Show that the ring $R = \mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1)$ is a field. Find the multiplicative inverse of the element [x^2+1] in that field.

From what I know,

$$\mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1) = \{ a_0 + a_1x+a_2x^2+a_3x^3 + x^4 : a _i = \{0,1\}\}$$

To show that $R$ is a field, I would need to show that it is a commutative division ring.

How can I show that formally? Especially the fact that every non-zero element of $R$ has an inverse with respect to multiplication.

  • 4
    $\begingroup$ Do you know that irreducible elements generate maximal ideals? And $R/I$ is a field if $I$ is maximal? $\endgroup$ – Joe Johnson 126 Nov 3 '13 at 19:04
  • $\begingroup$ First, "from what you know", is wrong: the quotient ring is not a set of polynomials, but of equivalence classes of polynomials...and all the classes have a representative which is a polynomial of degree *at most * three, not four. $\endgroup$ – DonAntonio Nov 3 '13 at 19:35

Why don't prove that $x^4+x+1$ is irreducible over $\Bbb Z/2\Bbb Z$?

Denote $a=[x]$. We have $a^4=a+1$, so $a^8=a^2+1$. But $a^{15}=1$ (since $R$ is a field with $2^4=16$ elements and $a\in R^{\times}$), so $a^7$ is the inverse of $a^2+1$. Now, from $a^4=a+1$ we find $a^7=a^3+a+1$.

  • $\begingroup$ Some care is needed here, I think: $\;a^{15}=1\;$ since $\;|R^*|=15\;$ and it is a multiplicative group. We don't need neither that this group is cyclic nor that it is generated by $\;a\;$ $\endgroup$ – DonAntonio Nov 3 '13 at 19:39
  • $\begingroup$ $|R^*|$ could at that point still be smaller and could even be not even a divisor of 15: if you'd started with the (reducible) polynomial $(x^2 + x + 1)^2$, $|R^*|$ would have been 9, since $R$ would have been isomorphic to ${\mathbb F}_4 \times {\mathbb F}_4$. So the given answer is incorrect, as it contains a circular reasoning. Still, the approach, showing that $x^4 + x + 1$ is irreducible, is by far the easiest way to go about this. $\endgroup$ – Magdiragdag Nov 3 '13 at 19:43
  • $\begingroup$ @Magdiragdag Your comment has no meaning to me. First I proved that $R$ is a field with $16$ elements since it is a field and an $\Bbb F_2$-vector space of dimension $4$. Then I computed the inverse. Where have you seen a "circular reasoning"? $\endgroup$ – user89712 Nov 3 '13 at 20:06
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    $\begingroup$ @user You are right; from your first sentence I thought you were going to prove that $x^4 + x + 1$ irreducible. You're not - of course - but you're suggesting the OP does so. After that, your computation of the inverse of $x^2+1$ is correct. $\endgroup$ – Magdiragdag Nov 3 '13 at 20:25

You are a little wrong. Since the ideal is generated by $x^4+x+1$, you are saying that $$ x^4+x+1=0 $$ in the quotient ring. In $\mathbb{Z}/2$ coefficients, this is the same as $$ x^4=x+1. $$ So, you have a generating set $1,x,x^2,x^3$. If you don't have any theorems to show that the quotient ring is a field, you need to check that each element is invertible. We can reduce our work a bit. Namely, if any nonzero element is a product of two invertible elements, then it is also invertible. First notice that $$ x(x^3+1)=x^4+x=x+1+x=1. $$ Thus $x$ is invertible. Then all powers of $x$ are invertible as long as they are nonzero. As in @user's answer, you can check, using $x^4=x+1$, that the set $\{x^n\,\,|\,\, n=0,\ldots 14\}$ generates all nonzero elements of the quotient ring.


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