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Birational Geometry of Algebraic Varieties

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Birational Geometry of Algebraic Varieties Birational geometry of algebraic varieties Caucher Birkar Cambridge University Rome, 2019 Algebraic geometry is the study of solutions of systems of polynomial equations and associated geometric structures. Algebraic geometry is an amazingly complex but beautiful subject. It is deeply related to many branches of mathematics but also to mathematical physics, computer science, etc. Algebraic geometry and associated geometric structures. Algebraic geometry is an amazingly complex but beautiful subject. It is deeply related to many branches of mathematics but also to mathematical physics, computer science, etc. Algebraic geometry Algebraic geometry is the study of solutions of systems of polynomial equations Algebraic geometry is an amazingly complex but beautiful subject. It is deeply related to many branches of mathematics but also to mathematical physics, computer science, etc. Algebraic geometry Algebraic geometry is the study of solutions of systems of polynomial equations and associated geometric structures. It is deeply related to many branches of mathematics but also to mathematical physics, computer science, etc. Algebraic geometry Algebraic geometry is the study of solutions of systems of polynomial equations and associated geometric structures. Algebraic geometry is an amazingly complex but beautiful subject. Algebraic geometry Algebraic geometry is the study of solutions of systems of polynomial equations and associated geometric structures. Algebraic geometry is an amazingly complex but beautiful subject. It is deeply related to many branches of mathematics but also to mathematical physics, computer science, etc. Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. Example: t2 − 3t + 1 has no solution in Q p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. Example: t2 + 1 has no solution in Q, not even in R p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Galois theory says: yes if deg f ≤ 4; not otherwise. One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? One variable Let k = Q, or R, or C. Let k[t] = polynomials in variable t with coefficients in k. Given f 2 k[t], we want to find its solutions. p Example: t2 − 3t + 1 has no solution in Q but has solutions in R: (3 ± 5)=2. p Example: t2 + 1 has no solution in Q, not even in R but has solutions in C: ± −1. p Example: t2 + bt + c has solutions in C: (−b ± b2 − 4c)=2. Question: Can we find solutions of a general f in terms of its coefficients (by radicals)? Galois theory says: yes if deg f ≤ 4; not otherwise. Examples: 2 2 t1 + t2 + 1 2 Q[t1; t2] has no solutions over Q but has solutions over C and F3. 2 2 t1 + t2 − 1 2 R[t1; t2] has lots of solutions in R. m m t1 + t2 − 1, for m ≥ 3, has only trivial solutions in Q: when t1 = 0 or t2 = 0. This is Fermat’s last theorem. 2 t2 − t1 and t3 − t1t2 have common solutions in C: 2 3 t1 = a; t2 = a ; t3 = a ; a 2 C: In general it is hard to see if a system of equations has a solution. But it is relatively easy when working with C. Question: can we visualise the set of solutions? Multi-variable 2 2 t1 + t2 + 1 2 Q[t1; t2] has no solutions over Q but has solutions over C and F3. 2 2 t1 + t2 − 1 2 R[t1; t2] has lots of solutions in R. m m t1 + t2 − 1, for m ≥ 3, has only trivial solutions in Q: when t1 = 0 or t2 = 0. This is Fermat’s last theorem. 2 t2 − t1 and t3 − t1t2 have common solutions in C: 2 3 t1 = a; t2 = a ; t3 = a ; a 2 C: In general it is hard to see if a system of equations has a solution.
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