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هل يمكن حل كل معادلة جبرية؟

هل تساءلت لماذا لا يمكن حل المعادلة (x^5 - x + 1 = 0) بالأساليب traditional مثل المعادلات من الدرجات أقل؟ نظرية غالوا توفّر الإجابة — عبر مفاهيم من الجبر abstract: الحقول (fields) والمجموعات (groups). غالوا، الذي مات في 20 سنة فقط، developed هذه النظرية لحل question: أي المعادلات يمكن حلها بالأساليب traditional (using radicals)؟

Definition: حقل (Field): مجموعة مع عمليات addition و multiplication، مثل الأعداد real أو complex. example: \(\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} | a,b \in \mathbb{Q}\}\).

أساسيات: الحقول والمجموعات

نبدأ بkey concepts:

  1. توسعات الحقول (Field extensions): إذا كان K حقل، و L حقل يتضمن K، ن say L هو extension من K. example: (\mathbb{Q}(\sqrt{2})) هو extension من (\mathbb{Q}).
  2. المجموعات الجالوية (Galois groups): مجموعة من automorphisms (تحويلات preserves operations) من حقل إلى نفسه، تترك fixed field fixed.

Key point: Galois group of a polynomial tells us whether it can be solved by radicals. If it's not solvable (like for degree 5), then the polynomial isn't solvable by radicals.

Example: Take \(f(x) = x^2 - 2\). Roots are \(\sqrt{2}, -\sqrt{2}\). Galois group is \(\mathbb{Z}/2\mathbb{Z}\) (swap the roots), which is solvable.

Polynomial Galois Group Solvable?
(x^2 - 2) (\mathbb{Z}/2\mathbb{Z}) Yes
(x^3 - 2) (S_3) No
(x^4 - 2) (D_4) Yes

Fundamental Theorem of Galois Theory

This theorem connects field extensions and group theory:

$$ \text{If } K \subseteq L \subseteq M \text{ are fields, then there's a bijection between subfields of } M \text{ containing } L \text{ and subgroups of } \text{Gal}(M/L). $$

Warning: Don't confuse Galois groups with permutation groups! Galois groups are groups of automorphisms, not just any permutations.

Practice: Find Galois group of (x^3 - 2)

  1. Roots are ( \sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2} ) where (\omega) is cube root of unity.
  2. The Galois group has 6 elements (like (S_3)), so it's not solvable. Thus, this equation can't be solved by radicals.

Summary

Key point: If the Galois group of a polynomial is not solvable, then the polynomial can't be solved by radicals. For example, degree 5 polynomials often have unsolvable groups.

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