Proof of Infinitely Many Units of Z[sqrt n] for Eacn Non-Square Natural Number

[Math Amateur]

In John Stillwell's book: Elements of Number Theory, Exercise 6.1.3 reads as follows:

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Show that \(\displaystyle \mathbb{Z} [ \sqrt{n}\) has infinitely many units for any non-square natural number n

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Can someone please help me get started on this exercise?

Peter

 

[Euge]

In John Stillwell's book: Elements of Number Theory, Exercise 6.1.3 reads as follows:

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Show that \(\displaystyle \mathbb{Z} [ \sqrt{n}\) has infinitely many units for any non-square natural number n

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Can someone please help me get started on this exercise?

Peter

Hint: Apply Dirchlet's unit theorem to $\Bbb Z[\sqrt{n}]$.

 

[mathbalarka]

Hint: Apply Dirchlet's unit theorem to $\Bbb Z[\sqrt{n}]$.

Another relatively diophantine approach would be to look at $x^2 - ny^2 = \pm 1$. Appropriate diophantine approximations on $\sqrt{n}$ can produce at least one solution to $x^2 - ny^2 = 1$, thus producing infinitely many of them by fundamental results on Pell equation.

 

[Math Amateur]

Hint: Apply Dirchlet's unit theorem to $\Bbb Z[\sqrt{n}]$.

Thanks Euge ...

Will have to read up on Dirichlet's Unit Theorem ...

Peter

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Another relatively diophantine approach would be to look at $x^2 - ny^2 = \pm 1$. Appropriate diophantine approximations on $\sqrt{n}$ can produce at least one solution to $x^2 - ny^2 = 1$, thus producing infinitely many of them by fundamental results on Pell equation.

Thanks Mathbalarka ...

Will have to read up on The Pell equation to be able to follow your advice ...

Peter

 

[CellCoree]

Scholze

To begin this exercise, we must first understand the definition of units in the context of number theory. In general, a unit is an element in a ring that has a multiplicative inverse. In the case of the ring \mathbb{Z} [ \sqrt{n} ], the units are elements that can be multiplied with another element in the ring to produce the multiplicative identity 1.

Now, let's consider the non-square natural number n. We know that n is not a perfect square, meaning it cannot be written as the square of any natural number. This implies that \sqrt{n} is irrational, and therefore, it cannot be written as a ratio of two integers.

Next, let's look at the elements in the ring \mathbb{Z} [ \sqrt{n} ]. We can write any element in this ring as a + b \sqrt{n}, where a and b are integers. Since \sqrt{n} is irrational, we know that a + b \sqrt{n} cannot be equal to 0 unless both a and b are 0. This means that every element in the ring except for 0 is a unit, as it has a multiplicative inverse.

But how do we show that there are infinitely many units in this ring? One way to do this is by showing that there are infinitely many irreducible elements in the ring. An irreducible element is an element that cannot be factored into smaller non-unit elements. In the ring \mathbb{Z} [ \sqrt{n} ], an irreducible element is of the form a + b \sqrt{n}, where a and b are integers and a^2 - nb^2 = \pm 1. This can be proven using the unique factorization property of the ring.

Now, since there are infinitely many irreducible elements in the ring, and each irreducible element is also a unit, we can conclude that there are infinitely many units in the ring \mathbb{Z} [ \sqrt{n} ]. This completes the proof that there are infinitely many units in the ring for any non-square natural number n.