What are the elements of $\mathbb{Z}_{2}[x]/I$? Is it a field?

Question

Define $\mathbb{F} = \mathbb{Z}_{2}[x]/I$, where $I$ is the ideal $(1+x+x^{2})$.

I am trying to find all the elements, and explicitly write up a multiplication table. First I notice that $f\in I \implies \deg(f)\geq 2$ or $f = 0$.This leaves us with the polynomials of the form $a+bx$.

So is $\mathbb{F} = \{0+I,1+I,x+I,1+x+I\}$?

If so then $(x+I)(x+I) = x^2 + I = 0+I$ since $x^2 \in I$? Would this mean there is a zero divisor and $\mathbb{F}$ is not a field(even though the question I am looking at claims it is)? I think that $x^2 \not\in I$ but does that means I missed some elements in $\mathbb{F}$?

Answer 1

So is $\mathbb{F} = \{0+I,1+I,x+I,1+x+I\}$?

Yes, that is correct but the reason you gave is a little shaky.

Since $I = (1 + x + x^2)$ it follows that

$$ x^2 \equiv 1 + x \pmod I $$

this is because $x^2 - (1 + x) = x^2 + (1 + x)$ (since $1 = -1$ in $\mathbf{Z}/2$).

The fact that $x^2$ can be replaced by a polynomial of smaller degree means that every element of $\mathbf{Z}/2[x]/I$ can be written as $a + bx$.

If you correct $x^2 \equiv 0$ to $x^2 \equiv 1 + x$ you should be able to determine the correct multiplication table.

Answer 2

The list of element you give is complete. Yet, as you suspect it is not true that $x^2 \in I$.

There might be a slight underlying misunderstanding how the list of elements is obtained.

Two polynomials $P$ and $Q$ are congruent modulo $I$ if their difference lies in $I$ that is if their difference is a multiple of $1+x + x^2$.

In other words every polynomial is congruent to its remainder upon division by $x^2 + x + 1$. This remainder has degree less than two and your list is thus complete.

Thus, $x^2 + x + 1$ is $0$ modulo $I$ but $x^2$ is equal to $x+1 + (x^2+x+1)$ thus $x^2$ is congruent to $x+1$ modulo $I$.