show that the extension $\mathbb{Z}_{2}(X)\supseteq \mathbb{Z}_{2}(X^2+X)$ is Galois

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let $K=\mathbb{Z}_{2}(X^2+X)$ and $L=\mathbb{Z}_{2}(X)$. I want to show that $L\supseteq K$ is a Galois extension, but I am stuck on finding a minimal polynomial of $X^2-X$ over $K=\mathbb{Z}_{2}(X^2+X)$.

First I want to construct a polynomial that will have all it's zeros in $L=\mathbb{Z}_{2}(X)$:

$p(t)=t(t+1)-X(X+1)\in \mathbb{Z}_{2}(X^2+X)[t]$ then it's zero is $X=-X$ in $\mathbb{Z}_{2}$

but $X\not\in\mathbb{Z}_{2}(X^2+X)$ so the above polynomial must be irreducible and $p(t)=min\;pol_{K} (X)$

$L\supseteq K$ is normal and $p^{\prime}(t)=2t+1=1$ and $(p(t),p^{\prime}(t))=1$ so $L\supseteq K$ is also separable. Hence the extension is Galois.

Is this okay?

Any help appreciated.

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Look's ok. Instead of going the route with $p'$ to show separability, I would just notice that the roots of $p$ are $X$ and $X+1$. This shows immediately that the extension is separable (because $p$ has no repeated roots) and that it's normal (because all roots are in $L$).