Is $\mathbb{Z}[\sqrt {14}]$ a UFD? I study the ring $\mathbb{Z}[\sqrt {14}]$
I  want to know that it is UFD OR NOT.
MY WORK
$\mathbb 14 = (\sqrt{14} ) (\sqrt 14)$
$\mathbb 14= (7)(2)$
$\mathbb=(7+2(\sqrt{14} )  )(-7+2(\sqrt{14} )   )(4+(\sqrt{14})  )(4- (\sqrt{14})  )$
Now I am unable to find that  √14  is reducible or irreducible
If √14 is irreducible ,then above ring is not UFD...
Give me suggestions about it..
 A: Let's prove that $\mathbb{Z}[\sqrt{14}]$ is a PID. This uses algebraic number theory, which i'm not sure you are familiar with.
Since $14\not\equiv 1 \ [4]$, the ring $\mathbb{Z}[\sqrt{14}]$ is the ring of integers of $\mathbb{Q}(\sqrt{14})$. Minkkowski's bound is $\sqrt{14}<4$, so any element of the class group is represented by an ideal of norm $1,2$ or $3$. Hence, the class group is generated by classes of prime ideals of norm $2$ or $3$.
By Dedekind's theorem, factorisation of a prime number $p$  is reflected by factorisation of $X^2-14$ modulo $p$. Thus $(2)=(2, \sqrt{14})^2$ and $(3)$ is prime since $X^2-14$ has no roots mod $3$. Hence the class group is generated by the classes of  $\mathfrak{p}_2=(2, \sqrt{14})$ and $(3)$. The second is a principal ideal, so it remains to shows that $\mathfrak{p}_2$ is principal. Since $\mathfrak{p}_2$ has norm $2$, a potential generator must have norm $\pm 2$, so we look for solutions of the equation $\pm 2= x^2-14y^2$.
One obvious solution is $4+\sqrt{14}$. We have obviously $4+\sqrt{14}\in\mathfrak{p}_2$, hence $(4+\sqrt{14})\subset \mathfrak{p}_2$. Since these ideals both have norm $2$, we have $(4+\sqrt{14})= \mathfrak{p}_2$.
So the class group is trivial, and $\mathbb{Z}[\sqrt{14}]$ is a PID.
Fun fact. One may show  that $\mathbb{Z}[\sqrt{14}]$ is Euclidean, but not for the norm function.
To go back to your original question: is $\sqrt{14}$ irreducible ? You can guess the answer is using number theory as follows. For similar reasons as above, we have $(7)=(7,\sqrt{14})^2=(7+2\sqrt{14})^2.$
Now $(14)=(\sqrt{14})^2=(4+\sqrt{14})^2(7+2\sqrt{14})^2$. Since factorization into prime ideals is unique, we get $(\sqrt{14})=(4+\sqrt{14})(7+2\sqrt{14})$. Hence $\sqrt{14}=u (4+\sqrt{14})(7+2\sqrt{14})$, where $u$ is a unit. Computations (just solve in $u$ !) show that $u=15-4\sqrt{14}$ (which has norm $1$, so it is indeed a unit)
Note that $4+\sqrt{14}$ and $7+\sqrt{14}$ are generators of prime ideals, so they are prime elements. In particular, they are irreducible.
All in all, a factorization of $\sqrt{14}$ into a product of a unit and irreducible elements is $\sqrt{14}=(15-4\sqrt{14})(4+\sqrt{14})(7+2\sqrt{14})$. Note this is the same factorization as @diracdeltafunk, since $(15-4\sqrt{14})(4+\sqrt{14})=4-\sqrt{14}$.
A: You're on the right track here! The trick is to use the norm map $N : \mathbb{Z}[\sqrt{14}] \to \mathbb{Z}$, defined by $N(a + b\sqrt{14}) = a^2 - 14b^2$. You can show that $N$ is multiplicative, meaning $N(xy) = N(x) N(y)$ for all $x,y \in \mathbb{Z}[\sqrt{14}]$.
In particular, if $x \in \mathbb{Z}[\sqrt{14}]$ is a unit, then $1 = N(1) = N(x x^{-1}) = N(x) N(x^{-1})$ so $N(x) = \pm 1$. In fact, the converse is true as well: if $N(a + b \sqrt{14}) = \pm 1$ then $\pm 1 = a^2 - 14b^2 = (a+b\sqrt{14})(a-b\sqrt{14})$, so $a + b\sqrt{14}$ is a unit. Now, we get an interesting consequence: if $N(x)$ is irreducible in $\mathbb{Z}$, then $x$ is irreducible. You should try to prove this (unless this is all very familiar to you, of course!)
Some obvious consequences of this are that $\pm 7 + 2\sqrt{14}$ and $4 \pm \sqrt{14}$ are irreducible because they have norms $-7$ and $2$, respectively. However, your hope that $\sqrt{14}$ will be irreducible is no good: indeed
$$(7+2\sqrt{14})(4-\sqrt{14}) = \sqrt{14}$$
and neither $7+2\sqrt{14}$ nor $4-\sqrt{14}$ are units (as just mentioned). Since $14 = 2 \cdot 7$, it might be more fruitful to try to factor $2$ or $7$ directly! Indeed, if you can find integers $a$ and $b$ such that $N(a + b\sqrt{14}) = 2$, then $(a + b\sqrt{14})(a - b\sqrt{14}) = 2$ gives a factorization of $2$ into irreducible elements. If you can find two distinct factorizations this way, you will have shown that $\mathbb{Z}[\sqrt{14}]$ is not a UFD!
Edit: Also, see this excellent expository paper by Keith Conrad for a related discussion.
Edit 2: Actually, I have no idea if this approach is fruitful here. I thought I had produced two distinct factorizations of $2$, but upon further inspection they were associates. I'll leave this answer up in case it helps, and apologies if I got your hopes up!
Edit 3: According to OEIS, $\mathbb{Z}[\sqrt{14}]$ is in fact a UFD. I don't know how to prove this.
