Minimal Polynomial Of √3 + √5: Irreducibility And Reducibility

Hey guys! Let's dive into an intriguing problem in field theory: exploring the minimal polynomial of $\sqrt{3} + \sqrt{5}$ over the rational numbers, $\mathbb{Q}$. We're going to show that this polynomial is irreducible in $\mathbb{Z}[t]$, meaning it can't be factored into non-constant polynomials with integer coefficients. But here's the twist: it becomes reducible modulo any prime $p$. Buckle up, it's going to be a fun ride!

Finding the Minimal Polynomial

First, let's find the minimal polynomial of $\alpha = \sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$. To do this, we'll manipulate the expression to eliminate the square roots.

1. Start with $\alpha = \sqrt{3} + \sqrt{5}$.
2. Square both sides: $\alpha^2 = (\sqrt{3} + \sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}$.
3. Isolate the remaining square root: $\alpha^2 - 8 = 2\sqrt{15}$.
4. Square both sides again: $(\alpha^2 - 8)^2 = (2\sqrt{15})^2 = 4 * 15 = 60$.
5. Expand and rearrange: $\alpha^4 - 16\alpha^2 + 64 = 60$, which simplifies to $\alpha^4 - 16\alpha^2 + 4 = 0$.

So, we have a candidate polynomial: $f(t) = t^4 - 16t^2 + 4$. Now, we need to show that this is indeed the minimal polynomial. We'll do this by proving it's irreducible over $\mathbb{Q}$.

To solidify our understanding, it's crucial to demonstrate that $f(t) = t^4 - 16t^2 + 4$ is genuinely the minimal polynomial for $\sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$. This involves not only finding a polynomial that $\sqrt{3} + \sqrt{5}$ satisfies but also confirming that it's the polynomial of the smallest degree with rational coefficients that does so. We've already shown that $f(\sqrt{3} + \sqrt{5}) = 0$, so the task now is to prove its irreducibility over $\mathbb{Q}$. If we can establish this, then $f(t)$ is guaranteed to be the minimal polynomial because any polynomial of smaller degree that annihilates $\sqrt{3} + \sqrt{5}$ would contradict the definition of a minimal polynomial. The approach we've taken, systematically eliminating square roots and arriving at a quartic polynomial, gives us a strong candidate. However, without proving irreducibility, we can't definitively claim it as the minimal polynomial. This step is not just a formality; it's a cornerstone of the argument, ensuring we have the simplest polynomial necessary to describe the algebraic nature of $\sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$. Thus, the subsequent proof of irreducibility is not merely an exercise but a crucial validation of our claim that $f(t)$ is indeed the minimal polynomial we seek. This careful and methodical approach underscores the rigor required in abstract algebra, where claims must be supported by solid proofs.

Irreducibility over \mathbb{Z}[t]

We can use Eisenstein's Criterion to prove the irreducibility of $f(t)$ over $\mathbb{Z}[t]$. Eisenstein's Criterion provides a powerful tool for determining the irreducibility of polynomials with integer coefficients. The criterion essentially states that if there exists a prime number that divides all the coefficients of the polynomial except the leading coefficient, and the square of the prime does not divide the constant term, then the polynomial is irreducible over the rational numbers. This criterion is particularly useful because it transforms the abstract question of irreducibility into a concrete check of divisibility properties of the polynomial's coefficients. In the case of our polynomial, $f(t) = t^4 - 16t^2 + 4$, applying Eisenstein's Criterion with the prime $p = 2$ allows us to bypass more complex irreducibility tests. This simplicity is one of the key reasons why Eisenstein's Criterion is a favorite among mathematicians when dealing with polynomials of this form. However, it's important to remember that while Eisenstein's Criterion is sufficient to prove irreducibility, it is not necessary. That is, if the criterion does not apply, it doesn't necessarily mean the polynomial is reducible; it simply means we need to explore other methods to determine its irreducibility. The elegance and efficiency of Eisenstein's Criterion make it a valuable asset in our algebraic toolbox, but understanding its limitations is equally important for a comprehensive approach to polynomial irreducibility.

• Consider the prime $p = 2$.
• 2 divides -16 and 4.
• 2 does not divide the leading coefficient (which is 1).
• $2^2 = 4$ does not divide 4.

