[Number theory] Calculate the Hilbert symbol

In summary: Your Name]In summary, the Hilbert symbol \left( \frac{2,0}{\mathbb F_{25}} \right) over the field with 5² elements is equal to 1. This can be proven by showing that the polynomial f(x) = 2x²-1 has a solution in the splitting field of F_5, which is isomorphic to F_25. This concludes the proof.
  • #1
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Homework Statement


Determine the Hilbert symbol [itex]\left( \frac{2,0}{\mathbb F_{25}} \right)[/itex] where the F denotes the field with 5² elements.

Homework Equations


[itex]\left( \frac{2,0}{\mathbb F_{5}} \right) = -1[/itex]

The Attempt at a Solution


Due to the formula that I put under "relevant equations", we know that the polynomial f(x) = 2x²-1 has NO solution in F_5, hence it is irreducible. Look at the splitting field of this polynomial, call it X. Then by construction [itex]\left( \frac{2,0}{X} \right) = 1[/itex]. Now note that since the degree of f is 2, X is a vector space over F_5 of dimension 2 and hence has to be isomorphic to F_25. This concludes the proof.

(The last part is what I'm unsure about; does it require more argumentation? And is there perhaps an even shorter way of showing that the Hilbert symbol is 1?)
 
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  • #2


Thank you for your question. The Hilbert symbol is a mathematical tool used to determine the solvability of quadratic equations over finite fields. In this case, we are looking at the field with 5² elements, denoted as F_25. The Hilbert symbol is defined as follows:

\left( \frac{a,b}{\mathbb F_{p^n}} \right) = \begin{cases} 1 & \text{if } ax^2+by^2 = 1 \text{ has a solution in } \mathbb F_{p^n} \\ -1 & \text{if } ax^2+by^2 = 1 \text{ has no solution in } \mathbb F_{p^n} \end{cases}

In this case, we are looking at the Hilbert symbol \left( \frac{2,0}{\mathbb F_{25}} \right). Using the relevant equation provided, we can see that \left( \frac{2,0}{\mathbb F_{5}} \right) = -1. This means that the polynomial f(x) = 2x²-1 has no solution in F_5, making it irreducible.

To determine the Hilbert symbol for \left( \frac{2,0}{\mathbb F_{25}} \right), we need to look at the splitting field of f(x) over F_5, denoted as X. Since X is a vector space over F_5 of dimension 2, it must be isomorphic to F_25. Therefore, \left( \frac{2,0}{X} \right) = 1.

This means that the polynomial f(x) = 2x²-1 has a solution in F_25, making the Hilbert symbol \left( \frac{2,0}{\mathbb F_{25}} \right) = 1. This concludes the proof.

I hope this helps to clarify your doubts. Let me know if you have any further questions. Keep up the good work!
 

What is the Hilbert symbol?

The Hilbert symbol is a mathematical function that is used to determine whether a number is a quadratic residue modulo a given prime number.

How is the Hilbert symbol calculated?

The Hilbert symbol is calculated using the Legendre symbol and the quadratic reciprocity law. It involves determining the value of the Legendre symbol for two given numbers and then applying the quadratic reciprocity law to obtain the final result.

What is the purpose of the Hilbert symbol?

The Hilbert symbol is primarily used in algebraic number theory to study quadratic fields. It can also be used to solve certain problems in algebraic geometry and algebraic topology.

What are the applications of the Hilbert symbol in real life?

The Hilbert symbol has applications in cryptography, particularly in the field of elliptic curve cryptography. It is also used in coding theory and error-correcting codes.

Are there any limitations to using the Hilbert symbol?

One limitation of the Hilbert symbol is that it can only be calculated for prime numbers. It also requires knowledge of the quadratic reciprocity law, which can be complex for some numbers.

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