Homomorphisms

Ring Homomorphism

Suppose that

(R, \oplus︎, \otimes)

and

(S, ⊞, ⊠)

are rings, and

ϕ : R \to S

is a function, then

ϕ

is a ring homomorphism if for any

a, b \in R

$ϕ (a \oplus︎ b) = ϕ (a) ⊞ ϕ (b)$
$ϕ (a \otimes b) = ϕ (a) ⊠ ϕ (b)$

Ring with Unity Homomorphism

Suppose that

R, S

are rings with unity, then we say that

ϕ

is a homomorphism between

R

and

S

when

ϕ

is a ring homomorphism and we have

ϕ (1_{R}) = 1_{S}

Crone Homomorphism

Suppose that

R, S

are crones, then if

ϕ

is a ring with unit homomorphism, then we say that

ϕ

is a crone homomorphism

We may also say that $ϕ$ is a homomorphism from $R$ to $S$ , or that $ϕ$ is a homomorphism of rings

Crone Homomorphism Between Two by Two Matices with 0 in Bottom Left and Integers

Let

R : = {(\begin{matrix} a & b \\ 0 & a \end{matrix}) : a, b \in ℤ}

along with the usual definition for matrix addition and matrix multiplication, and define

ϕ : R \to ℤ

be defined such that

ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix})) = a

Show the following

$ϕ$ is a crone homomorphism
Find $\ker (ϕ)$
Show $R / \ker (ϕ)$ is ismorphic to $ℤ$
is $\ker (ϕ)$ a prime ideal?

We start by showing $ϕ$ is a crone homomorphism, note that we already know that $R$ and $ℤ$ are crones, now note the following: $ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix}) (\begin{matrix} c & d \\ 0 & c \end{matrix})) = ϕ ((\begin{matrix} a c & a c + b d \\ 0 & a c \end{matrix})) = a c = ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix})) ϕ ((\begin{matrix} c & d \\ 0 & c \end{matrix}))$ and that $ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix}) + (\begin{matrix} c & d \\ 0 & c \end{matrix})) = ϕ ((\begin{matrix} a + c & b + d \\ 0 & a + c \end{matrix})) = a + c = ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix})) + ϕ ((\begin{matrix} c & d \\ 0 & c \end{matrix}))$

We recognize that the $2 \times 2$ identity is the identity in this ring as well, therefore we confirm that $ϕ ((\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) = 1$ Thus $ϕ$ is indeed a crone homomorphism

From the definition of $ϕ$ we can conclude that $ϕ ((\begin{matrix} a & b \\ 0 & a \end{matrix})) = 0$ iff $a = 0$ , therefore the kernel is all matrices of this form $\ker (ϕ) = {(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}) : b \in ℤ}$

Recall the definition of $R / \ker (ϕ)$ , it is $\begin{matrix} {M + \ker (ϕ) : M \in R} & = {{M + N : N \in \ker (ϕ)} : M \in R} \\ = {{(\begin{matrix} a & b \\ 0 & a \end{matrix}) + (\begin{matrix} 0 & c \\ 0 & 0 \end{matrix}) : c \in ℤ} : a, b \in ℤ} \\ = {{(\begin{matrix} a & b + c \\ 0 & a \end{matrix}) : c \in ℤ} : a, b \in ℤ} \\ = {{(\begin{matrix} a & d \\ 0 & a \end{matrix}) : d \in ℤ} : a \in ℤ} \end{matrix}$ Let $tl (M)$ be defined as the top left coefficent in $M$ , then for two matrices $M, N$ it should be clear that if $tl (N) = tl (M)$ , then they generate the same coset, also that if $tl (M) \neq tl (N)$ , then they generate different cosets. Also note that $tl (M N) = tl (M) \cdot tl (N)$ and that $tl (M + N) = tl (M) + tl (N)$ . Finally given any coset, since we know that $tl (X)$ agrees for every $X$ in the coset, we can justify the construction of $tl (N + \ker (ϕ)) = tl (N)$

We claim that $R / \ker (ϕ)$ is an crone isomorphism to $ℤ$ using $tl$ . We've actually already verified the that it respects addition and multiplication because of our discussion in the previous paragraph and also just note that if $I$ is the the identity then we know it's top left element is $1 \in ℤ$ so that $tl (I) = tl (I) = 1$ .

At this point we know that it's a homomorphism. So we just have to show it's a bijection to have an isomorphism so suppose that $M + \ker (ϕ) \neq N + \ker (ϕ)$ by the contrapositive of what we noted earlier since they generate different cosets, then $tl (N) \neq tl (M)$ , therefoer $tl (N + \ker (ϕ)) \neq tl (M + \ker (ϕ))$ , which shows that $tl$ is injective, it is clearly also surjective because since $R$ ranges over all matrices with integer coefficients, then if we pick a particular integer $k \in ℤ$ then there is always a matrix with that specific integer in it's top left component, say that matrix is $X$ , then $tl (X) = tl (X) = k$ as needed, so $tl$ is an isomorphism.

Now we verify that $\ker (ϕ)$ is a prime ideal, so suppose that $M N \in \ker (ϕ)$ , this means that $tl (M N) = 0$ , but as noted earlier $tl (M N) = tl (M) tl (N)$ , but we know that $ℤ$ so in this context we have $tl (M) tl (N) = 0$ , but $ℤ$ is a domain therefore we must have wlog $tl (M) = 0$ , but that implies that $M \in \ker (ϕ)$ which shows that $\ker (ϕ)$ is a prime ideal as neeeded.

Ring Isomorphism

Suppose that

ϕ

is a ring homomorphism and is a bijection, then we say that

ϕ

is a ring isomorphism

Homomorphisms carry Zero to Zero

Suppose that

ϕ : R \to S

is a homomorphism of rings, then

ϕ (0_{R}) = 0_{S}