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What are the zero divisors of $\mathbb F_2[x]/\langle x^2+1\rangle$?

I think it is not a field, since the polynomial is reducible over $\mathbb F_2$. And because the ring is finite one has to show that it has a zero divisor, is it:

$$\mathbb F_2[x]/\langle x^2+1\rangle =\mathbb F_2[x]/\langle (x+1)^2\rangle $$

I'm stuck here, does the latter imply that $x^2=x=1$, so are there less than $4$ elements, otherwise I cannot find a zero divisor, can you help ?

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    $\begingroup$ Hint: What happens if you multiply $x+1$ with itself in this quotient? $\endgroup$ – Tobias Kildetoft Sep 28 '16 at 8:37
  • $\begingroup$ @TobiasKildetoft but is it not already zero in our ring I mean, $x^2=x=1$ implies $x+1=0$ modulo $2$ ? $\endgroup$ – user1161 Sep 28 '16 at 8:38
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    $\begingroup$ Why do you think $x^2=x$? $\endgroup$ – Eric Wofsey Sep 28 '16 at 8:39
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    $\begingroup$ Where do you get $x^2 = x$ from? $\endgroup$ – Tobias Kildetoft Sep 28 '16 at 8:40
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    $\begingroup$ I think he gets $x^2 = x$ from the fact that $a^2 = a$ for all $a \in \mathbb{F}_2$. The misconception is the belief that a polynomial and the function it defines are the same. $\endgroup$ – Matthias Klupsch Sep 28 '16 at 8:42
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It is true that, over the field with two elements, $x^2+1=(x+1)^2$.

However, it is false that $x^2=x$, because $x$ is intended to be an indeterminate or, in other words, a transcendental element over the field.

You can avoid ambiguities if you try with $F[x]/\langle(x-r)^2\rangle$, where $F$ is any field and $r\in F$.

Any element of the quotient ring can be written in a unique way as $a+bx+\langle(x+1)^2\rangle$, so there are four of them $$ 0+I,\quad x+I,\quad 1+I,\quad 1+x+I $$ where $I=\langle(x+1)^2\rangle$; it shouldn't be difficult to do the check.

For instance, $(1+x+I)^2=0+I$. Is $x+I$ a zero divisor?

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  • $\begingroup$ Yes I thouth too much, $x$ is an indeterminate, you're correct. to your question $(x+I)(x+1+I)=(x^2+x+I)=(x^2+1+(x-1)+I)=x+1+I$ I don't see why it should be, thanks by the way $\endgroup$ – user1161 Sep 28 '16 at 9:30
  • $\begingroup$ @user1161 Consider that $(x+I)^2=1+I$ $\endgroup$ – egreg Sep 28 '16 at 9:42
  • $\begingroup$ apologize but where is $0$, what you've written means that $x+I$ is self inverse $\endgroup$ – user1161 Sep 28 '16 at 9:50
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    $\begingroup$ @user1161 Yes, it equals its inverse, so it can't be a zero divisor $\endgroup$ – egreg Sep 28 '16 at 9:52

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