Example: To show that the polynomials of a finite field is itself not a field, I need to show that a multiplicative inverse does not exist.

What are the general techniques that can be used to show that some ring does not have a multiplicative inverse?

I'm thinking the best way to go about it would be to assume that a multiplicative inverse exists by the definition, and then look for a contradiction.

What types of contradictions could help show that a multiplicative inverse doesn't exist?

Or are there any other ideas?

David Callanan
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  • In the case of the ring of polynomials in $n$ variables, over any ring, you can determine quite easily whether an element is invertible or not (look at exercise 1.2 and 1.3 on Atiyah-Macdonald) – Ezio Greggio Feb 24 '22 at 11:04
  • See the dupe for the case of polynomial rings (the question is far too broad for general rings). – Bill Dubuque Feb 24 '22 at 11:35

1 Answers1


Let's consider $\mathbb{Z}_2[X]$ and let $P(X)=X$. If this polynomial has an inverse $g(X)=\sum_{i=0}^{\infty}a_iX^i$ in $\mathbb{Z}_2[X]$, then $$P(X)G(X)=1=X(\sum_{i=0}^{\infty}a_iX^i)=\sum_{i=0}^{\infty}a_iX^{i+1}.$$

However, it is impossible that $\sum_{i=0}^{\infty}a_iX^{i+1}=1$ as this doesn't even contain a constant term.

There is a very general criterium that you can deduce here. Try the same with $P(X)=1+X$, is that one invertible?

Mathematician 42
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  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement [here](https://math.meta.stackexchange.com/q/33508/242). – Bill Dubuque Feb 24 '22 at 11:36