Determine whether a ring is local

Question

Is the ring $$ \text{Span}\left\{\text{id}_{2x2},\begin{pmatrix} 0 & t \\ 1 & 0\end{pmatrix}\right\}$$ a local ring on $\mathbb{F}_{2}(t)$? I'm having some troubles with this...

Answer 1

It's local but for the boring reason that it is a field.

If $X=\begin{bmatrix}0&t\\1&0\end{bmatrix}$

then $X^{-1}=\begin{bmatrix}0&1\\t^{-1}&0\end{bmatrix}$ is a unit.

Each element has the form $\begin{bmatrix}a&bt\\b&a\end{bmatrix}$ where $a,b\in F_2(t)$. Clearly if $a=0$ and $b\neq 0$, or $a\neq 0$ and $b=0$, this is a unit.

If $a\neq0\neq b$, then the determinant $a^2+b^2t\neq 0$ either. Let us see why. By multiplying both sides judiciously with elements of $F_2[t]$ we can replace the equation with one in which $a,b$ are nonzero elements of $F_2[t]$, and in that case $a^2=b^2t$ is clearly impossible because the degree of the left side is even and the right side is odd. Given that $a^2+b^2t\neq 0$ the adjugate inverse of $\begin{bmatrix}a&bt\\b&a\end{bmatrix}$ is available.

Therefore all nonzero elements are invertible, and you have a field.