Quadratic Integers: Understanding the Theorem and Proving 32 = ab in Q[sqrt -1]

In summary: Well if 2, b and sqrt d each are relatively prime to a, then I would say that the two factors are relatively prime but I am not sure. However, I think you have to look to cancel the quadratic part in another way since 4 + 4i and 4 - 4i are not relatively prime. Maybe try playing around with numbers 1,16 and i
  • #1
squelchy451
25
0
For the theorem that states that in quadratic field Q[sqrt d], if d is congruent to 1 mod 4, then it is in the form (a + b sqrt d)/2 and if it's not, it's in the form a + b sqrt d where a and b are rational integers, is it saying that if a and b are rational integers and the quadratic number are in the form according to its congruency mod 4, then the quadratic number is an integer?

Also, how would you prove that if 32 = ab where a and b are relatively prime quadratic integers in Q[sqrt -1], a = e(g^2) where e is a unit and g is a quadratic integer in Q [sqrt -1].
 
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  • #2
squelchy451 said:
For the theorem that states that in quadratic field Q[sqrt d], if d is congruent to 1 mod 4, then it is in the form (a + b sqrt d)/2 and if it's not, it's in the form a + b sqrt d where a and b are rational integers, is it saying that if a and b are rational integers and the quadratic number are in the form according to its congruency mod 4, then the quadratic number is an integer?

Also, how would you prove that if 32 = ab where a and b are relatively prime quadratic integers in Q[sqrt -1], a = e(g^2) where e is a unit and g is a quadratic integer in Q [sqrt -1].

When you have the product of two quadratic integers equal an integer, look first for two of the form A +/- B(sqrt -1). That is one use negative B and the other uses positive B so that the quadratic part cancels out.
 
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  • #3
On a side note, are a + b sqrt d and a - b sqrt d where a, b, and d are rational integers (and d is not a perfect square) relatively prime?
 
  • #4
squelchy451 said:
On a side note, are a + b sqrt d and a - b sqrt d where a, b, and d are rational integers (and d is not a perfect square) relatively prime?

Well if 2, b and sqrt d each are relatively prime to a, then I would say that the two factors are relatively prime but I am not sure. However, I think you have to look to cancel the quadratic part in another way since 4 + 4i and 4 - 4i are not relatively prime. Maybe try playing around with numbers 1,16 and i in lieu of 4 and i.
 
  • #5


The theorem that you mentioned is known as the "Gaussian Integers Theorem" and it states that in a quadratic field Q[sqrt d], where d is a positive integer, if d is congruent to 1 mod 4, then any element can be written in the form (a + b sqrt d)/2, where a and b are rational integers. If d is not congruent to 1 mod 4, then any element can be written in the form a + b sqrt d, where a and b are rational integers.

This theorem does not necessarily state that if a and b are rational integers and the quadratic number is in the form according to its congruency mod 4, then the quadratic number is an integer. It simply provides a way to express elements in a quadratic field in a specific form, which can be useful in certain calculations and proofs.

To prove that 32 = ab where a and b are relatively prime quadratic integers in Q[sqrt -1], we can use the unique factorization property of Gaussian integers. This property states that any Gaussian integer can be expressed as a product of a unit and prime Gaussian integers. In this case, we have 32 = ab, where a and b are relatively prime, which means that they do not share any common factors (other than units).

We can then use this property to write a and b as the product of units and prime Gaussian integers. Let a = e1p1p2...pn and b = e2q1q2...qm, where e1 and e2 are units, p1, p2, ..., pn are prime Gaussian integers and q1, q2, ..., qm are also prime Gaussian integers.

Since 32 = ab, we can equate the two expressions and get e1e2p1p2...pnq1q2...qm = 32. Since 32 is a rational integer, e1e2 must also be a rational integer. This means that e1 and e2 are actually units in the field Q[sqrt -1]. Therefore, we can write a = e(g^2) and b = g, where e is a unit and g is a quadratic integer in Q[sqrt -1].

In conclusion, the Gaussian Integers Theorem provides a useful way to express elements in a quadratic field and the unique factorization property can be used to prove statements about Gaussian integers.
 

1. What are quadratic integers?

Quadratic integers are numbers that can be expressed in the form of a+bi, where a and b are integers and i is the imaginary unit. They are a subset of complex numbers and play an important role in algebraic number theory.

2. How are quadratic integers different from regular integers?

Quadratic integers have the additional factor of the imaginary unit i, which regular integers do not have. This means that quadratic integers have a real and imaginary component, while regular integers only have a real component.

3. What are some properties of quadratic integers?

Quadratic integers have properties similar to regular integers, such as closure under addition, subtraction, and multiplication. However, they do not have the property of being ordered, as they cannot be compared on a number line.

4. How do I find the conjugate of a quadratic integer?

The conjugate of a quadratic integer a+bi is a-bi. This means that the sign of the imaginary component is flipped. For example, the conjugate of 3+2i is 3-2i.

5. Can quadratic integers be factored?

Yes, quadratic integers can be factored just like regular integers. However, the factors may also be quadratic integers and not just regular integers. For example, the quadratic integer 6+5i can be factored into (2+3i)(3+2i).

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