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In algebra, exponential rings are commutative rings with an operation that mimics exponentiation. Exponential rings are commonly used by model theorists who study exponential field theory and transcendental number theory.

All exponential fields are exponential rings, but not all exponential rings are exponential fields.

=Definition= Let $$R$$ be a commutative ring. An exponential on $$R$$ is a function $$E: R \rightarrow R$$ such that An exponential ring $$(R,E)$$ is a commutative ring $$R$$ with an exponential $$ E$$ on $$R$$.
 * 1) $$E(x+y) = E(x)\cdot E(y) $$
 * 2) $$E(0)=1$$

An exponential on a ring $$R$$ is a group homomorphism from the additive group $$(R, +,0)$$ to the group of units $$(R^*,\cdot ,1)$$. On any ring $$R$$, there exists an exponential $$\mathfrak{t}$$ such that $$\mathfrak{t}(x) = 1$$ for every $$x$$ in $$R$$. This is called the trivial exponential.

=Examples=
 * Let $$\mathfrak{t}$$ be the trivial exponential (ie $$\mathfrak{t}(x)=1$$) on the ring $$R$$. Then $$(R,\mathfrak{t})$$ is an exponential ring, and hence any ring can be made into an exponential ring.
 * Consider the exponential $$E:\mathbb{Z}\rightarrow \mathbb{Z}$$ given by $$E(x)=(-1)^x$$. This is the only non-trivial exponential on $$\mathbb{Z}$$. The pair $$( \mathbb{Z},E)$$ is an exponential ring.
 * Let $$E: \mathbb{R} \rightarrow \mathbb{R} $$ be the exponential given by $$E(x)=e^x$$. The pair $$(\mathbb{R},E)$$ is an exponential ring.
 * Consider $$ \mu:\mathbb{Z}[\sqrt{3}] \rightarrow \mathbb{Z}[\sqrt{3}] $$ where $$\mu(a+b\sqrt{3}) = (-1)^{b}$$ for every $$a+b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$$. The function $$ \mu $$ is an exponential function, and the pair $$ (\mathbb{Z}[\sqrt{3}], \mu) $$ is an exponential ring.
 * Let $$ E $$ be the exponential on $$ \R[x]$$ given by $$ E(\sum_i a_i x^i) = \prod_j E(a_{2j+1}) $$ for every polynomial $$\sum_i a_i x^i $$ in $$ \R[x]$$. Examples like this are useful when trying to disprove facts about morphisms of exponential rings.

=Basic Facts=
 * If $$\lambda$$ and $$\mu$$ are exponentials on a ring $$R$$, then $$\lambda\cdot \mu$$ is an exponential on $$R$$. For any $$x\in R$$ there is an exponential $$(\lambda . x)$$ defined by $$(\lambda. x)(y) = \lambda(xy)$$. In this way, the set of exponentials on $$R$$ is a right $R$-module (and therefore an abelian group). This can be generalized: if $$\phi: R\rightarrow R$$ is an additive function on $$R$$ then there is an exponential $$(\lambda \circ \phi)$$ defined by $$(\lambda\circ \phi)(x)=\lambda(\phi(x))$$, and hence the set of exponentials on a ring $$R$$ form a right module over the (non-commutative) ring of additive functions on $$R$$. Using this, it becomes easy to create many new examples of exponentials.


 * The only exponential on $$\mathbb{Q}$$ is the trivial exponential.


 * Let $$(A,\alpha)$$ and $$(B,\beta)$$ be exponential rings. The product exponential ring $$(A,\alpha) \oplus (B,\beta) = (A\oplus B, \alpha\oplus \beta)$$ is an expoential ring with the exponential $$\alpha\oplus \beta$$ defined by $$\big(\alpha\oplus \beta\big)(x\oplus y) = \big(\alpha(x)\big)\oplus \big(\beta(y)\big)$$ for every $$x \in A$$ and $$y \in B$$. Given an infinite family of exponential rings $$\{(A_i,\alpha_i)\}_{i\in I}$$ the product exponential ring $$\prod_{i\in I}(A_i,\alpha_i) = (\prod_{i\in I} A_i, \prod_{i\in I} \alpha_i)$$ is defined similarily.

=Exponential Morphisms= Let $$(R,\alpha)$$ and $$(S,\mu)$$ be exponential rings. The mapping $$\phi:(R,\lambda)\rightarrow (S,\mu)$$ is called a an exponential morphism if
 * 1) $$\phi:R\rightarrow S$$ is a ring homomorphism.
 * 2) $$\phi\big(\lambda(x)\big) = \mu\big(\phi(x)\big)$$ for every $$x$$ in $$R$$.

If $$\phi:(R,\lambda)\rightarrow (S,\mu)$$ and $$\psi:(S,\mu)\rightarrow (T,\gamma)$$ are exponential morphisms, then $$\psi \circ \phi:(R,\lambda)\rightarrow (T,\gamma)$$ is an exponential morphism. The image of any exponential morphism is an exponential subring of the codomain.

=Tensor Products of Exponential Rings= Suppose $$(R,\lambda)\rightarrow (A,\alpha)$$ and $$(R,\lambda)\rightarrow (B,\beta)$$ are exponential morphisms. Then the tensor product exponential ring $$(A,\alpha)\otimes_{(R,\lambda)} (B,\beta) = (A\otimes_R B, \alpha\otimes_\lambda \beta)$$ is an exponential ring with exponential defined by$$\big(\alpha\otimes_\lambda \beta\big)(x\otimes_R y) = \big(\alpha(x)\big)\otimes_R \big(\beta(y)\big)$$ for every $$x\in A$$ and $$y\in B$$. It can be shown that tensor product exponential ring satisfies the usual universal property of tensor products.

we can be an exponential ring, and suppose $$(A,\alpha)$$ and $$(B,\beta)$$ are exponential rings such that $$A$$ and $$B$$ are contain the ring $$R$$, and the image of $$\alpha$$ and the image of $$\beta$$ are contained in $$R$$, and $$\alpha|_R = \beta|_R$$. Then, we have the

=The Free Exponential Ring=

=Exponential Polynomial Rings=