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Chapter 2: Problem 33

Factor \(Y^{3}-2 X Y^{2}+2 X^{2} Y+X^{3}\) into linear factors in \(\mathbb{C}[X, Y]\)

Short Answer

Expert verified
Question: Factor the polynomial \(Y^3 - 2XY^2 + 2X^2Y + X^3\) into linear factors in \(\mathbb{C}[X, Y]\). Answer: \((Y + X)(Y^2 - XY + X^2)\)

Step by step solution

01

Identifying the Cubes

Examine the given polynomial \(Y^3 - 2XY^2 + 2X^2Y + X^3\) closely to identify the two terms that are cubed. In this case, the terms are: 1. \(Y^3\) - which means \(Y\) is a cube, i.e. \(Y=Y^1\). 2. \(X^3\) - which means \(X\) is a cube, i.e. \(X=X^1\). Now we rewrite the polynomial as a sum of cubes, so it looks like this: \((Y^3 + X^3)\).
02

Apply the Sum of Cubes Formula

Recall the sum of cubes formula: \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\). Apply this formula to our polynomial \((Y^3 + X^3)\): 1. \((a, b) = (Y, X)\) 2. \(a^3 + b^3 = Y^3 + X^3\) 3. \((Y + X)(Y^2 - X * Y + X^2)\)
03

Factor the Polynomial

Now, we can write the factored polynomial by substituting the terms and applying the sum of cubes formula: \(Y^3 - 2XY^2 + 2X^2Y + X^3 = (Y + X)(Y^2 - XY + X^2)\) So, the given polynomial \(Y^3 - 2XY^2 + 2X^2Y + X^3\) factors into linear factors \((Y + X)\) and \((Y^2 - XY + X^2)\) in \(\mathbb{C}[X, Y]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Geometry
Algebraic Geometry explores the fascinating connection between algebra and geometry by studying polynomial equations and their solutions. In this branch, polynomials are used to define geometric shapes, called algebraic varieties. These can exist in both real and complex dimensions. The link between algebra and geometry allows mathematicians to visualize and solve complex problems more intuitively. Understanding how polynomial factorization works within this field is vital. It helps describe the structure and properties of these varieties.

When we factor a polynomial like the one given, we are breaking down its structure into simpler components. These components, or factors, might correspond to lines or curves on a graph. By working within \(\mathbb{C}[X, Y]\), we access a realm where complex solutions provide new insights into the geometry of the polynomial. This bridges the gap between pure algebraic equations and tangible geometric figures.
Sum of Cubes
The sum of cubes is a useful identity for factoring expressions involving cubed terms. The formula is \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\). It provides a way to factor certain polynomials into a product of simpler expressions. This is particularly helpful in algebraic geometry and other mathematical fields.

In the original problem, identifying \((Y^3 + X^3)\) allowed us to use this important factorization technique. By applying the sum of cubes formula, we break down the expression into linear and quadratic factors. This makes it easier to manipulate and further analyze the polynomial. Understanding this process gives a new perspective on how expressions interconnect and simplifies the solving process.
Complex Polynomial
Complex polynomials involve variables raised to whole number powers with coefficients that may include complex numbers. Working with these can seem challenging, but they provide powerful insights into mathematical problems in fields such as algebraic geometry.

Factoring complex polynomials requires understanding roots and their multiplicity. In \mathbb{C}[X, Y]\, we consider both real and complex solutions. This broadens the potential solutions and connections to geometric interpretations. By factoring the given polynomial as \((Y + X)(Y^2 - XY + X^2)\), each factor represents different dimensions of solutions in the complex plane.

The interplay of complex numbers provides a richer framework, allowing us to explore intricate patterns within algebraic structures.

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Most popular questions from this chapter

Let \(f\) be a rational function on a variety \(V\). Let \(U=\\{P \in V \mid f\) is defined at P\\}. Then \(f\) defines a function from \(U\) to \(k\). Show that this function determines \(f\) uniquely. So a rational function may be considered as a type of function, but only on the complement of an algebraic subset of \(V\), not on \(V\) itself.

Let \(F\) be an irreducible polynomial in \(k[X, Y]\), and suppose \(F\) is monic in \(Y\) : \(F=Y^{n}+a_{1}(X) Y^{n-1}+\cdots\), with \(n>0 .\) Let \(V=V(F) \subset \mathbb{A}^{2}\). Show that the natural homomorphism from \(k[X]\) to \(\Gamma(V)=k[X, Y] /(F)\) is one-to-one, so that \(k[X]\) may be regarded as a subring of \(\Gamma(V)\); show that the residues \(\overline{1}, \bar{Y}, \ldots, \bar{Y}^{n-1}\) generate \(\Gamma(V)\) over \(k[X]\) as a module.

(a)] Let \(\varphi: \mathbb{A}^{1} \rightarrow V=V\left(Y^{2}-X^{3}\right) \subset \mathbb{A}^{2}\) be defined by \(\varphi(t)=\left(t^{2}, t^{3}\right)\). Show that although \(\varphi\) is a one-to-one, onto polynomial map, \(\varphi\) is not an isomorphism. (Hint:: \(\left.\tilde{\varphi}(\Gamma(V))=k\left[T^{2}, T^{3}\right] \subset k[T]=\Gamma\left(\mathbb{A}^{1}\right) .\right)\) (b) Let \(\varphi: \mathbb{A}^{1} \rightarrow V=V\left(Y^{2}-X^{2}(X+1)\right)\) be de fined by \(\varphi(t)=\left(t^{2}-1, t\left(t^{2}-1\right)\right)\). Show that \(\varphi\) is one-to-one and onto, except that \(\varphi(\pm 1)=(0,0)\)

Let \(I=(X, Y) \subset k[X, Y]\). Show that \(\operatorname{dim}_{k}\left(k[X, Y] / I^{n}\right)=1+2+\cdots+n=\frac{n(n+1)}{2}\).

Let \(W\) be a subvariety of a variety \(V\), and let \(I_{V}(W)\) be the ideal of \(\Gamma(V)\) corresponding to \(W\). (a) Show that every polynomial function on \(V\) restricts to a polynomial function on \(W\). (b) Show that the map from \(\Gamma(V)\) to \(\Gamma(W)\) defined in part (a) is a surjective homomorphism with kernel \(I_{V}(W)\), so that \(\Gamma(W)\) is isomorphic to \(\Gamma(V) / I_{V}(W)\)
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