Therefore, by Eisenstein's Criterion, $f(t)$ is irreducible over $\mathbb{Q}[t]$. Since $f(t)$ is a monic polynomial in $\mathbb{Z}[t]$, it is also irreducible over $\mathbb{Z}[t]$ (by Gauss's Lemma). This step is critical in establishing the fundamental algebraic properties of $f(t)$. By demonstrating that $f(t)$ is irreducible over $\mathbb{Z}[t]$, we're asserting that it cannot be factored into the product of two non-constant polynomials with integer coefficients. This fact is not just an isolated observation; it has far-reaching implications for understanding the algebraic structure of the extension field $\mathbb{Q}(\sqrt{3} + \sqrt{5})$. Specifically, the irreducibility of $f(t)$ over $\mathbb{Z}[t]$ confirms that $f(t)$ is indeed the minimal polynomial of $\sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$, as any polynomial of smaller degree with $\sqrt{3} + \sqrt{5}$ as a root would contradict this irreducibility. Moreover, the irreducibility over $\mathbb{Z}[t]$ is a stronger statement than irreducibility over $\mathbb{Q}[t]$ because it takes into account the finer structure of integer coefficients. This distinction is crucial when considering the behavior of the polynomial modulo primes, which is the next fascinating aspect we'll explore. So, this proof of irreducibility over $\mathbb{Z}[t]$ sets the stage for understanding how the polynomial's properties change when viewed through the lens of modular arithmetic.

Reducibility Modulo Any Prime p

Now comes the really cool part! We want to show that $f(t)$ becomes reducible modulo any prime $p$. This means that for any prime number $p$, the polynomial $f(t)$ can be factored into non-constant polynomials when we consider its coefficients modulo $p$.

This part requires a bit more work and casework.

Case 1: p = 2

Modulo 2, $f(t) \equiv t^4 - 16t^2 + 4 \equiv t^4 \equiv (t^2)^2 \pmod{2}$. Clearly, $f(t)$ is reducible modulo 2.

Case 2: p = 3

Modulo 3, $f(t) \equiv t^4 - 16t^2 + 4 \equiv t^4 - t^2 + 1 \pmod{3}$. We can check that $f(t)$ has no roots in $\mathbb{Z}_3$ (by plugging in 0, 1, and 2). If $f(t)$ were reducible, it would have to factor into two irreducible quadratic polynomials. The only irreducible quadratic polynomial in $\mathbb{Z}_3[t]$ is $t^2 + 1$. However, $(t^2 + 1)^2 = t^4 + 2t^2 + 1 \equiv t^4 - t^2 + 1 \pmod{3}$. So, $f(t) \equiv (t^2 + 1)^2 \pmod{3}$, and $f(t)$ is reducible modulo 3. This reducibility modulo primes is a fascinating phenomenon that highlights the interplay between number theory and polynomial algebra. When we say $f(t)$ is reducible modulo 3, we mean that when we consider the coefficients of $f(t)$ in the field of integers modulo 3, the polynomial can be factored into lower-degree polynomials. This is a significant shift from its irreducibility over $\mathbb{Z}[t]$, where no such factorization is possible. The fact that the minimal polynomial of $\sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$ exhibits this behavior is not just a curiosity but a reflection of deeper algebraic properties. It suggests that the structure of the field extension $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ has certain characteristics that manifest differently when viewed through the lens of modular arithmetic. This is a common theme in algebraic number theory, where the behavior of algebraic objects modulo primes often reveals crucial information about their global structure. In this specific case, the reducibility of $f(t)$ modulo 3, and indeed modulo any prime, points to the way the prime ideals in the ring of integers of $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ behave, offering insights into the arithmetic of this field. Thus, the observation that $f(t) \equiv (t^2 + 1)^2 \pmod{3}$ is not just a computational result but a window into the intricate algebraic landscape of number fields.

Case 3: p = 5

Modulo 5, $f(t) \equiv t^4 - 16t^2 + 4 \equiv t^4 + 4t^2 + 4 \equiv (t^2 + 2)^2 \pmod{5}$. Again, $f(t)$ is reducible modulo 5.

Case 4: p > 5

This is the trickiest case. We'll use the fact that if $f(t)$ has a root modulo $p$, then it's reducible. Also, if $f(t)$ factors into two irreducible quadratics modulo $p$, it's reducible.

We know that $\alpha = \sqrt{3} + \sqrt{5}$ is a root of $f(t)$. Let's consider the roots of $f(t)$: $\pm \sqrt{3} \pm \sqrt{5}$.

