Multiplicative inverse in ${GF}(2^4)$ I want to create a $4\times4$ multiplicative inverse table in $GF(2^4)$. The primitive polynomial given is $P(x)= x^4+x+1$
(NOTE: the values in the table need to be in hexadecimal format, hence I'll be using both polynomial and hexadecimal notations in the question henceforth).
Now, I was able to compute multiplicative inverse for the first row of the matrix i.e. (00,01,02,03). The inverse of 03 or $(x+1)$ comes out to be 0E or $(x^3+x^2+x)$.
However, when I try to compute the inverse of 10 or $x^4$, it again comes out to be 0E or $(x^3+x^2+x)$. Is it possible that two polynomials have exactly the same inverse? If not, I'm unable to figure out where I'm going wrong. Please help.
The Galois Field $\operatorname{GF}(2^4)$ (also represented $\mathbb{F_{2^4}}$) contains $16 = 2 ^4$ elements. The formal definition is;
$\mathbb{F_{2^4}}$ is the quotient ring $\mathbb{F_{2}}[X]/(x^4 = x + 1)$ of the polynomial ring $\mathbb{F_{2}}[X]$ by the ideal generated by $(x^4 = x + 1)$ is a field of order $2^4$.
We can list the elements of $\operatorname{GF}(2^4)$ on the polynomial representation with the defining primitive polynomial, namely $$a_3 x^3+a_2 x^2+a_1 x+a_0$$ where $a_i \in \operatorname{GF}(2)$ for $i=0,1,2,3$.
$\operatorname{GF}(2^4)$ is a Field therefore every element has a unique multiplicative inverse, except the zero.
$x^4$, as we can see, is not an element of the field, however, we can reduce it with the help of the defining polynomial's equation $x^4 = x + 1$. Therefore it has the same representation with $x+1$ in the field, so the inverse is the same.
Also, the multiplication inverse table has $2\times 16$ size, so there is only one row (or column ) to calculate.
\begin{array}{|c|c|}\hline
p(x) \in GF(2^4) &amp; inverse \\ \hline
1  &amp;  1  \\\hline
x  &amp;  x^3 + 1  \\\hline
x + 1  &amp;  x^3 + x^2 + x  \\\hline
x^2  &amp;  x^3 + x^2 + 1  \\\hline
x^2 + 1  &amp;  x^3 + x + 1  \\\hline
x^2 + x  &amp;  x^2 + x + 1  \\\hline
x^2 + x + 1  &amp;  x^2 + x  \\\hline
x^3  &amp;  x^3 + x^2 + x + 1  \\\hline
x^3 + 1  &amp;  x  \\\hline
x^3 + x  &amp;  x^3 + x^2  \\\hline
x^3 + x + 1  &amp;  x^2 + 1  \\\hline
x^3 + x^2  &amp;  x^3 + x  \\\hline
x^3 + x^2 + 1  &amp;  x^2  \\\hline
x^3 + x^2 + x  &amp;  x + 1  \\\hline
x^3 + x^2 + x + 1  &amp;  x^3  \\\hline
\end{array}
The non-zero elements of the field, usually represented by adding a star on the upper right $\mathbb{F}^*_{2^4} = \mathbb{F}_{2^4}- \{0\}$ form a multiplicative cyclic group. $\mathbb{F}^*_{2^4}$ can be generated by $x$, i.e. $\mathbb{F}^*_{2^4} = \langle x \rangle$. The powers of the generator;
\begin{array}{|c|c|}\hline
i &amp; x^i \\ \hline
x^ 1  &amp;  x  \\ \hline
x^{ 2 } &amp;  x^2  \\ \hline
x^{ 3 } &amp;  x^3  \\ \hline
x^{ 4 } &amp;  x + 1  \\ \hline
x^{ 5 } &amp;  x^2 + x  \\ \hline
x^{ 6 } &amp;  x^3 + x^2  \\ \hline
x^{ 7 } &amp;  x^3 + x + 1  \\ \hline
x^{ 8 } &amp;  x^2 + 1  \\ \hline
x^{ 9 } &amp;  x^3 + x  \\ \hline
x^{ 10 } &amp;  x^2 + x + 1  \\ \hline
x^{ 11 } &amp;  x^3 + x^2 + x  \\ \hline
x^{ 12 } &amp;  x^3 + x^2 + x + 1  \\ \hline
x^{ 13 } &amp;  x^3 + x^2 + 1  \\ \hline
x^{ 14 } &amp;  x^3 + 1  \\ \hline
x^{ 15 } &amp;  1  \\ \hline
x^{ 16 } &amp;  x  \\ \hline
\end{array}
$p(x) = 0$ is not included since it has no multiplicative inverse.

Below is the SageMath code used in this answer.
#Base field
R.&lt;y&gt; = PolynomialRing(GF(2), 'y')

#Defining polynomial
G = y^4+y+1

#The field extension
S.&lt;x&gt; = QuotientRing(R, R.ideal(G))
S.is_field()

for p in S:
    if ( p != 0 ):
        print( p, &quot; - &quot;, 1/p )