1
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 $a^7=a^3+a+1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.