# Division Rings and RIng Homomorphisms .... A&W Corollary 2.4 ....

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In summary: So $\ker f$ is trivial if and only if $f$ is injective.In summary, we discussed Corollary 2.4 which states that if the kernel of a homomorphism is the trivial group, then the homomorphism is injective. This can be proven by showing that if the homomorphism is not injective, then there exists a non-zero element in the kernel. Conversely, if the homomorphism is injective, the kernel is trivial. Therefore, the kernel being trivial is equivalent to the homomorphism being injective.
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I am reading "Algebra: An Approach via Module Theory" by William A. Adkins and Steven H. Weintraub ...

I am currently focused on Chapter 2: Rings ...

I need help with an aspect of the proof of Corollary 2.4 ... ...

Corollary 2.4 and its proof read as follows:View attachment 7933
In the above proof of Corollary 2.4 we read the following:

" ... ... If $$\displaystyle \text{Ker} (f) = \{ 0 \}$$ then $$\displaystyle f$$ is injective ... ... "
Can someone please explain exactly how/why $$\displaystyle \text{Ker} (f) = \{ 0 \}$$ implies that $$\displaystyle f$$ is injective ... ?
Help will be appreciated ...

Peter

Hi Peter,

If $f$ is not injective, there are distinct elements $a$ and $b$ such that $f(a)=f(b)$. Because $f$ is a homomorphism, this implies $f(a-b)=0$, and, by definition, $a-b\in\ker f$. If $a\ne b$, this gives you a non-zero element of $\ker f$.

Note that this is more generally true for any group homomorphism $f$: $\ker f$ is trivial if and only if $f$ is injective.

castor28 said:
Hi Peter,

If $f$ is not injective, there are distinct elements $a$ and $b$ such that $f(a)=f(b)$. Because $f$ is a homomorphism, this implies $f(a-b)=0$, and, by definition, $a-b\in\ker f$. If $a\ne b$, this gives you a non-zero element of $\ker f$.

Note that this is more generally true for any group homomorphism $f$: $\ker f$ is trivial if and only if $f$ is injective.
Thanks castor28 ...

You seem to have proved the contrapositive ... but then that proves it!

Peter

Peter said:
Thanks castor28 ...

You seem to have proved the contrapositive ... but then that proves it!

Peter
Hi Peter,

In fact, the converse is also true. If $f$ is injective, the equation $f(x)=0$ has no other solution than $0$, and the solution set is $\ker f$ by definition.

## 1. What is a division ring?

A division ring, also known as a skew field, is a mathematical structure that satisfies all the properties of a field except for commutativity. This means that in a division ring, multiplication is not necessarily commutative.

## 2. What is a ring homomorphism?

A ring homomorphism is a function between two rings that preserves the operations of addition and multiplication. This means that for two elements x and y in the first ring, the homomorphism will satisfy f(x+y) = f(x) + f(y) and f(xy) = f(x)f(y).

## 3. What does A&W Corollary 2.4 state?

A&W Corollary 2.4 states that if a ring homomorphism is surjective, then the image of the homomorphism is also a ring. In other words, the image of a surjective ring homomorphism is a subring of the codomain ring.

## 4. How is A&W Corollary 2.4 useful?

A&W Corollary 2.4 is useful because it allows us to prove that certain structures are rings by showing that they are isomorphic to another ring. This can simplify proofs and provide a deeper understanding of the structure being studied.

## 5. What are some examples of division rings?

Some examples of division rings include the real numbers, the complex numbers, and the quaternions. These are all examples of non-commutative division rings, as they do not satisfy the commutativity property of multiplication.

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