Polynomial quotients and number fields

The symbol $\sqrt2$ is a real number, but it can also be understood structurally as an element satisfying the relation $\alpha^2=2$. Polynomial quotients make that relation into a rigorous construction and explain why a new field appears. The same method also shows why some apparently similar relations produce zero divisors instead.

1 Arithmetic in a quadratic field

Consider $K=\{a+b\sqrt2:a,b\in\mathbb Q\}\subset\mathbb R$. Addition and multiplication stay in this set:

$$(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2.$$

The expression is unique. If $a+b\sqrt2=c+d\sqrt2$ and $b\ne d$, then $\sqrt2=(c-a)/(b-d)$ would be rational. The standard parity proof excludes this: writing $\sqrt2=p/q$ in lowest terms forces both $p$ and $q$ even from $p^2=2q^2$.

For a nonzero $a+b\sqrt2$, its conjugate $a-b\sqrt2$ satisfies

$$(a+b\sqrt2)(a-b\sqrt2)=a^2-2b^2.$$

This rational denominator cannot be zero unless $a=b=0$: if $b\ne0$, it would imply $a/b=\pm\sqrt2$. Hence

$$(a+b\sqrt2)^{-1}=\frac{a-b\sqrt2}{a^2-2b^2}\in K.$$

With the inherited real operations, $K$ is therefore a field. For instance, $(3+2\sqrt2)^{-1}=3-2\sqrt2$, because $9-8=1$.

2 A polynomial construction of the same object

In $\mathbb Q[x]$, say $f\equiv g\pmod{x^2-2}$ when $f-g$ is a polynomial multiple of $x^2-2$. This is an equivalence relation. Addition and multiplication respect it because multiples of $x^2-2$ are closed under sums and absorb multiplication by any polynomial. Such an additive subgroup with absorption is called an ideal, denoted $(x^2-2)$.

Division by the monic polynomial $x^2-2$ gives a unique remainder $a+bx$. Thus every quotient class has a unique representative of degree below 2. In the quotient, $[x]^2=[2]$, so multiplication of remainders is

$$[a+bx][c+dx]=[(ac+2bd)+(ad+bc)x].$$

The map $[a+bx]\mapsto a+b\sqrt2$ is well defined, bijective and preserves both operations, so the quotient is isomorphic to $K$. We have constructed a field through a relation rather than simply assuming its inverse formulas.

The evaluation map $E:\mathbb Q[x]\to K$, $f\mapsto f(\sqrt2)$, has kernel exactly $(x^2-2)$. To prove the nontrivial inclusion, divide $f=q(x^2-2)+(a+bx)$. If $f(\sqrt2)=0$, then $a+b\sqrt2=0$, and uniqueness gives $a=b=0$. This is a concrete ring version of the first isomorphism theorem.

3 Why irreducibility matters

A polynomial over a field is irreducible when it has positive degree and cannot factor into two polynomials of smaller positive degree. A quadratic is irreducible precisely when it has no root in that field. Thus $x^2-2$ is irreducible over $\mathbb Q$.

More generally, $F[x]/(p)$ is a field for nonconstant irreducible $p$. If $[f]\ne0$, then $\gcd(f,p)=1$. The polynomial Euclidean algorithm gives $uf+vp=1$, so $[u]$ is the inverse of $[f]$. Conversely, a factorization $p=ab$ of smaller positive degrees supplies nonzero quotient classes $[a],[b]$ with product zero, preventing a field.

For $\mathbb Q[x]/(x^2-1)$, the classes $[x-1]$ and $[x+1]$ are nonzero but multiply to zero. The formal relation $x^2=1$ therefore does not produce a field. A relation by itself is not enough; the factorization behavior over the chosen field is decisive.

Making the concrete quotient map fully explicit

In the irrationality argument, an odd integer $2k+1$ has odd square $4k(k+1)+1$. Thus $p^2=2q^2$ makes $p=2r$ even, after which $q^2=2r^2$ makes $q$ even. This contradicts a lowest-terms fraction with positive denominator.

The set $K$ contains 0 and 1 and is closed under subtraction as well as the displayed multiplication. Associativity, commutativity and distributivity are inherited from $\mathbb R$; the nonzero-denominator inverse formula therefore verifies all field requirements. For the quotient map, each class has one representative $a+bx$. Uniqueness of $a+b\sqrt2$ proves injectivity, its definition of $K$ proves surjectivity, and the displayed product formula plus coefficientwise addition prove both operation laws and preservation of 1. The zero-divisor classes in the $x^2-1$ example are nonzero because a nonzero degree-one polynomial cannot be a multiple of a degree-two polynomial. These degree arguments are part of the construction, not optional numerical checks.

4 Graded practice

E1 Foundation. Multiply $(1+\sqrt2)(2-\sqrt2)$ and find the inverse of $1+\sqrt2$.

E2 Reduction. Reduce $x^5+3x^2+1$ modulo $x^2-2$.

E3 Fields. Decide whether $\mathbb Q[x]/(x^2+1)$ and $\mathbb R[x]/(x^2-2)$ are fields.

E4 Proof. Explain why every nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots.

5 Full solutions

S1. The product is $2-\sqrt2+2\sqrt2-2=\sqrt2$. The inverse of $1+\sqrt2$ is $(1-\sqrt2)/(1-2)=\sqrt2-1$, verified by multiplication.

S2. The relation gives $x^4=4$ and $x^5=4x$, while $3x^2=6$. The remainder is $7+4x$, of degree below 2. Polynomial division gives the same unique result.

S3. $x^2+1$ has no rational root, so it is irreducible quadratic and the first quotient is a field. Over $\mathbb R$, $x^2-2=(x-\sqrt2)(x+\sqrt2)$, so the second quotient has nonzero zero divisors and is not a field. The same polynomial can behave differently over different fields.

S4. If $a$ is a root, division by $x-a$ gives zero remainder, so $f=(x-a)g$. Every other root $b$ has $(b-a)g(b)=0$; since the coefficient system is a field and $b\ne a$, $g(b)=0$. Induction on degree proves the bound. A zero polynomial is excluded because it vanishes everywhere.

Summary

Quadratic-number arithmetic can be built from polynomial classes. Ideals make quotient operations independent of representatives, and irreducibility guarantees inverses through Bézout's identity. The coefficient field and the distinction between a domain and a ring with zero divisors must remain explicit.