Question

Let \(F\) be a field. (a) Show that for all \(b \in F,\) we have \(b^{2}=1\) if and only if \(b=\pm 1\). (b) Show that for all \(a, b \in F,\) we have \(a^{2}=b^{2}\) if and only if \(a=\pm b\). (c) Show that the familiar quadratic formula holds for \(F\), assuming \(F\) has characteristic other than 2 , so that \(2_{F} \neq 0_{F}\). That is, for all \(a, b, c \in F\) with \(a \neq 0,\) the polynomial \(g:=a X^{2}+b X+c \in F[X]\) has a root in \(F\) if and only if there exists \(e \in F\) such that \(e^{2}=d,\) where \(d\) is the discriminant of \(g,\) defined as \(d:=b^{2}-4 a c,\) and in this case the roots of \(g\) are \((-b \pm e) / 2 a\)

Short answer

Question: Prove that elements with square equal to 1 are precisely ±1 in fields. Answer: If a field element has a square equal to 1, it means that either the element is 1 or its additive inverse -1. This follows from the fact that (b - 1)(b + 1) = b² - 1² = 0, and since F is a field, either b - 1 = 0, meaning b = 1, or b + 1 = 0, meaning b = -1.

Step-by-step solution

(a) Show that b² = 1 if and only if b = ±1.

To show this, we'll first assume that \(b^2 = 1\) and show that \(b\) must be either \(1\) or \(-1\). Then, we'll show that if \(b\) is either \(1\) or \(-1\), then \(b^2 = 1\). Assume \(b^2 = 1\). We know that \((-1)^2 = 1\) as well, so \((b - 1)(b + 1) = b^2 - 1^2 = 0\). Since F is a field, either \(b - 1 = 0\), meaning \(b = 1\), or \(b + 1 = 0\), meaning \(b = -1\). Now, we'll show the converse. If \(b = 1\), then \(b^2 = 1^2 = 1\). If \(b = -1\), then \(b^2 = (-1)^2 = 1\). Therefore, \(b^2 = 1\) if and only if \(b = 1\) or \(b = -1\).

(b) Show that \(a^2 = b^2\) if and only if \(a = ±b\).

To show this, we'll first assume that \(a^2 = b^2\) and show that \(a\) must be either \(b\) or \(-b\). Then, we'll show that if \(a\) is either \(b\) or \(-b\), then \(a^2 = b^2\). Assume \(a^2 = b^2\). Then \((a - b)(a + b) = a^2 - b^2 = 0\). Since F is a field and \(0\) has a unique inverse, either \(a - b = 0\), meaning \(a = b\), or \(a + b = 0\), meaning \(a = -b\). Now, we'll show the converse. If \(a = b\), then \(a^2 = b^2\). If \(a = -b\), then \(a^2 = (-b)^2 = b^2\). Therefore, \(a^2 = b^2\) if and only if \(a = b\) or \(a = -b\).

(c) Show that the quadratic formula holds for F.

Assuming \(F\) has characteristic other than 2, this means that \(2_F \neq 0_F\). Firstly, let's show that the discriminant of \(g\) defined as \(d:= b^2 - 4ac\) exists in the field: \(d = b^2 - 4ac = b^2 + (-2a)(-2c)\). Since \(F\) is a field and \(2_F \neq 0_F\), \(-2a\) and \(-2c\) also have unique inverses, so \(d\) exists in \(F\). Now assuming that there exists \(e \in F\) such that \(e^2 = d\), we can write the roots of \(g\) as: \(x = \frac{-b \pm e}{2a}\). To prove that these are indeed the roots of \(g\), we must show that: \(g(x) = ax^2 + bx + c = 0\) Plug in the values of \(x\): \(g(x) = a\left(\frac{-b \pm e}{2a}\right)^2 + b\left(\frac{-b \pm e}{2a}\right) + c\). Now, let's simplify by factoring out \(a\) and combining terms: \(g(x) = \frac{(-b \pm e)^2}{4a} + \frac{(-b \pm e)b}{2a} + c\) Let's simplify by dividing by \(2a\): \(g(x) = \frac{b^2 \pm 2eb + e^2}{4a} + \frac{-2b^2 \pm 2eb}{4a} + \frac{4ac}{4a}\) Note that \(e^2 = d = b^2 - 4ac\). Combining terms: \(g(x) = \frac{b^2 \pm 2eb + e^2 - 2b^2 \pm 2eb + 4ac}{4a}\) \(g(x) = \frac{-b^2 + e^2 + 4ac}{4a} = \frac{-b^2 + b^2 - 4ac + 4ac}{4a} = \frac{0}{4a} = 0\) So, the roots of \(g\) are indeed \((-b \pm e) / 2a\) when there exists \(e \in F\) such that \(e^2 = d\).

