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Basically are the integers mod 4 a field? I want to know because I am reading a text and it has a problem assuming the integers modulo any number are a field

ziggurism
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John doeee
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    No. $2\times 2=0$ and a field can't have zero divisors. – lulu Oct 23 '16 at 12:53
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    Out of curiosity, what is the problem? The intention might be to obtain an absurdity from the assumption that $\mathbb Z/n\mathbb Z$ is always a field. –  Oct 23 '16 at 12:55
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    The problem involves an integer q where q is equal to p^m for some prime p. later on the problem has a symbol F subscript q, which I think is the integers modulo q, but I dont know what that means now – John doeee Oct 23 '16 at 12:58
  • In that case $\mathbb{F}_q$ is (isomorphic to) the field $\mathbb{F}_p[x]/(f(x))$ where $f(x)\in\mathbb{F}_p[x]$ is irreducible and of degree $m$. Alternatively let $\overline{\mathbb F_p}$ be the algebraic closure of $\mathbb F_p$, then $\mathbb F_q={a\in\overline{\mathbb{F}_p}\mid a^q=a}$. If your textbook doesn't cover this then you should look up finite fields. –  Oct 23 '16 at 13:35
  • is the fact that Fq={a∈Fp∣a^q=a} a definition or is there a proof for that? – John doeee Oct 23 '16 at 14:13
  • @Johndoeee It depends who you ask. $\mathbb F_q$ is the finite field with $q$ elements. After that, you need to prove that that field exists and is unique up to isomorphism. Once you know that, all ways of defining a finite field of $q$ elements have to be equivalent. – Jack M Oct 23 '16 at 16:33
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    Next time, ask about the notation if you don't know the meaning, instead of assuming it means something and ending up asking a question with a false premise... For the same reason, you should have also edited this question and it would be eligible for re-opening. – user21820 Feb 03 '17 at 10:55

3 Answers3

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No.

The ring of integers mod $p$ for $p$ a prime is indeed a field, in fact the unique field of $p$ elements. If you heard that there is also a unique field with higher prime power $p^n$ elements, it would seem natural to guess that it's again the ring of integers modulo the number of elements. But it is only the case for the primes, not for the powers of primes. It is the case that the field of two elements is the ring of integers mod two, but it is not the case that the field of four elements is the ring of integers mod four.

To explain, addition and multiplication mod $n$ are well defined, so $\mathbb{Z}/n$, the integers mod $n$, is always a ring, but not a field in general unless $n$ is a prime. In particular, the integers mod 4, (denoted $\mathbb{Z}/4$) is not a field, since $2\times 2 = 4 = 0\mod 4$, so $2$ cannot have a multiplicative inverse (if it did, we would have $2^{-1}\times 2\times 2 = 2 = 2^{-1}\times 0 = 0$, an absurdity. $2$ is not equal to $0$ mod $4$). For this reason, $\mathbb{Z}/p$ a field only when $p$ is a prime.

(Note that sometimes people denote the ring of integers mod $n$ as $\mathbb{Z}_n$, but I prefer not to, and reserve that notation for the $p$-adic integers. So I prefer to only write it as $\mathbb{Z}/n$ (or $\mathbb{Z}/(n)$ or $\mathbb{Z}/n\mathbb{Z}$ if you're not into the whole brevity thing). However if you're more familiar with the $\mathbb{Z}_n$ notation, feel free to substitute.)

However, while $\mathbb{Z}/4$ is not a field, there is a field of order four. In fact there is a finite field with order any prime power, called Galois fields and denoted $\mathbb{F}_q$ or $GF(q)$, or $GF_q$ where $q=p^n$ for $p$ a prime. If $q$ is not a prime power, say a composite with distinct primes like $6=2\times 3,$ then there is no field of order $q$.

But $\mathbb{F}_q$ is not $\mathbb{Z}/q$ unless $n=1$ so $q=p$ is itself a prime. So $\mathbb{Z}/2$ and $\mathbb{F}_2$ are the same ring, but $\mathbb{Z}/4$ and $\mathbb{F}_4$ are not. Nor are $\mathbb{Z}/8$ and $\mathbb{F}_8.$ Similarly, $\mathbb{Z}/3$ and $\mathbb{F}_3$ are the same ring, but $\mathbb{Z}/9$ and $\mathbb{F}_9$ are not.

