# Does a Finite Field with Four Elements Actually Exist?

dear StackExchange community, so once again I'm confused with this problem concerning finite fields. The given field has four elements with the addition table as

$$\begin{array}{c|c|c|} + & 0 & 1 & x & x + 1 \\ \hline 0 & 0 & 1 & x & x + 1 \\ \hline 1 & 1 & 0 & x + 1 & x \\ \hline x & x & x + 1 & 0 & 1 \\ \hline x + 1 & x + 1 & x & 1 & 0 \\ \hline \end{array}$$

and multiplication table as

$$\begin{array}{c|c|c|} \times & 0 & 1 & x & x + 1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & x & x + 1 \\ \hline x & 0 & x & x + 1 & 1 \\ \hline x + 1 & 0 & x + 1 & 1 & x \\ \hline \end{array}$$

Now the question I'm supposed to answer is "Does a finite field with four elements actually exist?". I know the answer to this question must be yes (via looking at multiple threads here) and I could argue the existence by proving every field axiom, but I don't think that is the wanted solution (proving distribution law would also take forever).

I'm just very confused about the question itself. My course is still very elementary and I don't understand the meaning of a field to exist. Maybe the question is poorly worded? I just don't know. Any help would be nice and thank you in advance.

• Your multiplication table can't be right since $x + 1 \ne 0 \Longrightarrow (x + 1)^2 \ne 0$. How about $(x + 1)^2 = x$? – Robert Lewis Nov 14 '17 at 1:51
• I made a mistake. I corrected it. Thanks. – user396246 Nov 14 '17 at 1:54
• Checking for distributivity and associativity isn't too bad. You can quickly show that $a(b+c)=ab+ac$ is true when any of $a,b,c$ are zero, and that it's true when $a=1$. By commutativity of addition (which is easy to check at a glance on the table), if it's true for $a,b,c$ then it's true for $a,c,b$. So now you only have to check $2\cdot6=12$ possibilities - $2$ possibilities for $a$ other than $0$ and $1$, and $6$ possible choices for $b,c$. – Jack M Nov 14 '17 at 9:18
• Why did you accept an answer that does not address your question, nor shed any light on the mathematics behind finite fields? Are you actually interested in the mathematics? – user21820 Nov 20 '17 at 15:16
• @user21820 I thought it was helpful at least in that moment. Maybe I should have thought about it more carefully. I'm outside right now, but I'll try to look at the answers till tomorrow. I'm sorry if it wasn't the correct choice. – user396246 Nov 20 '17 at 17:57

If you're asked whether something exists, and you want to assert that it does, an effective way is to probably just construct it!

What I mean by this is that you have been asked whether a certain thing exists (a field with $4$ elements), and you want to assert that it does. What you can do is say yes, and prove that what you are asserting is a field with $4$ elements actually is one.

The meaning of "does a field exist?", is exactly what it says. They want you to prove it by giving them an example of one, or disprove it by providing a proof to the contrary.

• @MattS - $\Bbb Z_2 \times \Bbb Z_2$ is used to denote the ring where both multiplication and addition are defined coordinate-wise. Since $(0,1)\cdot(1,0) = (0,0)$, this is not a field. Thus the 4-element field, whose multiplication is different, is commonly denoted by $F_4$. – Paul Sinclair Nov 14 '17 at 3:37
• -1: this answer fails to address the OP's concern. The OP expresses that he could check the axioms, but very reasonably notes that checking associativity and/or distributivity is a huge pain. This answer offers nothing in that light. – Martin Argerami Nov 14 '17 at 5:16
• I agree with you, @MartinArgerami. This is generic advice, not an answer, that could have been more succinct, and best posted as a comment. – Namaste Nov 14 '17 at 16:50
• You may disagree with two commenters, @MattS, who find little value in your answer. Indeed, it looks like, given your absurd comment below the first comment, that you know little on the topic. It's not difference than if I were to say: "Reread your question; read the definition you're learned as to how a field is defined. Make up something so it satisfies the definitions of a field." Substitute (where "field" appears, the word "group" or "determinate ", or "equivalence relation",) and you've got an algorithm to write an unhelpful answer almost anywhere on this site. – Namaste Nov 14 '17 at 18:48
• The intrusion of pop psychology ("irate", "calm down") as a substitute for rational dscussion is rarely a good sign for the validity of the arguments surrounding it. – Did Nov 15 '17 at 9:25

