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To construct the finite field $GF(2^{3})$ we need to choose an irreducible polynomial of degree $3$. Why we should choose an irreducible polynomial? I don't understand this lemma

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up vote 4 down vote accepted

If $K$ is any field and $P$ is a reducible polynomial in $K[X]$, then $K[X]/(P)$ is not an integral domain, and so certainly not a field. Indeed if $P=QR$ with non-constant $Q,R\in K[X]$, then the images of $Q$ and $R$ in $K[X]/(P)$ are nonzero (since their degrees are striclty less than $P$, they cannot reduce modulo $P$). But the product of those images is the image of $P$, which by definition is $0\in K[X]/(P)$, so those images are zero divisors in $K[X]/(P)$. This is why one should always choose irreducible polynomials to construct extension fields.

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