Now, consider the quadratic fields $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{5})$, and $\mathbb{Q}(\sqrt{15})$. For $f(t)$ to be irreducible modulo $p$, both 3 and 5 must be quadratic non-residues modulo $p$ (otherwise, $\sqrt{3}$ or $\sqrt{5}$ would exist in $\mathbb{Z}_p$, making $f(t)$ have a root). Moreover, 15 must also be a quadratic non-residue modulo $p$. This part of the analysis represents a significant step up in complexity, as it requires us to delve into the world of quadratic residues and non-residues modulo primes. The core idea here is to understand how the existence of square roots of 3, 5, and 15 modulo $p$ impacts the reducibility of $f(t)$. If either 3 or 5 is a quadratic residue modulo $p$, it means that there exists an integer $x$ such that $x^2 \equiv 3 \pmod{p}$ or $x^2 \equiv 5 \pmod{p}$. In this case, $\sqrt{3}$ or $\sqrt{5}$ effectively exists in the finite field $\mathbb{Z}_p$, which makes $f(t)$ have a root modulo $p$ and thus reducible. Similarly, if 15 is a quadratic residue modulo $p$, it implies the existence of $\sqrt{15}$ in $\mathbb{Z}_p$, which also leads to the reducibility of $f(t)$. The condition that all three, 3, 5, and 15, must be quadratic non-residues for $f(t)$ to potentially remain irreducible is a stringent requirement. It highlights the delicate balance between the arithmetic of the integers and the structure of polynomials modulo primes. This connection is not merely a technical detail; it's a manifestation of the deep relationships that exist between number theory and abstract algebra. The use of quadratic residues and non-residues provides a powerful lens through which to examine the reducibility of polynomials, revealing how the seemingly simple question of factorization is intertwined with the fundamental properties of prime numbers and modular arithmetic.

By the properties of Legendre symbols, we have $\left( \frac{15}{p} \right) = \left( \frac{3}{p} \right) \left( \frac{5}{p} \right)$. So, if both 3 and 5 are quadratic non-residues, then 15 is a quadratic residue. This is a crucial observation that dramatically simplifies our analysis. The Legendre symbol, denoted as $\left( \frac{a}{p} \right)$, is a powerful tool in number theory that concisely captures whether an integer $a$ is a quadratic residue modulo a prime $p$. It takes the value 1 if $a$ is a quadratic residue modulo $p$, -1 if $a$ is a quadratic non-residue, and 0 if $p$ divides $a$. The multiplicative property of the Legendre symbol, which states that $\left( \frac{15}{p} \right) = \left( \frac{3}{p} \right) \left( \frac{5}{p} \right)$, is a direct consequence of the definition and properties of quadratic residues. This property is particularly useful in our case because it allows us to relate the quadratic residuosity of 15 to that of 3 and 5. Specifically, if both 3 and 5 are quadratic non-residues modulo $p$, then their Legendre symbols are both -1. When we multiply these symbols together, we get $(-1) \times (-1) = 1$, which means that 15 is a quadratic residue modulo $p$. This result is not just a mathematical curiosity; it has profound implications for the reducibility of our polynomial $f(t)$. It tells us that we cannot simultaneously have 3 and 5 as quadratic non-residues without 15 being a quadratic residue. This realization effectively closes the door on the possibility of $f(t)$ remaining irreducible modulo $p$ under these conditions. Thus, the multiplicative property of the Legendre symbol provides a concise and elegant way to demonstrate the reducibility of $f(t)$ for all primes $p > 5$, showcasing the deep and interconnected nature of number theory.

Therefore, it is impossible for 3 and 5 to both be quadratic non-residues modulo $p$ while also having 15 as a quadratic non-residue. This means that for any prime $p > 5$, at least one of 3, 5, or 15 is a quadratic residue, so $f(t)$ has a root in some quadratic extension of $\mathbb{Z}_p$ and is therefore reducible modulo $p$.

Conclusion

We've shown that the minimal polynomial of $\sqrt{3} + \sqrt{5}$ over $\mathbb{Q}$ is $f(t) = t^4 - 16t^2 + 4$. This polynomial is irreducible in $\mathbb{Z}[t]$ but becomes reducible modulo any prime $p$. This fascinating example highlights the interplay between field theory, Galois theory, and number theory. It demonstrates how a polynomial can have drastically different properties depending on the field over which it is considered. Isn't that just awesome?