Ideals and the polynomial Euclidean algorithm

Modulo arithmetic works because the elements declared zero absorb multiplication by every ring element. That condition defines an ideal. For polynomial rings over a field, division with remainder makes every ideal especially simple and turns inverse construction into a finite algorithm.

1 Ring conventions and ideals

Use commutative rings with identity and homomorphisms preserving identity. A ring has an abelian additive group and associative multiplication distributing over addition. A nonzero commutative ring without zero divisors is an integral domain; a field additionally gives every nonzero element a multiplicative inverse. A subring need not absorb multiplication by elements outside it, so it need not be an ideal.

An ideal $I\subseteq R$ is an additive subgroup satisfying $ri\in I$ for all $r\in R$, $i\in I$. The ideal generated by one element $a$ is $(a)=\{ra:r\in R\}$, called principal. For several elements, $(a,b)$ consists of all $ra+sb$.

Quotient construction. Declare $x\sim y$ if $x-y\in I$. Its additive classes form $R/I$, with $(x+I)(y+I)=xy+I$. Replacing $x,y$ by $x+i,y+j$ changes the product by $xj+yi+ij\in I$, proving well-definedness. If a proposed zero set is only an additive subgroup, this step may fail.

Example distinguishing a subring from an ideal. Inside $\mathbb R[x]$, constant polynomials form a subring isomorphic to $\mathbb R$. It is not an ideal: it contains 1 but not $x\cdot1=x$. In fact an ideal containing 1 is the entire ring, since $r=r1$ for every $r$.

2 Division and uniqueness

Let $F$ be a field and $g\in F[x]$ nonzero. Given $f$, repeatedly cancel its leading term using a multiple of $g$. Division by the leading coefficient is possible because it is nonzero in a field. Degree strictly decreases at every step, so the process ends with

$$f=qg+r,\qquad r=0\ \text{or}\ \deg r<\deg g.$$

If $f=q'g+r'$ is another such expression, then $(q-q')g=r'-r$. If $q\ne q'$, the left side has degree at least $\deg g$ while the right has smaller degree, a contradiction. Thus quotient and remainder are unique. Over a general ring the leading coefficient need not be invertible; the field hypothesis cannot simply be removed.

Principal-ideal theorem for $F[x]$. Every ideal is generated by one polynomial.

Proof. The zero ideal is $(0)$. Otherwise choose a nonzero element $p$ of least degree, scaled to be monic. For any $f\in I$, divide $f=qp+r$. The remainder $r=f-qp$ lies in $I$ and has smaller degree than $p$, so minimality forces $r=0$. Therefore $I=(p)$. Uniqueness of a monic generator follows from mutual divisibility and equal leading coefficients. $\square$

3 Euclidean algorithm and Bézout coefficients

A greatest common divisor of nonzero polynomials $f,g$ is a common divisor divisible by every common divisor, normalized to be monic. If $f=qg+r$, then a polynomial divides both $f,g$ if and only if it divides both $g,r$. Repeating division terminates because nonzero remainder degrees strictly decrease. The last nonzero remainder, normalized, is the gcd. Back substitution expresses it as $uf+vg$.

This proves both existence and the Bézout identity. It also identifies $(f,g)$ with $(\gcd(f,g))$: every linear combination is divisible by the gcd, and the identity places the gcd itself in $(f,g)$.

Worked calculation over $\mathbb Q$. Let $f=x^3-2$ and $g=x^2-1$. Then

$$f=xg+(x-2),\qquad g=(x+2)(x-2)+3.$$

The last nonzero remainder is 3, so the monic gcd is 1. Back substitution gives

$$3=g-(x+2)(f-xg)=(x+1)^2g-(x+2)f.$$

Hence in $\mathbb Q[x]/(x^3-2)$ the inverse of $[x^2-1]$ is $[(x+1)^2/3]$. Verification reduces their product to 1 using $x^3=2$. The algorithm supplies an inverse with a proof, not a guess.

