Find the multiplicative inverse of $a+1$ in $\mathbb{Z_2}(a)$.

Question

Let $a$ be a zero of $x^3+x^2+1$ in some extension of $\mathbb{Z}_2$. Find the multiplicative inverse of $a+1$ in this extension.

Attempt: we know that if $F$ is a field and $p(x)$ is irreducible in $F[x]$ and if $a$ is a zero of $p(x)$ then $F(a)\simeq F[x]/\langle p(x)\rangle$. After that how can I proceed?

I believe that you are seeking for $(a+1) ^{-1}$ in the same extension? Otherwise, who guarantees that such an inverse exists in $\Bbb Z _2$? — Alex M., Jun 08 '15 at 12:24
If everything else fails, the field has only $2^3=8$ elements, so you can just try them successively. — hmakholm left over Monica, Jun 08 '15 at 12:24
Not in $\mathbb{Z}_2$, in the extension. Note that $a^2(a+1)=-1$. — André Nicolas, Jun 08 '15 at 12:25
... and the field has characteristic 2, so $-1$ is actually ... — hmakholm left over Monica, Jun 08 '15 at 12:26

score 1 · Answer 1 · answered Jun 08 '15 at 12:37

1

The answer has been posted in the comments($a^2$) but here is a general method for finding inverses:

Let us find the inverse of $a(x)$ in $F[x]/(p(x)$. Assume the inverse is some polynomial $f(x)$ of degree less than $p(x)$. Then solve for $a(x)f(x) = 1$ by reducing the powers of $x$ greater than $d = \text{deg }p(x)$ in the product and use liner algebra to solve. You will have d coefficients and $d$ equations.

answered Jun 08 '15 at 12:37

Asvin

8,229

The inverse is $,-a^2,,$ not $,a^2.,$ The computation is usually done by the extended Euclidean algorithm, not linear algebra - see my answer. – Bill Dubuque Jun 08 '15 at 15:20
1

@BillDubuque (We're working in characteristic $2$.) – Travis Willse Jun 08 '15 at 15:46
@Travis Right, but it works much more generally if you keep the minus sign! – Bill Dubuque Jun 08 '15 at 15:48

Bill Dubuque · Answer 2 · 2020-04-27T20:40:44.197

Hint $\ $ Dividing a polynomial $\,f(x)\,$ by $\, x\!+\!1\,$ yields (via Polynomial Remainder Theorem)

$\qquad\qquad\ \ f(x) - (x\!+\!1) q(x) = f(-1)\ [\, =\color{#c00} 1\ \ {\rm for}\ f = x^3+x^2+1]$

Eval at $\,a$ $\overset{f(a)\,=\,0}\Longrightarrow -(a\!+\!1)\color{#0a0}{q(a)}\, = f(-1)\, =\, \color{#c00} 1\,$ in $\,\Bbb Z_2[a]\ $ so $\,(a\!+\!1)^{-1} = \,\color{#0a0}{-q(a)}$

Generally we see $\ (a\!+\!1)^{-1} = \dfrac{-q(a)}{f(-1)}\ $ if $\,f(-1)\,$ is a unit (invertible), where

$$ q(x)\, =\, \frac{f(x)-f(-1)}{x+1}\, \left[ = \frac{x^3+x^2}{x+1} = x^2\rm\ \ in\ \ OP\right]\qquad$$

Remark $\ $ Generally for coprime polynomials $\,f,g\,$ over a field we can use the extended Euclidean algorithm to obtain the Bezout identity $\,a f + b g = 1\,$ hence $\, g^{-1}\!\equiv b \pmod f.\ $ The above is simply the special case when the Euclidean algorithm terminates in a single step, i.e. when $\, f\ {\rm mod}\ g = c\,$ is constant, i.e. $\ f = qg + c,\,\ \deg c = 0.$

Find the multiplicative inverse of $a+1$ in $\mathbb{Z_2}(a)$.

2 Answers2

Linked

Related