By Ivan Soprunov

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FIELDS AND POLYNOMIAL RINGS 31 Proof. Consider f, g, h as elements of F(x)[y]. By Gauss’s lemma f is irreducible in F(x)[y]. Now F(x)[y] is the ring of univariate polynomials over a field, so by the usual Euclid’s lemma f |gh implies f |g or f |h in F(x)[y]. 2) g(x, y) = f (x, y)q(x, y) for some q ∈ F(x)[y]. Let’s clear the denominators: there exists c(x) ∈ F[x] such that c(x)q(x, y) ∈ F[x, y]. As before, factor out the gcd of its coeﬃcients to make it primitive: c(x)q(x, y) = d(x)q1 (x, y) for a primitive q1 ∈ F[x][y].

4) (Euclid’s lemma) If p ∈ F[x] is irreducible and p|f g then either p|f or p|g. Moreover when R = Z we have (5) (Gauss’ lemma) If p ∈ Z[x] factors in Q[x] then it factors in Z[x]. How do we define F[x, y], the ring of polynomials in two variables? One way would be to say that F[x, y] consists of finite linear combinations of monomials xi y j with coeﬃcients in F and i ≥ 0, j ≥ 0: � f (x, y) = ai,j xi y j , aij ∈ F, all but finitely many aij are zero. i,j≥0 Another way is to set R = F[x], which is a commutative ring with 1, and define F[x, y] = R[y] = F[x][y].

Recall that in Uz we have u = xz , v = yz , so the two equations become yz − 2 xz = 0 and yz − 2 xz − 2 = 0. If we clear the denominators we obtain y − 2x = 0 and y − 2x − 2z = 0. Notice that now they make sense for z = 0 as well. Thus they define two lines in the projective space: ¯ 1 = {(x : y : z) ∈ P2 | y − 2x = 0}, L ¯ 2 = {(x : y : z) ∈ P2 | y − 2x − 2z = 0}, L which coincide with L1 and L2 on Uz . However each of them has one extra point ¯ 1 and L ¯ 2 intersect in P2 at one point (1 : 2 : 0).

### Algebraic Curves and Codes [Lecture notes] by Ivan Soprunov

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