Does the binomial theorem hold for a ring without unity?

Question

I'm reading Theorem 8.4 and its proof from textbook Analysis I by Amann/Escher.

From this Wikipedia page, I got

From this Wikipedia page, semiring mentioned above is defined to have a multiplicative identity.

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So both my textbook and Wikipedia assert that the binomial theorem holds for a ring with unity. On the other hand, I do not see that the proof in my textbook utilizes this property.

My question:

Does the binomial theorem hold for a ring without unity?

Answer 1

The right-hand side of the equation $$(a+b)^n = \sum_{k=0}^n \binom nk a^k b^{n-k}$$ is not well defined in a ring without unity because $a^0$ and $b^0$ are not defined. You can claim instead that $$(a+b)^n = a^n + b^n + \sum_{k=1}^{n-1} \binom nk a^k b^{n-k}$$ which can be obtained either by the same proof, or by simply expanding the product and counting the terms as @Captain Lama suggested.

Answer 2

Yes, all of the computations still hold.

No identity is ever mentioned. You simply have a sum of finite products of $a$ and $b$ and nothing else.

Answer 3

The answer is YES.

All ring without unity can be canonically embedded into a ring containing unity (Dorroh extension). Let $A$ be a ring without unity. Consider the product ring $A\times \mathbb Z $ of elements $(x,m)\in A\times \mathbb Z $. The function $$f(x)=(x,0)$$ is an homomorphism injective of $A$ into $A\times \mathbb Z $ and this ring have the unity $(0,1)$ where $0\in A$ and $1\in \mathbb Z $.