Key concepts

Field Theory

Field theory is a fascinating area of algebra that deals with the study of fields - mathematical structures in which we can perform addition, subtraction, multiplication, and division without leaving the set. A field, denoted in our exercise as 'F', has certain properties that make it similar to standard number systems like the integers, rationals, or reals.

For instance, within a field, we have notions of '0' and '1', known as the additive and multiplicative identities, respectively. These elements behave in the way you would expect: any number added to '0' or multiplied by '1' in the field remains unchanged. Additionally, each element (except for '0') has an inverse, meaning that for any non-zero element 'a' in the field, there exists an element 'b' such that 'a * b = 1'.

The concept of field theory extends to various mathematical systems and is crucial for understanding polynomial equations. In field theory, we can discuss quadratic formulas, solution sets, and other properties in a broad sense, applying to different kinds of fields beyond the familiar real numbers.

Polynomial Roots

Speaking of polynomial roots, these roots are solutions to polynomial equations when set equal to zero. Specifically, in the context of a quadratic polynomial 'g' with coefficients in a field 'F', we are interested in values 'x' in 'F' that satisfy the equation 'g(x) = 0'.

These roots have particular importance because they represent the points at which the graph of the polynomial would intersect the horizontal axis when plotted. The roots can be real or complex numbers, but in a general field setting, there's no guarantee that roots exist within the field unless specific conditions are met. It's interesting to note that in the quadratic case, if the field 'F' has a characteristic different from 2, the usual quadratic formula can help us find the roots of the polynomial within the field, assuming a solution exists based on the discriminant 'd'.

Discriminant of a Quadratic

The discriminant of a quadratic is a key concept that tells us about the nature of the roots of a quadratic polynomial without actually solving the equation. In essence, the discriminant 'd', given by the formula 'd = b^2 - 4ac' where 'a', 'b', and 'c' are coefficients of the quadratic polynomial 'ax^2 + bx + c', can indicate the number and type of roots.

If the discriminant is positive, there are two distinct roots, and if it's zero, there is exactly one root (a repeated root). A negative discriminant suggests that there are no real roots, but rather two complex conjugate roots. However, this interpretation depends on whether our field 'F' accommodates complex numbers or not. It is this discriminant that must be a perfect square in the field for the quadratic formula to provide solutions within the field, as seen in the exercise.

Characteristic of a Field

The characteristic of a field is an intrinsic property that informs us about the behavior of the number '1', or more precisely, how many times one must add '1' to itself to get '0'. This is a critical concept because the charaacteristic affects the structure of the field and influences which algebraic properties hold.

Fields can have either characteristic 0, which means that no matter how many times you add '1' to itself, you won't get '0', much like the rational or real numbers. Otherwise, they have a positive characteristic 'p', which is always a prime number. This tells us that adding '1' to itself 'p' times yields '0'. In our problem, it was essential that the field in question did not have characteristic 2 because that would interfere with the solution's reliance on division by '2', a crucial step in the quadratic formula.