I think you are supposed to verify the axioms. The ones for addition and multiplication are easy-you can just point to the groups that they represent. As you say, distributivity is the hard one. Nominally, given that addition is comutative you have $24$ cases to check, four multiplies times six sums. Half of them are trivial because they are multiplying by $0$ or $1$. That only leaves $12$, which isn't so many. I would do them and declare victory.

A field with $4$ elements is easily built: consider the quotient field $\;\mathbf Z/2\mathbf Z[X]/(X^2+X+1)$. It is a field because the polynomial $X^2+X+1$ is irreducible over $\mathbf Z/2\mathbf Z$, as it is a quadratic polynomial with no root inAlso, as a $\mathbf Z/2\mathbf Z$-vector space, it has dimension $2$, hence its cardinality is $4$.

The tables you provide reflect the laws on this quotient. For instance $x(x+1)=x^2+x\equiv -1$ since $x^2+x+1\equiv0$, and $-1\equiv1$ in $\mathbf Z/2\mathbf Z$.

• I suspect you are using machinery not available to OP. I would guess that the question comes up early in a course in field theory. – Ross Millikan Nov 14 '17 at 1:54
• I suspect that the phrase "My course is still very elementary" (among others) unfortunately implies that this proof will be of little assistance. I find it difficult to believe that OP will understand practically anything that you've provided. – Matt Nov 14 '17 at 1:56
• Maybe you're right. However the argument are very elementary: they suppose the notion of irreducible polynomial, dimension of a vector space and the high school $\mathbf Z/2\mathbf Z$. – Bernard Nov 14 '17 at 1:57
• I completely forgot about modular! Thanks, I think I get it now. I'll try for myself. – user396246 Nov 14 '17 at 2:05
• You had a really nice high school if they got anywhere near the concept of Z/2Z. – user2357112 supports Monica Nov 14 '17 at 7:22

Simply observe that addition and multiplication as defined in the table is equivalent to addition and multiplication of integer polynomials in $x$ modulo $2$ and modulo $x^2+x+1$, and then observe that they commute in a certain sense. $\def\zz{\mathbb{Z}}$

Namely, for any $f,g \in \zz[x]$ define:

$f \oplus g = (f+g) \bmod 2 \bmod (x^2+x+1)$.

$f \odot g = (f·g) \bmod 2 \bmod (x^2+x+1)$.

Then prove that for any $f,g,m \in \zz[x]$ we have:

$f \bmod m \bmod m = f \bmod m$.

$(f+g) \bmod m = ( f \bmod m + g \bmod m ) \bmod m$.

$(f·g) \bmod m = ( ( f \bmod m ) · ( g \bmod m ) ) \bmod m$.

Together these can be easily used to show (for you to do!) that associativity and commutativity and distributivity of $\oplus,\odot$ on the set of polynomials $F = \{0,1,x,x+1\}$ is equivalent to associativity and commutativity and distributivity of $+,·$ on integer polynomials, which is very easy to understand intuitively and prove.

• Note that $F$ is simply the result of taking the whole set of integer polynomials in $x$ modulo $2$ and modulo $x^2+x+1$. This clearly shows that the ring structure is preserved under modulo. So the only thing we will need to check if we want the resulting structure to be a field is whether every element has a multiplicative inverse, since we already started with a commutative ring. And in fact this is the crucial issue. – user21820 Nov 14 '17 at 6:50