function in the example cannot be injective because $\mathbb Z$ is infinite but $\mathbb Z_6$ is finite. – lhf May 24 '11 at 12:59
Just to clarify: a ring cannot be "isomorphic". Two rings can be isomorphic. – Fredrik Meyer May 24 '11 at 13:19
up vote 3 down vote accepted

A function f(x) has one value for each x you plug into it, right? And it's valid for every x you put into it (it's defined over the whole domain). Well, injective and surjective are saying these things about the range. Injective means that for every value that you can get out of f(x), there is only one x you can use to get that value. Surjective means that you can take any value in the set where f(x) is defined (the codomain) and get some f(x) that maps to that value (there is some x where f(x) equals that value). In other words, it says the range spans the whole codomain.

I hope this informal explanation is helpful.

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thanks, that was very helpful. But what is a codomain? – Tyler Hilton Jan 25 '11 at 1:34
No, the codomain is the set that the function maps to, so it's a superset of the range. If you have $f: A \rightarrow B$, then $B$ is the codomain. The range doesn't necessarily cover the whole set $B$, but if it does, it's called "surjective". – Matt Gregory Jan 25 '11 at 1:44

For injectivity: For $a,b \in R$, $f(a) = f(b) \implies a = b$.

Surjectivity: For each $s \in S$ there exists $r \in R$ such that $f(r) = s$.

$f: \mathbb{Z} \to \mathbb{Z}_6$ defined by $f(x) = [x]$ is not injective because $f(6) = f(12)$ but $ 6 \neq 12$. Or in other words, the kernel is nontrivial.

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So does injectivity imply that the two rings are the same (ie, their elements are the same?) also I am editing my original question. – Tyler Hilton Jan 25 '11 at 1:11
No it means the following: IF f(a) = f(b), THEN a = b. – PEV Jan 25 '11 at 1:13
The contrapositive is the following: IF $a \neq b$ then $f(a) \neq f(b)$. – PEV Jan 25 '11 at 1:16

Injective means one-to-one, while surjective means onto. I am assuming you know what one-to-one and onto functions are; you can check out these articles if not: One-to-one functions and Onto functions.

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Do you want to get clear about what injective and surjective mean, or do you want to get clear about what isomorphic means? If you mainly want to get clear about what isomorphic means, you can just say that "iii*) $f : R \to S$ such that $f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$ for all $a,b \in R$, such that there exists an inverse function $f^{-1} : S \to R$, $f^{-1}(a+b)=f^{-1}(a)+f^{-1}(b)$ and $f^{-1}(ab)=f^{-1}(a)f^{-1}(b)$. That is, the composition of $f$ and $f^{-1}$ yields the identity function for the object in question."

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