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Z-Algebras

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
After defining an R-algebra at the bottom of page 5 (see attachment from Sharp), on the top of page 6 we find the following:

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"We should point out at once that the concept of an R-algebra introduced in 1.9 above occurs very frequently in ring theory, simply because every ring is a Z-algebra. We explain in 1,10 why this is the case."

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Sharp then proceeds as follows:


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"Let \(\displaystyle R \) be a ring. Then the mapping \(\displaystyle F \ : \ \mathbb{Z} \to R \) defined by \(\displaystyle f(n) = n(1_R) \) for all \(\displaystyle n \in \mathbb{Z} \) is a ring homomorphism and, in fact, is the only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \).

Here

\(\displaystyle n(1_R) = 1_R + 1_R + ... \ ... + 1_R \) (n-terms) ... for n > 0

\(\displaystyle n(1_R) = 0_R \) for n = 0

and

\(\displaystyle n(1_R) = (-1_R) + (-1_R) + ... \ ... + (-1_R) \) (n-terms) ... for n < 0

It should be clear from 1.5 (The subring criterion - see attachment) that the intersection of any non-empty family of subrings of a ring R is again a subring of R, This observation leads to the following Lemma (Lemma 1.10 - see attachment, page 6). ... ... ... "

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My first problem with the above is this:

Sharp writes: "Let \(\displaystyle R \) be a ring. Then the mapping \(\displaystyle F \ : \ \mathbb{Z} \to R \) defined by \(\displaystyle f(n) = n(1_R) \) for all \(\displaystyle n \in \mathbb{Z} \) is a ring homomorphism and, in fact, is the only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \)."

But why is this the only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \)?

My second problem is as follows: Sharp connects establishing a Z-Algebra (section 1,10 page 6) with establishing the structure of a polynomial ring - but what is the relationship and big picture here?

My third problem is the following:

Dummit and Foote on page 339 define Z-Modules (see attachment) , but seem to structure them slightly differently using a an element a from an abelian group (and not the identity of the group) whereas Sharp uses the multiplicative identity of a ring in establishing a Z-algebra. I presume this is because the abelian groups are being treated as additive and one cannot use the additive identity - can someone please confirm this please - or clarify and explain the links between Z_algebras and Z-modules.

I would really appreciate some help and clarification of these issues.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
First off, let's just say we have any old homomorphism:

$f:\Bbb Z \to R$.

Since for integers $k,m$ we have:

$f(km) = f(k)f(m)$

we have:

$f(m) = f(1\ast m) = f(1)f(m)$ and:

$f(k) = f(k \ast 1) = f(k)f(1)$

so we see that $f(1)$ is a multiplicative identity for $f(\Bbb Z)$ (even if $R$ is not commutative).

Moreover, since $f(k)f(m) = f(km) = f(mk) = f(m)f(k)$, we see that the image of $f$ is a commutative sub-ring of $R$ with unity.

If $R$ is a ring with unity, the unity of $R$ is unique, and if we insist that a ring homomorphism preserve the ring unity, then we have no choice but to set:

$f(1) = 1_R$

But then the homomorphism property:

$f(k+1) = f(k) + f(1) = f(k) + 1_R$

yields (via induction on $n$):

$f(n) = nf(1) = n(1_R)$, for all $n \in \Bbb Z^+$.

Also:

$f(0) = f(0+0) = f(0) + f(0)$ shows that $f(0) = 0_R$.

Finally, for $n \in \Bbb Z^+$:

$0_R = f(0) = f(n + (-n)) = f(n) + f(-n)$, which means that:

$f(-n) = -f(n) = n(-1_R)$.

This means that $f$ is completely determined by $f(1)$, and we only have one choice in the matter, so only one homomorphism is possible (this is NOT true if we do not require homomorphisms to preserve unity:

Consider $f:\Bbb Z \to \Bbb Z \times \Bbb Z$ given by $f(k) = (k,0)$ and

$g:\Bbb Z \to \Bbb Z \times \Bbb Z$ given by $g(k) = (k,k)$).