So then what is $\mathbb{F}_4$, the finite field of order four? Well wikipedia has some details, but one construction in brief is $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2+x+1) = \{0,1,x,x+1\}$. So it's residues of polynomials with coefficients in $\mathbb{F}_2$ modulo an irreducible quadratic (of which $x^2+x+1$ is the only one).

One needs to ask whether these rings are isomorphic, meaning the same rings dressed up in different notations. After all they both have four elements, and being the same size is a requirement for isomorphism. But they are not isomorphic.

Note that the characteristic (number of times you add one to itself to get zero) of $\mathbb{Z}/4$ is 4, while the characteristic of $\mathbb{F}_4$ is 2. So all elements of $\mathbb{F}_4$ are of order 2 under addition, while $\mathbb{Z}/4$ has two elements of order 4. The rings' additive groups are not isomorphic.

The multiplicative group of nonzero elements of $\mathbb{F}_4$ is the cyclic group of three elements, whereas the nonzero elements of $\mathbb{Z}/4$ do not form a group at all, since they are not closed under multiplication. In $\mathbb{Z}/4$, $2\times 2 = 0$, whereas in $\mathbb{F}_4$, $x\cdot x=x^2=x+1$ and $(x+1)\cdot (x+1)=(x+1)^2=x^2+2x+1 = (x^2+x+1)+x = x.$ So there is no nilpotent element in $\mathbb{F}_4$, as we discussed in the third paragraph cannot happen in a field. To say it another way, the group of units of $\mathbb{Z}/4$ has cardinality two, $(\mathbb{Z}/4)^\times=\{1,3\},$ while the group of units of $\mathbb{F}_4$ has cardinality three, $\mathbb{F}_4^\times=\{1,x,x+1\},$ and so they are not isomorphic.

So neither the underlying additive groups, nor the multiplicative monoids of these rings are isomorphic, so of course the rings are not isomorphic.

ziggurism
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    $\mathbb{Z}/4$ and $\mathbb{F}_4$ don't have the same additive group either: In the first one we have $1 + 1 = 2 \neq 0$, in the second one we have $1 + 1 = 0$. – Paŭlo Ebermann Oct 23 '16 at 18:19
  • @PaŭloEbermann good point. – ziggurism Oct 23 '16 at 19:23
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    $GF(4)$ == $GF(2^2)$ can be defined as a normal basis tower field:, 1 bit coefficients to $GF(2^8)$ constants. https://math.stackexchange.com/q/4647432/327047 . In this variation of $GF(2^2)$, (1)(1) = (x), (1)(x) = (x+1), (x)(x) = (1), (x+1)(x+1) = (x+1). This variation of $GF(2^2)$ is isomorphic to a conventional $GF(2^2)$. To map from a conventional $GF(2^2)$ to the tower field, multiply elements by the 2 by 2 matrix {{0 1}{1 1}}. To map back, multiply elements by the 2 by 2 matrix {{1 1}{1 0}}. – rcgldr Oct 02 '24 at 02:06
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It is not a field, because $2$ has no inverse for the multiplication law:

$$2\times 0=0,\quad 2\times 1=2,\quad 2\times 2=0,\quad 2\times 3=2.$$

The integers mod $n$ will be a field if, and only if, $n$ is a prime number.

Kan't
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E. Joseph
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If $q=p^n$, then $\mathbf F_q$ denotes the field with $q$ elements.

You have to know that for any integer $n\ge 1$, there exists a finite field with $p^n$ elements, and this field is unique up to an isomorphism.

It is even unique in the still more restrictive sense: a field with $p^n$ elements is unique within a given algebraic closure of the prime field $\mathbf F_p$. Furthermore, for any two such finite fields, $$\mathbf F_{p^m}\subseteq\mathbf F_{p^n}\iff m\mid n.$$

Bernard
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