Question

Show that \(V\left(Y^{2}-X(X-1)(X-\lambda)\right) \subset A^{2}(k)\) is an irreducible curve for any algebraically closed field \(k\), and any \(\lambda \in k\).

Short answer

Question: Prove that the algebraic variety \(V\left(Y^{2}-X(X-1)(X-\lambda)\right) \subset A^{2}(k)\) is an irreducible curve in an algebraically closed field \(k\) for any \(\lambda \in k\). Answer: The polynomial f(X,Y) = \(Y^2 - X(X-1)(X-\lambda)\) is irreducible in the algebraically closed field k, according to Eisenstein's criterion. Therefore, the algebraic variety \(V\left(Y^{2}-X(X-1)(X-\lambda)\right) \subset A^{2}(k)\) is an irreducible curve in k for any \(\lambda \in k\).

Step-by-step solution

Write down the polynomial in the variety

The polynomial in the given variety is: $$ f(X,Y) = Y^2 - X(X-1)(X-\lambda) $$

Perform some factorisation

Let's expand this polynomial: $$ f(X,Y) = Y^2 - (X^3 - (\lambda+1)X^2 + \lambda X) $$

Apply Eisenstein's Criterion

To apply Eisenstein's criterion to the polynomial f(X,Y), we consider the polynomial g(Y) = f(X,0) = -(X^3 - (\lambda+1)X^2 + \lambda X). Then g(Y) has the following coefficients: $$ a_0 = \lambda,\; a_1 = -(\lambda + 1),\; a_2 = 1 $$ Now, let's choose a prime number p=2. Clearly, p does not divide the leading coefficient \(a_2\). Moreover, \(p^2 = 4\) does not divide the constant term \(a_0\) for any \(\lambda\) in k, since k is an algebraically closed field. Thus, using Eisenstein's criterion, we can conclude that g(Y) is an irreducible polynomial in k.

Conclusion

Since g(Y) is an irreducible polynomial in k, the polynomial f(X,Y) must also be irreducible. Therefore, \(V\left(Y^{2}-X(X-1)(X-\lambda)\right) \subset A^{2}(k)\) is an irreducible curve in algebraically closed field k for any \(\lambda \in k\).

Key concepts

Eisenstein's Criterion

Understanding the concept of Eisenstein's criterion is crucial for students dealing with the question of polynomial irreducibility. To give you a clear foundation, Eisenstein's criterion is a tool that helps to determine if a polynomial is irreducible over the field of rational numbers. Let's simplify this complex definition with an easy to follow explanation.

Imagine you have a polynomial with integer coefficients, and you want to check if it can be factored into simpler polynomials also with integer coefficients. Eisenstein's criterion provides a simple set of checks - if a certain prime number divides certain coefficients of the polynomial and does not divide the leading coefficient or the coefficient squared of the constant term, then the polynomial is irreducible.

To apply this to our exercise, consider a polynomial like the one given: If there is a prime number, such as 2 in the step-by-step solution, that matches Eisenstein's criteria, then the polynomial is irreducible over the rationals. In the context of an algebraically closed field, this can sometimes indicate irreducibility as well, as shown in the solution. This is an intuitive way to gauge a polynomial's irreducibility without extensive computation.

Polynomial Factorization

Polynomial factorization is the process of decomposing a polynomial into a product of simpler polynomials that, when multiplied together, give the original polynomial. It's similar to breaking down a number into its prime factors. In algebra, factorizing polynomials is vital since it simplifies many problems, from finding roots to integrating functions.

However, not all polynomials can be easily factored. An irreducible polynomial is like a prime number—it cannot be factored into the product of other polynomials of lower degree with coefficients in the same field. The exercise illustrates a situation where we aim to prove the irreducibility of a polynomial, implying that it cannot be factorized further within the context of the field we're considering.

If we successfully prove that a polynomial is irreducible using Eisenstein's criterion or other methods, this means that in the algebraic field in question, the polynomial cannot be broken down into simpler components. In our case, when we demonstrate that the polynomial defining the curve cannot be factored, it confirms that the curve itself is irreducible.

Irreducibility in Algebraic Geometry

In algebraic geometry, irreducibility has a slightly different meaning than in basic algebra, though it's based on a similar principle. Here, we are concerned with algebraic varieties, which are essentially the sets of solutions to polynomial equations. An irreducible algebraic variety is one that cannot be represented as the union of two smaller nonempty algebraic varieties.

In simpler terms, just as a polynomial is irreducible if it can’t be factored into simpler polynomials, an algebraic curve is irreducible if it can’t be 'split' into simpler curves. The exercise presented is centered around proving the irreducibility of a curve given by a specific polynomial equation. By demonstrating that the polynomial is irreducible in the algebraic field, we are thus also showing that the curve itself is irreducible in algebraic geometry terms.

This property is significant because irreducible curves have nicer properties and are often easier to study, just as prime numbers are fundamental building blocks in number theory. The proof of irreducibility of the curve for any algebraically closed field and any λ in that field underscores the curve's robust and unbreakable nature within the landscape of algebraic geometry.