To construct the finite field GF(23) we need to choose an irreducible polynomial of degree 3. Why we should choose an irreducible polynomial? I don't understand this lemma
Answer
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∈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∈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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