# Novikov Ring Proof

I have a four part question involving the Novikov ring.

For part (1), I have:

$$(\Sigma_{\gamma\in \tau}a_{\gamma}t^\gamma) +_{N} (\Sigma_{\gamma\in \tau}b_{\gamma}t^\gamma) =\Sigma_{\gamma\in \tau}(a_{\gamma}+b_{\gamma})t^\gamma$$

We know that $$(a_{\gamma}+b_{\gamma})\in \mathbb{Z}$$ because $$a_{\gamma} \in \mathbb{Z}$$ and $$b_{\gamma} \in \mathbb{Z}$$

But I am unsure how to show that it satisfies the condition:

Similarly, I am unsure about how to prove closure for $$\cdot_{N}$$

For part (2), I just need to prove that it is an integral domain, so I must show that it is commutative, and I must show that it has no zero divisors. Again, I am unsure how to start the zero divisors portion of this part.

For part (3), I know the units in this ring must act as "multiplicative inverses", but I am unsure on how to express them.

For part (4), I must show that the Novikov ring can be generated by a single element. I am unsure how to approach this as well.

Any help would be appreciated.

• But I am unsure how to show that it satisfies the condition: that condition is give to you as a hypothesis. You don't have to prove it. – rschwieb Jun 6 '16 at 4:10
• For part (4), I must show that the Novikov ring can be generated by a single element. that is not what a principal ideal ring is. You'd have to show every ideal is generated by a single element. – rschwieb Jun 6 '16 at 4:12