# Is $\mathbb{Z}_2$ mathematically the same as $\mathbb{F}_2$? [closed]

Is $$\mathbb{Z}_2$$ mathematically the same as $$\mathbb{F}_2$$? Or are they different in any way?

When I mention $$\mathbb{Z}_2$$, I am referring to the integers modulo 2, which are {0, 1}, with addition and multiplication done modulo 2.

For example, is $$\mathbb{Z}_2[x]/(x^{256}-1)$$ the same as $$\mathbb{F}_2[x]/(x^{256}-1)$$?

• Perhaps you could define your terms. $\mathbb Z_2$ has more than one standard meaning. My guess is that you are simply describing the unique field with $2$ elements in two different ways, but why should i guess? And your question about polynomials makes no sense at all. I could make more guesses, but I won't.
– lulu
Commented Sep 5 at 3:24
• As what? What's your definition of $\mathbb Z_2$? Commented Sep 5 at 3:24
• When I mention $\mathbb{Z}_2$, I am referring to the integers modulo 2, which are {0, 1}, with addition and multiplication done modulo 2. Commented Sep 5 at 3:32
• $\mathbb{F}_2$ is defined as the (unique, as finite fields are unique up to cardinality) field with two elements, which is then $\mathbb{Z}_2$ indeed. On a similar note, $\mathbb{F}_p = \mathbb{Z}_p$ for any prime $p$, whereas $\mathbb{F}_4$ for example is not the same as $\mathbb{Z}_4$ as the latter is no longer a field. Commented Sep 5 at 3:34
• I would say they are intentionally different, but turn out they coincide. $\mathbb{Z}_2$ just randomly describing the equivalence class/ elements of modulo $2$. While $\mathbb{F}_2$ specify the set is a field with $2$ elements, and we can identify them with a map with $\mathbb{Z}_2$ also. Commented Sep 5 at 5:49

In this specific case, $$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \F_2 = \Z_2$$. This is because $$\F_n$$, more generally, is the field of $$n$$ elements, and the finite fields of a fixed size are unique up to isomorphism.

However, one should not extrapolate this to the logic of "okay, so $$\F_n = \Z_n$$ for every $$n$$."

• For one, the only allowed $$n$$ are the prime powers. For instance, there is no $$\F_{10}$$, since $$10 = 2 \cdot 5$$ is not a prime or a power of a single prime.

• For another, the structure of the corresponding field might not mirror the ring structure in $$\Z_n$$; that is, $$\Z_4$$ (as an example) may behave differently than $$\F_4$$ does. To construct $$\F_{p^n}$$ (up to isomorphism) where $$p$$ is prime, you need to find $$f \in \F_p[x]$$ which is irreducible with $$\deg f = n$$, so $$\F_{p^n} = \frac{\F_p[x]}{(f(x))}$$ From this you can easily see how $$\F_{p^n}$$ and $$\Z_{p^n}$$ might have different structures. (In general, too, $$\Z_p$$ is a field if and only if $$p$$ is a prime. We will only have $$\F_p = \Z_p$$ for primes $$p$$ as a result.)

The pivotal part of the question is

mathematically the same as

Based on the phrase I'd conclude you aren't completely clear on what that might mean.

What is helpful is that you also elaborated that you meant you were thinking of $$\mathbb Z_2$$ as a ring. When talking about two structures of a comparable type (one might say they are from the same category) then the right notion of "sameness" is isomorphism.

It's true that in the category of rings, the ring of integers modulo $$2$$ is isomorphic to the field of 2 elements.

Had you not mentioned that the multiplication operation behaves that way, one could have chosen a different multiplication on the group of integers modulo 2, for example, the one in which $$xy=0$$ for every pair $$x,y$$. This would be nonisomorphic to the field of two elements in the category of rngs, which is like the category of rings except you don't necessarily have identity.

And if you hadn't mentioned multiplication at all, then someone could object that $$\mathbb Z_2$$ could mean just the group of two elements, in which case there isn't a way to say they are "mathematically the same" at all. Coming from separate categories, the field of two elements and the group of two elements are like apples and oranges. They conform to separate collections of axioms that define them, and that makes an inherent difference.

So I hope this helps you see why "mathematically the same as" is somewhat vague. It could be asking about isomorphism, or alternatively if they are even comparable in some category.