# $\mathbb{F}_2[\alpha] \cong \mathbb{F}_2[x]/(x^2+1)$?

My question involves part (b) of Chapter 11 problem 6.4 in Artin's Algebra textbook.

In each case, describe the ring obtained from $\mathbb{F_2}$ by adjoining an element $α$ satisfying the given relation:

(a) $α^2+α+1=0$

(b) $α^2+1=0$

(c) $α^2+α=0$

Now, I obtained that the ring in part (a) is isomorphic to $\mathbb{F_4}$ and that the ring in part (c) is isomorphic to $\mathbb{F_2}\times\mathbb{F_2}$.

It seems to me that the ring in part (b) would be isomorphic to $\mathbb{F}_2[x]/(x^2+1)$, but my teacher doesn't agree.

He said,

"Be careful: notice the polynomial $x^2+1$ is not irreducible over $\mathbb{F}_2$. Adjoining a root of a reducible polynomial is not the same as taking the quotient $\mathbb{F}_2[x]/(x^2+1)$"

So, is my teacher right, or am I? And why?

• I don't see why your teacher disagrees on $\mathbb{F}[\alpha]=\mathbb{F}[\alpha]/(\alpha^2+1)$. $\alpha$ is already in $\mathbb{F}$ and $\alpha=1$ so $\mathbb{F}[\alpha]/(\alpha^2+1)\cong\mathbb{F}$ – cansomeonehelpmeout Feb 19 '18 at 21:05
• @cansomeonehelpmeout I shall disagree, $\mathbb{F}[a]/(a^2+1)$ has an element $a$ in it and hence $\not = \mathbb F$. – mike239x Feb 19 '18 at 21:07
• @cansomeonehelpmeout But $\Bbb{F}_2[x]/(x^2+1)$ is a ring with four elements (by the division algorithm of polynomials). It cannot be the same ring as $\Bbb{F}_2$. – Jyrki Lahtonen Feb 19 '18 at 21:08
• Yes, and the elements are $0+(x^2+1),1+(x^2+1),x+(x^2+1),1+x+(x^2+1)$ – Fakemistake Feb 19 '18 at 21:09
• It seems the teacher is right, since $x^2+1 = (x+1)^2$ in $\mathbb F_2$ (so all its roots are already in $\mathbb F_2$), but I forgot all the corresponding theory and can't really explain why he is right. – mike239x Feb 19 '18 at 21:13

The answer depends on how you interpret "adjoining an element $\alpha$ to $F$ satisfying $f(\alpha) = 0$".

The standard interpretation is: consider the algebraic closure $\bar F$ of $F$, take an element $\alpha \in \bar F$ satisfying $f(\alpha) = 0$ and look at the smallest field containing both $F$ and $\alpha$, i.e, $F(\alpha)$. If $f$ is irreducible, this is isomorphic to $F[x]/(f)$. This makes the answer to (a) ${\Bbb F}_4$ and to (b) and (c) both ${\Bbb F}_2$.

Since Artin says "the ring obtained from ${\Bbb F}_2$ $\dots$", you could also interpret it as: "add a new free element $\alpha$ to $F$, subject only to the constraint $f(\alpha) = 0$". That is, by definition, consider $F[x]/(f)$. With this interpretation, the answer to (a) is still ${\Bbb F}_4$, (b) is (isomorphic to) ${\Bbb F}_2[x]/(x^2)$ and (c) is ${\Bbb F}_2^2$.

• In the second interpretation, why would it be isomorphic to $\mathbb{F}_2[x]/(x^2)$ and not to $\mathbb{F}_2[x]/(x^2+1)$? – Pascal's Wager Feb 19 '18 at 21:46
• @Pascal Those are isomorphic, by sending $x$ to $x + 1$; note that $x^2 + 1 = (x+1)^2$ over ${\Bbb F}_2$. – Magdiragdag Feb 20 '18 at 7:14

Yes, $x^2+1=(x+1)^2$ in $\mathbb{F_2}[x]$ you deduce that $\mathbb{F_2}[x]/(x^2+1)$ is a $2$ dimensiobal vector space over $\mathbb{F_2}$.

Adjoining an element (of the algebraic closure) which satisfies $\alpha^2+1=0$ is adjoining $1$ and the result is $\mathbb{F_2}$.

• Who is right? The teacher or I? – Pascal's Wager Feb 19 '18 at 21:01
• $\frac{\mathbb{F}_2[x]}{(x+1)^2}$ has a nonzero nilpotent (namely $x+1$), but $\mathbb{F}_2 \times \mathbb{F}_2$ doesn't, so they can't be isomorphic. – Richard D. James Feb 19 '18 at 21:23