Herstein's Definition: In the Euclidean ring $R$, a nonunit $\pi$ is said to be a prime element of $R$ if whenever $\pi=ab$, where $a,b \in R$, then one of $a$ or $b$ is a unit in R.
$\mathbb Q$ is a field and hence a Euclidean ring. In $\mathbb Q$ every element except $0$ is a unit. So there are no prime elements in $\mathbb Q$?
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Yes, that's right: $\mathbb{Q}$ has no prime elements besides $0$ (and $0$ may or may not be a prime, depending on your conventions). This shouldn't be too surprising, though. Prime elements generate prime ideals. But since $\mathbb{Q}$ is a field, its only ideals are $(0)$ and $\mathbb{Q}$.
By the way, your definition of a prime element is not standard. The definition you've stated is usually that of an irreducible element. The linked article specifically discusses the difference between the two.
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Yes, that's right. There are no primes in $\mathbb{Q}$.
On a different note, many people would say that what you call a prime is an irreducible.
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Yes there are no primes in a field. $5$ is a prime in $\mathbb{Z}$ but not in $\mathbb{Q}$
More generally if you have two rings $A\subset B$, being a prime in $A$ does not imply being a prime in $B$. As an example take $\mathbb{Z}\subset \mathbb{Z}[\sqrt{-2}]$. Then we have $3$ is a prime in $\mathbb{Z}$ while $3=(1+\sqrt{-2})(1-\sqrt{-2})$ is not a prime in $\mathbb{Z}[\sqrt{-2}]$
Actually yes, There is no prime elements in $\mathbb Q$.
Every element in $\mathbb Q$ is divisible by another element in $\mathbb Q$
Think about any element in $\mathbb Q$.
For example, $3 \in \mathbb Q$ is not prime because we have that $\frac{1}{3} \times {9} = 3$
In general, an element $\alpha \in \mathbb Q$ is in the form of $\frac{a}{b}$ where $a,b$ are both integers and $b \neq 0$.
Now $\frac{1}{b} \in \mathbb Q$ and $a \in \mathbb Q$ and $\frac{1}{b} \times a = \frac{a}{b}$ and so you can't have a prime element in $\mathbb Q$
However, That definition is actually the definition for an irreducible element. However, sometimes irreducibility implies prime and in a euclidean domain, it's true because a euclidean domain is also a unique factorization domain.
However, In a general integral domain, Irreducible doesn't imply prime.
The definition for a prime element is the following.
$p$ is said to be a prime element , if $p$ is a positive non unit element and if $p \mid ab$ then $p \mid a$ or $p \mid b$
Now to give you an example where there is an irreducible element which is not prime.
Consider the integral domain $\sqrt{-5}$
Now $2$ is an irreducible element that divides the product $(1+\sqrt{-5})(1-\sqrt{-5}) = 6$.
However $2$ does not divide any of the factors and hence it's not prime.