4 Kernels and ring isomorphism

For a ring homomorphism $\phi:R\to S$, the kernel is an ideal: additive closure follows from the additive homomorphism law, and $\phi(ri)=\phi(r)\phi(i)=0$ gives absorption. The map $r+\ker\phi\mapsto\phi(r)$ is a ring isomorphism from $R/\ker\phi$ onto the image. Equality of images is equivalent to the representative difference lying in the kernel, which proves both well-definedness and injectivity; the map laws give both operation-preservation properties.

Evaluation at $a\in F$ is onto $F$ because constants hit every target value. Division by $x-a$ gives remainder $f(a)$, so its kernel is $(x-a)$. Therefore $F[x]/(x-a)\cong F$. This is the factor theorem as a quotient statement.

Zero and unit cases in the ideal statements

The principal-ideal theorem includes $I=(0)$ separately; only a nonzero ideal has a monic generator. If $I=(p)=(q)$ is nonzero and both generators are monic, write $p=aq$, $q=bp$. Degree addition gives $\deg a=\deg b=0$, and monicity then gives $a=b=1$, so $p=q$. A nonzero constant generator gives the unit ideal $F[x]$; its quotient is the zero ring, not a field. A nonconstant generator gives a proper ideal because a nonzero multiple cannot equal the constant 1.

For the additive quotient, reflexivity, symmetry and transitivity follow from $0\in I$, closure under negatives and closure under sums. Replacing $x,y$ by $x+i,y+j$ also changes their sum by $i+j\in I$. Ring associativity, distributivity, additive inverse and the identity $1+I$ descend from the representative formulas. This completes the addition and identity checks alongside the displayed multiplication check. Every use of one-sided absorption here retains the commutative-ring hypothesis stated at the start.

5 Graded practice

E1 Foundation. Prove $I\cap J$ and $I+J=\{i+j:i\in I,j\in J\}$ are ideals of a commutative ring.

E2 Euclidean arithmetic. Find the monic gcd of $x^4-1$ and $x^3-1$ over $\mathbb Q$, with a Bézout identity.

E3 Inverse. Find the inverse of $[x+1]$ in $\mathbb Q[x]/(x^2+1)$.

E4 Field hypothesis. Explain why the equation $x=q(x)\,2+r(x)$ with remainder degree below zero cannot always be solved in $\mathbb Z[x]$.

E5 Ideal versus subgroup. The even-degree-support polynomials $\{a_0+a_2x^2+a_4x^4+\cdots\}$ form an additive subgroup of $\mathbb R[x]$. Are they an ideal?

6 Full solutions

S1. Both contain zero. Differences of two elements remain in the respective set by additive closure. For an intersection, $r$ times an element remains in both ideals. For the sum, $r(i+j)=ri+rj\in I+J$. Thus the subgroup and absorption conditions hold. Commutativity makes the one-sided absorption convention sufficient here.

S2. $f-xg=x-1$. Both original polynomials are divisible by $x-1$, so their monic gcd is $x-1$, and $1\cdot f+(-x)\cdot g=x-1$ is a Bézout identity. No larger common divisor can exist because it would divide this degree-1 combination.

S3. $(x+1)(1-x)=1-x^2$, which becomes 2 modulo $x^2+1$. Thus the inverse is $[(1-x)/2]$. The division by 2 is valid over $\mathbb Q$; changing to characteristic 2 changes this conclusion.

S4. A nonzero constant divisor has degree 0, so a remainder of smaller degree must be zero. The equation would require $q=x/2$, which is not in $\mathbb Z[x]$. The leading coefficient 2 is not a unit, so the field division algorithm is unavailable.

S5. No. The set contains 1 but not $x$, so multiplying its element 1 by the ambient ring element $x$ leaves the set. It is a subring but not an ideal. This failure would obstruct quotient multiplication.

Summary

Ideals encode exactly the absorption needed by ring quotients. Over a field, polynomial division proves every polynomial ideal is principal. The Euclidean algorithm gives gcds and Bézout identities, which in turn calculate inverses and expose the structure of evaluation maps and quotient rings.