# What is the abstract algebraic structure of “Elementary Algebra”?

From what I understand the Set of Complex Numbers is Closed under Addition, Subtraction, Multiplication, Division, Exponentiation, Radicals and Logarithms.

I understand this to mean that these binary operations applied to any two elements of this Set will always result in a third element also contained within this set (the complex numbers). (However upon further reflection perhaps division by zero creates an exception to this)

My question is: what sort of algebraic structure does this create? I've been learning about Rings, Groups, Fields etc. But have always been left wondering what structure arises from all the standard tools of Elementary Algebra taken as a whole. That is the set of Complex Numbers and the standard operations I listed at the start.

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$0$ can not be a unit. $\mathbb{C}$ is algebraically-closed. –  user2468 Mar 24 '12 at 18:09
Note that general exponentiation (that is, $a^b$ rather than $\exp(b)$), radicals and logarithms are not very well-behaved on $\mathbb C$ -- they are either multi-valued or discontinuous along an arbitrary chosen branch cut. –  Henning Makholm Mar 24 '12 at 18:10
In terms of this table: wiki/Group_(mathematics)#Generalizations, are we dealing with violation of closure, or violation of totality? –  user2468 Mar 24 '12 at 18:29
@J.D. My informal description of Algebraic Closure contains no mention of a "Unit" and I'm unsure of the meaning of your usage of it. Could you please clarify? (I apologize in advance for the myriad of basic concepts I am no doubt still unaware of. I have spent a lot of time and effort attempting to research such things on my own only to find there are questions I still can't answer myself!) –  jcelios Mar 24 '12 at 18:59
I was referring to unit and zero divisor. Zero, by definition, has no multiplicative inverse (i.e., you can not divide by zero is not an anomaly). –  user2468 Mar 24 '12 at 19:01

This isn't a trivial question, and in answering it I hope to convince you that "exponential" actually means at least two different things, so I'll answer it in stages.

Groups, or maybe abelian groups, give you addition.

Rings give you addition and multiplication. Sometimes an element in a ring will be invertible, in which case you get a limited notion of division. But this is subtle: for noncommutative rings, elements can have right inverses $a a^{-1} = 1$ without having left inverses $a^{-1} a = 1$, and vice versa. Furthermore, right inverses and left inverses are not unique. However, any left inverse must equal any right inverse (proof: if $ba = ac = 1$ then $bac = b(ac) = b = (ba)c = c$), so if an element has both a left and a right inverse then it has a two-sided inverse and this two-sided inverse is unique.

In any monoid written multiplicatively it is possible to make sense of the exponential $a^n = a \cdot a \cdot ... \cdot a$ for non-negative integers $n$ (and for integers $n$ if $a$ is invertible, in particular if the monoid is a group). This exponential satisfies $a^{n+m} = a^n a^m$ and $(a^n)^m = a^{nm}$ as expected, and satisfies $(ab)^n = a^n b^n$ if $a$ and $b$ commute, but not in general. If the monoid is multiplication in a ring, then the binomial theorem $$(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}$$

again holds if $a$ and $b$ commute, but not in general. (If $a$ and $b$ are matrices and $n = 2$, for example, then $(a + b)^2 = a^2 + ab + ba + b^2$.)

An important point, which I cannot stress enough, is that in this general form of exponentiation the base and the exponent are different types of things. One is an element of a monoid or a ring and the other is an integer. This is generally the expected behavior of various types of exponentiation and the fact that it appears to be false for ordinary numbers is deeply misleading.

Again, in any monoid written multiplicatively it is possible to make sense of radicals: we say that $\sqrt[n]{a} = b$ if $a = b^n$. Radicals neither exist nor are unique in general, but you already see this behavior for real numbers so it shouldn't be too surprising. So, to summarize:

In any ring $R$ we can make sense of addition, subtraction, multiplication, division, exponentials where the base is an element of the ring and the exponent is a non-negative integer, and radicals. Division is only partially defined, and radicals are both partially defined and multi-valued. Some nice properties fail when $R$ is noncommutative.

