# What is the difference between $\Bbb F_4$ and $\Bbb Z/4\Bbb Z$

And more generally what is he difference between $\Bbb F_n$ and $\Bbb Z/n\Bbb Z$ when we look at them as rings with multiplication and addition? Aren’t the tables of multiplication and addition the same?

• Do you mean "the" finite field of order 4? Because $\mathbb{Z}/4\mathbb{Z}$ is not a field, so they're definitely different. May 2, 2018 at 15:56
• Yes sorry, I guess F is a field May 2, 2018 at 15:57
• They line up if and only if $n$ is a prime. May 2, 2018 at 15:57
• And how can it be a field if 2 has no multiplicative inverse? May 2, 2018 at 15:57
• It's not. There IS a field of order 4, but it is not $\mathbb{Z}_4$. May 2, 2018 at 15:58

$$\mathbb{F}_4$$ means the finite field with $$4$$ elements, but $$\mathbb{Z}/4\mathbb{Z}$$ is not a field, since $$\bar 2\times \bar2=\bar 0$$ in $$\mathbb{Z}/4\mathbb{Z}$$.

Generally, $$p$$ is prime if and only if $$\mathbb{Z}/p\mathbb{Z}$$ is a field and the only field with $$p$$ elements up to isomorphism. Otherwise, any finite field with $$q=p^r$$ elements where $$r>1$$ is not isomorphic to the quotient ring of $$\mathbb{Z}$$.

$\mathbb{F}_n$ is typical notation for "the" finite field of $n$ elements. It is a theorem that such a thing is unique up to isomorphism. The notation $GF(n)$ is also used. Also, $n$ must be a power of a prime $n=p^d$ so you may also see $GF(p^d)$. (Consequence of another theorem: $GF(p^d)$ is vector space of dimension $d$ over $\mathbb{Z}_p$.)

To your question, $\mathbb{Z}_p$ is a field when $p$ is prime, so by the uniqueness result, $GF(p)$ "is" $\mathbb{Z}_p$. When $n$ is NOT prime, $\mathbb{Z}_n$ is certainly not a field, so it cannot be $GF(n)$. To see this, if $n$ is not prime then $n = ab$ for integers $1< a< b <n$. But then $[a][b]= [n] =[0]$ in $\mathbb{Z}_n$ so we have zero divisors. Thus $\mathbb{Z}_n$ is not a field.

To get $GF(p^d)$ when $d > 1$ you need to know a good deal about ideal theory and polynomials. There is a theorem that $\mathbb{Z}_p[x]/I$ is a field when $I$ is a maximal ideal in $\mathbb{Z}_p[x]$. You then classify all such maximal ideals $I$ as the principals $I = (g(x))$ where $g(x)$ is irreducible. A simple count then shows that $\mathbb{Z}_p[x]/I$ has order $p^d$ where $I = (g(x))$ and $g(x)$ has degree $d$.

• Can you explain a little bit the last sentence please? May 2, 2018 at 16:32
• Well, it's a lot to bite off. If the irreducible $g(x)$ has degree $d$ then every coset $f(x) +I$ of the quotient ring can be uniquely written--using long division---as $b_0 + b_1x + \cdots + b_{d-1}x^{d-1} + I$. Now count up the number of ways to choose the coefficients and you get $p^d$ different cosets/elements. May 2, 2018 at 16:35
• I probably shouldn't've said it was a "simple" count. May 2, 2018 at 16:37