Number fields and rings of integers

Knowing exactly what the integers of $K$ are is the mandatory first step for doing arithmetic in $K$: factoring elements or ideals, computing class groups, or deciding which primes ramify all require an honest description of $\mathcal O_K$, not the naive guess $\mathbb Z[\sqrt d]$, which is wrong exactly half the time.

Write $\alpha=a+b\sqrt d$ with $a,b\in\mathbb Q$. If $b=0$, $\alpha=a\in\mathbb Q$ is an algebraic integer iff $a\in\mathbb Z$ (rational root theorem), which is the case $b=0$ of $\mathbb Z[\sqrt d]$. If $b\neq0$, the minimal polynomial of $\alpha$ over $\mathbb Q$ is $x^2-2ax+(a^2-db^2)$ (its trace is $2a$, its norm $a^2-db^2$), so $\alpha$ is integral iff both $2a\in\mathbb Z$ and $a^2-db^2\in\mathbb Z$.

Write $u=2a\in\mathbb Z$. From $a^2-db^2\in\mathbb Z$, multiplying by $4$ gives $u^2-4db^2\in4\mathbb Z\subset\mathbb Z$, so $4db^2\in\mathbb Z$. Write $b=p/q$ in lowest terms ($\gcd(p,q)=1$); then $4db^2=4dp^2/q^2\in\mathbb Z$ forces $q^2\mid4dp^2$, and since $\gcd(p,q)=1$ this forces $q^2\mid4d$. Because $d$ is squarefree, the largest perfect square dividing $4d$ is at most $4$, so $q^2\mid4$, i.e. $q\in\{1,2\}$; either way $v:=2b\in\mathbb Z$.

Now substitute $u=2a$, $v=2b$ (both integers) into $a^2-db^2\in\mathbb Z$: this becomes $\tfrac{u^2-dv^2}{4}\in\mathbb Z$, i.e. $u^2\equiv dv^2\pmod4$. Squares mod $4$ are only $0$ (even) or $1$ (odd). If $v$ is even, $dv^2\equiv0\pmod4$ forces $u$ even too (since $u^2$ odd would give $1\not\equiv0$); so $u,v$ both even, meaning $a,b\in\mathbb Z$ — the "trivial" solution, always available, giving $\mathbb Z[\sqrt d]\subseteq\mathcal O_K$ always.

If $v$ is odd, $dv^2\equiv d\pmod4$ (since $v^2\equiv1$), so we need $u^2\equiv d\pmod4$. If $d\equiv2,3\pmod4$ (i.e. $d\equiv2,3\pmod4$), no square $u^2\in\{0,1\}\pmod4$ can equal $2$ or $3\pmod4$, so this case is impossible — only the trivial $u,v$ both even survives, giving exactly $\mathcal O_K=\mathbb Z[\sqrt d]$. If instead $d\equiv1\pmod4$ ($d\equiv1\pmod4$), we need $u^2\equiv1\pmod4$, satisfied whenever $u$ is odd — and since $v$ is already odd, $u\equiv v\pmod2$ is exactly the condition for $\alpha=\tfrac{u+v\sqrt d}2$ with $u,v$ both odd, which is precisely $\mathbb Z+\mathbb Z\cdot\tfrac{1+\sqrt d}2$ (check: $\tfrac{1+\sqrt d}2$ itself has $u=v=1$, both odd), so in this case $\mathcal O_K=\mathbb Z\!\left[\tfrac{1+\sqrt d}2\right]$, strictly larger than $\mathbb Z[\sqrt d]$ (index $2$).

This is what justifies calling $\mathcal O_K$ a ring of integers at all: without closure under multiplication in particular, one could not multiply two integers of $K$ and stay inside $\mathcal O_K$, which would make ring-theoretic arithmetic (factorization, ideals) impossible.

Case $d\equiv2,3\pmod4$ ($\mathcal O_K=\mathbb Z[\sqrt d]$, i.e. all elements $a+b\sqrt d$ with $a,b\in\mathbb Z$): addition is componentwise, $(a+b\sqrt d)+(c+e\sqrt d)=(a+c)+(b+e)\sqrt d$, clearly again of this form. For multiplication, $(a+b\sqrt d)(c+e\sqrt d)=(ac+bed)+(ae+bc)\sqrt d$, and since $a,b,c,e,d\in\mathbb Z$, both $ac+bed$ and $ae+bc$ are integers, so the product stays in $\mathbb Z[\sqrt d]$.

Case $d\equiv1\pmod4$ ($\mathcal O_K=\mathbb Z[\omega]$ with $\omega=\tfrac{1+\sqrt d}2$): first compute $\omega^2$ directly. $\omega^2=\left(\tfrac{1+\sqrt d}2\right)^2=\tfrac{1+2\sqrt d+d}4=\tfrac{1+d}4+\tfrac{\sqrt d}2$. Since $\sqrt d=2\omega-1$, this is $\tfrac{1+d}4+\omega-\tfrac12=\tfrac{d-1}4+\omega$. Because $d\equiv1\pmod4$, $t:=\tfrac{d-1}4$ is an integer, so $\omega^2=t+\omega$: a genuine integer relation showing $\omega$ satisfies the monic integer polynomial $x^2-x-t$, confirming $\omega$ itself is integral (as it must be, matching Theorem 1), and crucially expressing $\omega^2$ again as an integer combination of $1$ and $\omega$.

Now take general elements $a+b\omega$ and $c+e\omega$ with $a,b,c,e\in\mathbb Z$; addition is again componentwise and clearly closed. For the product, $(a+b\omega)(c+e\omega)=ac+(ae+bc)\omega+be\,\omega^2=ac+(ae+bc)\omega+be(t+\omega)=(ac+bet)+(ae+bc+be)\omega$. Both coefficients $ac+bet$ and $ae+bc+be$ are integers (products and sums of integers), so the product lies in $\mathbb Z+\mathbb Z\omega=\mathbb Z[\omega]$.

In both cases, closure under addition and multiplication is verified by direct, elementary computation, confirming $\mathcal O_K$ is genuinely a subring of $K$ containing $\mathbb Z$ in every quadratic field, exactly as the name "ring of integers" promises.