Is $a + b5^{1/4} ; a,b \in \Bbb{Z}$ an Integral domain? Closed under multiplication?

Question

Problem 6d in Section 1.2 from "A Survey of Modern Algebra" - Birkhoff, Mac Lane

Are the following sets of real numbers integral domains? (d) all real numbers $a + b5^{1/4} $ where a and b are integers.

My question is more specifically about what would prove the closure of multiplication if at all. It seems like $a + b5^{1/4} ; a,b \in \Bbb{Z}$ isn't closed under multiplication.

Here is what I've got thus far.

Zero

$(a + b5^{1/4}) + (0 + 0*5^{1/4}) = (a + b5^{1/4})$

Unit

$(a + b5^{1/4})(1 + 0*5^{1/4}) = (a + b5^{1/4})$

Additive Inverse

$(a + b5^{1/4}) + (-a + [-b]5^{1/4}) = (a-a) + (b-b)5^{1/4} = (0 + 0*5^{1/4})$

Closure Under Addition

$x,y \in \Bbb{Z}$

$(a + b5^{1/4}) + (x + y5^{1/4}) = (a+x) + (b+y)5^{1/4}$

$ (a + x) \in \Bbb{Z}, (b+y) \in \Bbb{Z}$

Closure Under Multiplication

$ (a + b5^{1/4})(x + y5^{1/4}) = ax + (ay + xb + by5^{1/4})5^{1/4}$

$(ay + xb + by5^{1/4}) \notin \Bbb{Z}$

I'm fairly confident that it isn't closed for multiplication and therefor not an integral domain but if history has proven anything it's that my level of confidence and level of correctness are not usually correlated. Thank you in advance for taking the time to read my question and answer.

Note: My original post title had incorrect notation. I believed $a + b5^{1/4} ; a,b \in \Bbb{Z}$ to be equivalent to $\Bbb{Z}[5^{1/4}]$

Answer 1

You're right: the set $\;\{a+b\, 5^{1/4}\mid a, b\in\mathbf Z\}$ isn't closed under multiplication: $$(a+b \,5^{1/4})(a'+b' \,5^{1/4})=aa'+(ab'+a'b)\,5^{1/4}+bb'5^{1/2}.$$

Remark: $\;\mathbf Z\bigl[5^{1/4}\bigr]$ is not $\;\{a+b\, 5^{1/4}\mid a, b\in\mathbf Z\}$. Actually, it is $$\{a+b\, 5^{1/4}+c\,5^{1/2}+d\, 5^{3/4}\mid a, b,c,d\in\mathbf Z\}$$ and this one is indeed closed under multiplication, hence an integral domain since it is a subring of the field $\mathbf R$.

Answer 2

You're on the right track, but your proof isn't quite correct. Indeed, the subset $R=\left\{a+b\cdot5^{1/4}\mid a,b\in\mathbb{Z}\right\}$ is not an integral domain because it's not closed under multiplication. By that last part of your work doesn't really demonstrate that.

Note that to demonstrate that some property doesn't always hold true, a specific counterexample (just one!) would be sufficient. So, can you come up with two specific elements $x,y\in R$ such that their product $xy\notin R$? Hint: maybe use $x=5^{1/4}$ and $y=\ldots$

Answer 3

Let $\theta=5^{\frac{1}{4}}$.

The notation $Z[\theta]$ refers to the subring of $\mathbb{C}$ generated by $\mathbb{Z}$ and $\{\theta\}$, or equivalently, the intersection of all subrings of $\mathbb{C}$ which have $\theta$ as an element. Thus, $Z[\theta]$ is automatically a ring, and since it's a subring of $\mathbb{C}$, it's automatically an integral domain ($\mathbb{C}$ is a field, so has no nonzero zero divisors).

But the exercise refers to the subset $S$ of $\mathbb{C}$ defined as $$S = \{a+b\theta\mid a,b\in\mathbb{Z}\}$$ which, as you surmised, is not closed under multiplication.

It suffices to show $\theta^2\notin S$.

Suppose instead that $\theta^2=a+b\theta$ for some $a,b\in\mathbb{Z}$.

Let $f(x) = x^4-5$, and let $g(x)=x^2-bx-a$.

By Eisenstein's criterion, $f$ is irreducible in $\mathbb{Z}[x]$, hence $f$ is also irreducible in $\mathbb{Q}[x]$.

Since $\theta$ is a root of $f$, it follows that $\theta$ is not a root of any nonzero polynomial in $\mathbb{Q}[x]$ with degree less than $4$.

But $\theta$ is also a root of $g$, so we have a contradiction, since $g\in\mathbb{Q}[x]$ and $\deg(g)=2$.

Therefore $\theta^2\notin S$, so $S$ is not closed under multiplication.