Problem 3 Find the minimum polynomial of t... [FREE SOLUTION]

Chapter 27: Problem 3

Find the minimum polynomial of the following numbers over the indicated fields: $$ \begin{array}{ll} \sqrt{3}+i & \text { over } \mathbb{R} ; \text { over } \mathbf{Q}: \text { over } \mathbb{Q}(i) ; \text { over } Q(\sqrt{3}) \\ \sqrt{i+\sqrt{2}} & \text { over } \mathbb{R} ; \text { over } Q(i) ; \text { over } \mathbb{Q}(\sqrt{2}) ; \text { over } \mathbb{Q} \end{array} $$

Short Answer

Expert verified

Minimal polynomials are: over $\mathbb{R}$ is $x^2-2\sqrt{3}x+4$, over $\mathbb{Q}$ is $x^4-14x^2+16$, over $\mathbb{Q}(i)$ is $x^2-3$, over $\mathbb{Q}(\sqrt{3})$ is $x^2+1$ for both terms.

Step by step solution

Identify the polynomial with root in $\sqrt{3}+i$ over $\mathbb{R}$

The number $\sqrt{3}+i$ is complex, meaning its conjugate $\sqrt{3}-i$ must also be considered. The polynomial with roots $\sqrt{3}+i$ and $\sqrt{3}-i$ is $(x-(\sqrt{3}+i))(x-(\sqrt{3}-i)) = (x-\sqrt{3})^2 - i^2 = x^2 - 2\sqrt{3}x + 3 + 1 = x^2 - 2\sqrt{3}x + 4.$ This polynomial is irreducible over $\mathbb{R}$.

Identify over $\mathbb{Q}$

In $\mathbb{Q}$, both $\sqrt{3}$ and $i$ are irrational. Therefore, the polynomial with roots $\sqrt{3}+i$, $\sqrt{3}-i$, $-\sqrt{3}+i$, and $-\sqrt{3}-i$ is $(x^2-2\sqrt{3}x+4)(x^2+2\sqrt{3}x+4)=(x^4-14x^2+16).$ This polynomial is irreducible over $\mathbb{Q}$.

Identify over $\mathbb{Q}(i)$

Considering $\mathbb{Q}(i)$, the real part $\sqrt{3}$ is the only consideration for irrationality. We can treat $i$ as a coefficient within the polynomial over this field. Thus, $(x-\sqrt{3})(x+\sqrt{3}) = x^2-3$ is irreducible over $\mathbb{Q}(i)$.

Identify over $\mathbb{Q}(\sqrt{3})$

In $\mathbb{Q}(\sqrt{3})$, the imaginary unit $i$ becomes the concern. Consequently, $(x-i)(x+i) = x^2+1$ is irreducible over $\mathbb{Q}(\sqrt{3})$.

Analyze $\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$

$\sqrt{i+\sqrt{2}}$ is complex, thus its conjugate needs consideration. The corresponding polynomial is $(x-\sqrt{i+\sqrt{2}})(x+\sqrt{i+\sqrt{2}}) = x^2-(i+\sqrt{2}).$ This is reducible and requires further investigation.

Simplify over $\mathbb{Q}(i)$

Given $\mathbb{Q}(i)$, rationalizing involves breaking down $(x-\sqrt{z})(x+\sqrt{z})=x^2-z.$ Since $(x^2-(i+\sqrt{2}))$ does not simplify directly, further transformations may apply depending on the internal square root.

Simplify over $\mathbb{Q}(\sqrt{2})$

Because only $i$ is irrational over this field, we have $(x-\sqrt{z})(x+\sqrt{z})=x^2-(z),$ analogous to $x^4 - 2x^2+2-x\sqrt{2},$ implying a nested transformation results from double-square terms.

Simplify over $\mathbb{Q}$

Given $\mathbb{Q}$, include comprehensive root separation explore after squaring, leading to $(x^2-(i+\sqrt{2}))^2.$ Transform joint conjugates accordingly for minimal polynomial in fractions and base roots.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Complex Numbers

Complex numbers are an extension of the real number system and include all numbers that can be written in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit, defined by $i^2 = -1$. These numbers allow us to solve equations that have no solutions in the real numbers, such as $x^2 + 1 = 0$. When working with complex numbers in polynomials, as seen in our exercise, we often introduce their conjugates.

The conjugate of a complex number $a + bi$ is $a - bi$.
This is helpful because the product of a complex number and its conjugate yields a real number: $(a + bi)(a - bi) = a^2 + b^2$.

We use these properties to form polynomials that are "closed" and can be factored in certain fields, which play a critical role in finding minimum polynomials for complex numbers.

Real Numbers

Real numbers are the numbers commonly used in everyday life, encompassing both rational and irrational numbers. They form a continuum, filling out the number line from negative to positive infinity without gaps. In the context of polynomials, real coefficients imply that all non-real roots must appear as conjugate pairs to ensure the polynomial is defined over real numbers.

In the given exercise, the polynomial $x^2 - 2\sqrt{3}x + 4$ with complex roots is an example of how polynomial equations can have real coefficients even when the roots are complex.
This duo ensures any polynomial of degree higher than one has matching degree and symmetry in its roots.

Thus, while complex numbers might appear as roots in the polynomial, the properties of real numbers ensure the polynomial adheres to the constraints of the field it is defined across.

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. These numbers are dense in the real number line, meaning between any two rational numbers, there is another rational number.

In our exercise, when we consider the field $\mathbb{Q}$, we must consider the irrational components $\sqrt{3}$ and $i$ separately.
This results in a polynomial that captures all interactions between these components鈥攔esulting in higher degrees to accommodate irrationality, as seen in $x^4 - 14x^2 + 16$.

Working within the constraints of rational numbers requires understanding how irrational components affect polynomial expressions and calls for the inclusion of their conjugates and opposite pairs.

Field Extensions

Field extensions are a way to expand a given field to include solutions to equations that couldn't originally be solved within that field. They allow us to work with new elements while preserving the structure of a field.

For example, moving from $\mathbb{Q}$ to $\mathbb{Q}(i)$ involves extending the field to include the unit $i$, allowing solutions involving complex numbers.
This is seen in our exercise where the irreducible polynomial changes based on the field's ability to contain its roots, such as $x^2 - 3$ being irreducible over $\mathbb{Q}(i)$.

The principles behind field extensions help us determine the minimum polynomial by identifying the smallest field in which the polynomial can be factored completely, therefore integrating complex interactions in a logical framework.

91影视

Short Answer

Step by step solution

Identify the polynomial with root in \(\sqrt{3}+i\) over \(\mathbb{R}\)

Identify over \(\mathbb{Q}\)

Identify over \(\mathbb{Q}(i)\)

Identify over \(\mathbb{Q}(\sqrt{3})\)

Analyze \(\sqrt{i+\sqrt{2}}\) over \(\mathbb{R}\)

Simplify over \(\mathbb{Q}(i)\)

Simplify over \(\mathbb{Q}(\sqrt{2})\)

Simplify over \(\mathbb{Q}\)

Key Concepts

Complex Numbers

Real Numbers

Rational Numbers

Field Extensions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Mechanics Maths

Geometry

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.