Operation Table for Field $F=\{0,1,a,b\}$

Question

I'm taking a mathematics course in my university that relies on past knowledge of an older course I've taken. It's been a long time, and I don't remember much from the other course.

One thing I'm struggling on is coming up with the operation tables for the elements in a field. I know they have something to do with modular arithmetic, but I'm also new to that. Anyway here's what my professor came up with for the addition table:

\begin{array}{|c|c|c|c|} \hline +& 0 & 1 & a & b \\ \hline 0 & 0 & 1 & a & b\\ \hline 1& 1 & 0 & b & a\\ \hline a& a & b & 0 & 1\\ \hline b& b & a & 1 & 0 \\ \hline \end{array}

And this for the multiplication table:

\begin{array}{|c|c|c|c|} \hline * & 0 & 1 & a & b \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & a & b \\ \hline a & 0 & a & b & 1 \\ \hline b & 0 & b & 1 & a \\ \hline \end{array}

I get that adding anything to zero is that element, and multiplying by one is also that element, and multiplying by 0 is zero. What I don't get is how to get the other answers. I'm a bit stuck and any help in the right direction would be greatly appreciated.

Answer 1

We know the multiplication table looks like \begin{array}{|c|c|c|c|} \hline * & 0 & 1 & a & b \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & a & b \\ \hline a & 0 & a & ? & ? \\ \hline b & 0 & b & ? & ? \\ \hline \end{array} so far.

$a^2$ is not $a$, since $a^2 - a = a(a-1)$ is the product of two nonzero elements. Similarly, $b^2$ is not $b$. It is impossible for $a^2$ and $b^2$ to both be $1$: in that case we could argue that $$0 = \underbrace{a^2 - b^2}_{(a-b)(a+b)} = \underbrace{a^2 - 1}_{(a-1)(a+1)} = \underbrace{b^2 - 1}_{(b-1)(b+1)}$$ and the fact that $a-b, a-1, b-1$ are not zero implies $a+1 = b+1 = a+b = 0$ and $a = b = -1$, contradiction.

This lets you complete the multiplication table with $ab = 1$ and so $a^2 = b$ and $b^2 = a.$

Now in the addition table

\begin{array}{|c|c|c|c|} \hline +& 0 & 1 & a & b \\ \hline 0 & 0 & 1 & a & b\\ \hline 1& 1 & ? & ? & ?\\ \hline a& a & ? & ? & ?\\ \hline b& b & ? & ? & ? \\ \hline \end{array}

you can use the multiplication table: since $a^2 - 1 = (a-1)(a+1)$ and $b^2 - 1 = (b-1)(b+1)$ are not zero, it follows that $1+a,1+b$ are not zero, so the additive inverse of $1$ must be $1$ itself.

This implies $a+a = a(1+1) = 0$ and $b+b = b(1+1) = 0.$

Since $1+a$ can't equal $0+a = a$, the only option left is $1+a = b$. Therefore, $1+b = 1+1+a = a$ and $a+b = a + 1+a = 1.$ This completes the table.

Of course there are more abstract ways to get this answer but it is a fun logic exercise this way.

Answer 2

First you should know a finite field has a non-zero characteristic, i.e. a smallest integer $p>1$ such that $p\cdot 1=0$, and this characteristic is a prime number. As a consequence, for any element $x$ in the field, we have $px=0$, since $$px=p(1\cdot x)=(p1)\cdot x=0\cdot.$$ This explains why all elements in diagonal of the addition table are $0$.

All fields of characteristic $p$ contains (a subfield isomorphic to) $\mathbf Z/p\mathbf Z$, usually denoted $\mathbf F_p$ (its prime subfield). A finite field is a vector space over its prime subfield, and thus its cardinality is a power of $p$.

Here, we read in the table the prime subfield is $\mathbf F_2$, and as it has $4$ elements, it is an $\mathbf F_2$-vector space of dimension $2$.

Now let's see why $1+a=b$: note that, since $1+1=0$, we have $-1=1$, more generally, $-x=x$. It can't be $0$ since it would mean $a=1$; nor can it be $1$, as it would mean $a=0$; nor $a$: it would imply $1=0$. The only possibility is $b$.

Let's see multiplication: for similar reasons, $a^2=b$: it can't be $0$ (would imply $a=0$) nor $1$: the polynomial $X^2+1$ ($=X^2+1=(X-1)^2$ already has $1$ as a double root, and in a field, the number of roots of a polynomial is at most its degree. Nor $a$: $a^2=a$, and $a\ne0$ implies $a=1$ by the cancellation law. So, once more there remains the unique possibility $a^2=b$.