Let $X$, $Y$ be indeterminates over $F_2$, the finite field with 2 elements. Let $L = F_2(X, Y )$ and $K = F_2(u, v)$, where $u = X + X^2$, $v = Y + Y^2$.

Explain why $L$ is a simple extension of $K$. Find an element $\gamma \in L$ such that $L = K(\gamma)$. [Hint: First show that $X, Y$ , and $X + Y$ are all algebraic of degree 2 over K.]

I have shown that it is a simple extension but I have trouble finding the primitive element. (I'm preparing for my prelims and I have been stuck at this for almost a week now. Please help me with a hint to solve this)

  • $\begingroup$ By "the primitive element" are you referring to "an element $y\in L$ such that $L = K(y)$"? If so, how did you establish that $L$ is a simple extension of $K$? $\endgroup$ – hardmath May 11 '15 at 18:56

Note that sending $X$ to $X+1$ is a $K$-automorphism of $L$ since this is always an automorphism of $K$ since $X$ is transcendental over $F_2$, and it fixes $u$ since $X+1+(X+1)^2=X+1+X^2+1=X+X^2$. Similarly sending $Y$ to $Y+1$ is a $K$-automorphism. Note that these automorphisms necessarily generate the Galois group of the extension since the extension has degree at most 4 over $K$ since $X$ and $Y$ each have degree 2 over the $K$. Note that $XY$ has four distinct images under the automorphism group, $XY$, $X(Y+1)=XY+1$, $(X+1)Y=XY+Y$, and $(X+1)(Y+1)=XY+X+Y+1$. Therefore $XY$ has minimal polynomial $(t-XY)(t-(XY+X))(t-(XY+Y))(t-(XY+X+Y+1))$ over $K$, and $XY$ is a primitive element for the extension.

To see this more directly note that if we let $w=XY$, then $$\frac{uv-w-w^2}{w}=\frac{XY+XY^2+X^2Y+X^2Y^2-XY-X^2Y^2}{XY}=X+Y,$$ so $X+Y\in K[w]$. In particular $X^2+X+X+Y=X^2+Y\in K[w]$. Multiplying by $X+Y$ gives $X^3+X^2Y+XY+Y^2\in K[w]$ subtracting $XY$ and $Y^2+X^2=(X+Y)^2$ gives $X^3+X^2Y+X^2=X(X^2+XY+X)\in K[w]$. Since $X^2+XY+X\in K[w]$, this implies $X\in K[w]$. Together with $X+Y\in K[w]$, we have $Y\in K[w]$.

  • $\begingroup$ Thanks! I had already shown that the degree of extension is less that or equal to 4 and that L is the splitting field of a separable polynomial over K, to prove that the extension is simple $\endgroup$ – Samantha May 11 '15 at 19:52
  • $\begingroup$ @Samantha Sure, glad to be helpful! :) $\endgroup$ – jgon May 11 '15 at 19:53

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