Compute the (multiplicative) inverse of $4x+3$ in the field $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$?

Question

So I am finding a polynomial $px+q$ ($p,q \in \Bbb F_{11}$) which is multiplicative inverse of $4x+3$ in $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$.

i.e. $[(4x+3)+\langle x^2+1 \rangle][(px+q)+\langle x^2+1 \rangle]=1+\langle x^2+1 \rangle$

$\Rightarrow$

$(4x+3)(px+q)+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$

$\Rightarrow$

$4px^2+(4q+3p)x+3q+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$.

We see that the remainder,when $(4x+3)(px+q)$ is divided by $x^2+1$, is $1$. So by Division algorithm,

$$ \require{enclose} \begin{array}{r} 4p \\[-3pt] x^2+1 \enclose{longdiv}{4px^2+(4q+3p)x+3q} \\[-3pt] \underline{4px^2+4p} \\[-3pt] (4q+3p)x+(3q-4p) \\[-3pt] \end{array} $$

So I equate $(4q+3p)x+(3q-4p)=1$ and solve the simultaneous linear equations

$4q+3p=0, 3q-4p=1$. I get $p=6,q=1$

Hence $6x+1$ is the required inverse.

I am pretty sure that the answer is correct but is the method to achieve it right?

Answer 1

Yes, that is very good. And you can check that $(4x+3)(6x+1) = 24x^2 + 22x + 3 = 2x^2 + 3 = 2(-1) + 3 = 1$ as required.

Answer 2

Here's an alternative approach, doable since the vector space here is only two-dimensional. Multiplication by $4x+3$ corresponds to the matrix $\begin{bmatrix}3&7\\4&3\end{bmatrix}$ using the basis $\{1,x\}$. This matrix has inverse $\frac{1}{-19}\begin{bmatrix}3&-7\\-4&3\end{bmatrix}\equiv\frac{1}{3}\begin{bmatrix}3&-7\\-4&3\end{bmatrix}\equiv4\begin{bmatrix}3&4\\7&3\end{bmatrix}\equiv\begin{bmatrix}1&5\\6&1\end{bmatrix}$, which corresponds to multiplication by $1+6x$ (upon inspecting the first column).