Question
Let $F$ be a field of order 32 . Show that the only subfields of $F$ are $F$ itself and $\{0,1\}$.
Step 1
First, recall that a field of order 32 must be an extension field of the form $\mathbb{F}_{2^n}$, where $2^n = 32$. In this case, $n = 5$, so $F = \mathbb{F}_{2^5}$. Show more…
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