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I'm having a little trouble understanding the finite fields theory, so I'm sorry if my question would seem a little stupid.

I wanted to know if a finite field of 36 elements could exist. Basically, I thought of the field $\mathbb{F}_{37}$, which has 37 elements. But does for example the invertibles of $\mathbb{F}_{37}$ form a field of 36 elements or not? Since all elements are invertible and we have 36 invertibles in $\mathbb{F}_{37}$.


marked as duplicate by Squeamish Ossifrage, kelalaka, Maeher, Maarten Bodewes Feb 27 at 15:13

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  • $\begingroup$ The invertibles of $\mathbb{F}_{37}$ is a group under multiplication, with 36 elements. But it is do not form a field, because it does not contain the neutral for addition (0), wich is (thus) not even an internal law. The present observation does not answer the question, but this does. $\endgroup$ – fgrieu Feb 25 at 11:31

Finite fields only exist for an order $q$ if $q$ is prime or a power of a prime (http://mathworld.wolfram.com/FiniteField.html). Since 36 is neither a prime itself, nor it is a power of a prime ($36 = 2^3 * 3^2$), there is no finite field with 36 elements.


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