Instead of asking for exponentials where the base is an object of the ring, we might ask for exponentials where the exponent is an element of the ring. More precisely, we might ask for the object $$e^a = \sum_{n=0}^{\infty} \frac{a^n}{n!}$$

to be well-defined. For this to happen, $R$ needs more structure. First, we need to be able to divide by $n!$, so $R$ needs to be a $\mathbb{Q}$-algebra. Second, we need to take infinite sums, so we need some notion of limits. (Actually, there is a small cop-out here: if $a$ is nilpotent then we can work with finite sums, but the resulting function will only be partially defined because the identity, for example, in a ring is never nilpotent.) So $R$ needs to be a topological $\mathbb{Q}$-algebra. Of course, this doesn't guarantee that the limit exists.

One setting in which the above limit always exists is when $R$ is a Banach algebra. In this setting $e^{a+b} = e^a e^b$ if $a$ and $b$ commute but, once again, not in general.

In a Banach algebra, we can also define logarithms of certain elements using the power series expansion $$\log(1 + a) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{a^n}{n}.$$

This only converges for $|a| < 1$ in general, but where defined it provides a suitable inverse to the exponential. Again, summarizing:

In a topological $\mathbb{Q}$-algebra we can make sense of addition, subtraction, multiplication, division, exponentials where the base is an element of the ring and the exponent is a non-negative integer, radicals, exponentials where the base is $e$ and the exponent is an element of the ring, and logarithms. The latter two are only partially defined, although the exponential is always well-defined in a Banach algebra.

I cannot stress this enough: there are at least two things that ought to be called the exponential, and the fact that they happen to agree for, say, positive reals is a red herring. In general the base and exponential are different types of objects and play different roles. Of course, one can always define $$a^b = e^{\log(a) b}$$

but owing to the partially-defined and multi-valued nature of the logarithm I am not convinced this is a productive thing to do. Better to think about exponentials and logarithms separately.

In closing I'll link to the Wikipedia article on exponential rings, but my impression is that this is not a formalism which sees much use outside of model theory. (Any Banach algebra equipped with the exponential $e^a$ above is an exponential ring, but it has much much more structure than this given by the holomorphic functional calculus.)

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The fundamental nature is one of a field, $(\mathbb F,+,\times,0,1)$ equipped with an onto group homomorphism $\exp:(\mathbb F,+)\rightarrow (\mathbb F^\times,\times)$.

The fact that $exp$ is onto means that there is a (possibly multi-valued) logarithm. In turn, that means that there is a more general (possibly multi-valued) exponentiation, $a^b$ with $a,b\in \mathbb F$ when $a\neq 0$.

If $\mathbb F$ has characteristic $0$, then it has all roots.

You can theoretically do the same for a more general commutative rings, $R$: $exp:(R,+)\rightarrow (R^{\times},\times)$, where $R^{\times}$ is the set of units of $R$.

Non-commutativity is an additional problem. For example, while we can define $exp$ in the ring of $n\times n$ matrices, it is not a group homomorphism since addition of matrices is commutative but multiplication in general is not. In this case, we still get $\exp(A)\exp(B)=\exp(A+B)$ if $AB=BA$. Also, we have that if $U$ is an invertible $n\times n$ matrix, then $exp(U^{-1}AU)=U^{-1}exp(A)U$.

Of course, you can also have fields where $\exp$ is not onto, such as $\mathbb R$ with $exp(x)=e^x$. Then you don't have all roots or logarithms, but only for the image of $\exp$.

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There is some sense in the non-commutative case, that the relationship between Lie Algebras and Lie Groups is related deeply to exponentiation, and that the Lie operator is naturally a measure of the extent to which an exponential operator fails to be commutative. But my understanding of this is very vague, at best, so maybe somebody else can elaborate. –  Thomas Andrews Mar 24 '12 at 19:10