What this means, in the language of category theory, is that the integers are an INITIAL OBJECT in the category of commutative rings with unity (with morphisms consisting of unity-preserving ring homomorphisms). Naively, the integers are your basic "starter ring", they have a special role to play in the theory of commutative rings.

To get a better understanding of what is going on, here, let's look at an alternate definition of $R$-algebra, and show our two definitions are equivalent.

Alternate definition of $R$-algebra:

A (left) $R$-module $M$ together with a bilinear ($R$-linear in both arguments) operation:

$[\cdot,\cdot]:M \times M \to M$

that possesses an identity for this operation, $1_M$.

Suppose now that we have an $R$-algebra in the sense of Sharp, so we have a commutative ring with unity $R$, a commutative ring with unity $S$, and a ring homomorphism (which preserves unity) $f:R \to S$.

For our $R$-module, we will take $M = S$. We define the "scalar multiplication" as:

$r(s) = f(r)s$, where the right-hand side is the ring multiplication in $S$. It should be (hopefully) clear that this indeed does produce an $R$-module:

$(S,+)$ is an abelian group
$(r + r')(s) = f(r + r')s = [f(r) + f(r')]s = f(r)s + f(r')s = r(s) + r'(s)$
$r(s+s') = f(r)(s + s') = f(r)s + f(r)s' = r(s) + r'(s)$
$r(r'(s)) = r(f(r')s) = (f(r))(f(r'))s) = (f(r)f(r'))s = f(rr')s = (rr')(s)$
$1_R(s) = f(1_R)s = 1_Ss = s$

For our binary operation $[\cdot,\cdot]$, we will use the multiplication in $S$. So all we have to verify is bilinearity:

$[s,t+t'] = s(t+t') = st + st' = [s,t] + [s,t']$
$[s+s',t] = (s+s')t = st + s't = [s,t] + [s',t]$
$r[s,t] = r(st) = f(r)(st) = (f(r)s)t = [f(r)s,t] = [r(s),t]$
$r[s,t] = r(st) = r(ts) = f(r)(ts) = (f(r)t)s = s(f(r)t) = [s,f(r)t] = [s,r(t)]$

and identity: clearly we may take $1_M = 1_S$.

Now, conversely, suppose we have the alternate definition, and we wish to prove this is an $R$-algebra in the sense of Sharp. So we need to exhibit a (unity-preserving) ring homomorphism $f:R \to M$. Define:

$f: R \to M$ by $f(r) = r(1_M)$.

The proof that this is such a homomorphism is left to you.

(Note: the requirement that $S$ is commutative can actually be relaxed to the condition that $f(S) \subseteq Z(S)$, the center of $S$, and the above proofs remain the same. Can you spot where this is required?).

********

By dint of the unique homomorphism from $\Bbb Z$ to $R$, we see immediately that any ring is a $\Bbb Z$-algebra, the $\Bbb Z$-module structure comes from the fact that the additive group of a ring is an abelian group. Some care must be taken, however, when writing $\Bbb Z$-linear combinations of elements of $R$, because $R$ may be of finite characteristic.

Every $\Bbb Z$-algebra *is* a $\Bbb Z$-module, but not every $\Bbb Z$-module has a bilinear product associated with it (in much the same way as some vector spaces have a multiplication defined on them, but not every vector space does).

The polynomial ring $R[x]$ is, in fact, the free $R$-algebra generated by the set $\{x\}$. This is what is expressed by the theorem:

Let $R$ be any commutative unital ring with $a \in R$. Then there is a unique ring homomorphism:

$\phi_a:R[x] \to R$

such that if $f:\{x\} \to \{a\}$ is defined the only way possible, and $\iota: \{x\} \to R[x]$ is the inclusion map, then:

$\phi_a \circ \iota = f$

Such a theorem is called a "universal property", because it says $R[x]$ represents the most efficient way of creating an $R$-algebra out of the single element set $\{x\}$.

The homomorphism $\phi_a$ is often called "the evaluation map at $a$" which returns the element of $R$, for a given polynomial $p(x) \in R[x]$ given by forming the expression $p(a)$. It may be an interesting exercise for you to show that $\phi_a$ is completely determined by its value at $x$.