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Reciprocal roots require $c_3\ne 0$, equivalently every original root is nonzero. Their monic polynomial is the normalized reciprocal polynomialIf $c_3=0$, at least one reciprocal in the source is undefined, so that part has no polynomial interpretation without the nonzero-root qualification.

Scaling every root scales its elementary symmetric polynomials by the corresponding powers of $d$. Therefore the monic polynomial isThis remains valid for $d=0$, when all three roots are zero; deriving it by symmetric coefficients avoids dividing by $d$.

For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves

The set ${\mathcal{O}}_K$ consists of the elements of $K$ integral over $\mathbb{Z}$, meaning roots of monic polynomials in $\mathbb{Z}[X]$.

First prove the finite-module criterion for integrality in a slightly stronger form, for any nonzero finitely generated $\mathbb{Z}$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\dots ,m_r$ and write $\alpha m_i={\sum }_j{a}_{ij}{m}_j$ with integer coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its adjugate matrix. This gives $\det (\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a field, so $\det (\alpha I-A)=0$. The polynomial $\det (XI-A)$ is monic with integer coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for ${\mathcal{O}}_K$.

Conversely, if $\alpha$ satisfies a monic polynomial of degree $d$, then $M=\mathbb{Z}[\alpha ]$ is generated by $1,\alpha ,\dots ,{\alpha }^{d-1}$ and is stable under $\alpha$. Every $\beta \in M$ also maps $M$ into itself because it is an integer polynomial in $\alpha$; the finite-module criterion for integrality just proved makes every such $\beta$ integral. Thus this $M$ really lies in ${\mathcal{O}}_K$, proving the stated equivalence without circularity.

For two integral elements $\alpha ,\beta$, the module spanned by ${\alpha }^i{\beta }^j$ for $i<d_{\alpha }$, $j<d_{\beta }$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha +\beta$ and $\alpha \beta$. The finite-module criterion for integrality proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{{\mathcal{O}}_K\text{ is a ring}}$.

The integral basis of the cubic field of discriminant minus 140 can be obtained as follows. For the specified field, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the polynomial is monic. The discriminant of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, soHere ${\Delta }^2$ is the printed notation for the signed discriminant, namely the square of a Vandermonde product of the conjugates; that product need not be real.

Suppose $\eta =(a+b\theta +c{\theta }^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\theta )=0$, $\mathrm{Tr}({\theta }^2)=-4$. Thus $\mathrm{Tr}(\eta )=(3a-4c)/2$ must be an integer, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta ={\theta }^2/2$. If $b=1$, multiply by $\theta$ and use ${\theta }^3=-2\theta -2$:Closure of the ring then again gives ${\theta }^2/2\in {\mathcal{O}}_K$.

But $N_{K/\mathbb{Q}}(\theta )=-2$, so the field norm of ${\theta }^{-1}$ is $-1/2$. The field norm of an algebraic integer is an integer, proving ${\theta }^{-1}\notin {\mathcal{O}}_K$. Dividing the defining cubic by $2\theta$ gives ${\theta }^2/2=-1-{\theta }^{-1}$. Thus ${\theta }^2/2$ cannot be integral either.

Finally the discriminant-index formula for an integral lattice says $-140=[{\mathcal{O}}_K:\mathbb{Z}[\theta ]{]}^2\mathrm{disc}(K)$. The square of the integer index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta +c{\theta }^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of ${\theta }^2/2$. Hence

Assume $f\in C^{n+1}[a,b]$ and set $\omega (t)={\prod }_{i=0}^n(t-x_i)$, a monic polynomial of polynomial degree $n+1$. Fix $x\in [a,b]$ distinct from all the interpolation nodes, and chooseThe auxiliary functionvanishes at the $n+2$ distinct points $x_0,\dots ,x_n,x$. Order those points and repeatedly apply Rolle's theorem. After $n+1$ applications, there is some $\xi$ strictly between their smallest and largest values, hence in $(a,b)$, such that $F^{(n+1)}(\xi )=0$.

Because $p_n$ has polynomial degree at most $n$ and $\omega$ is monic of polynomial degree $n+1$,Substituting the resulting value of $C$ proves the polynomial interpolation error formulaAt an interpolation node, both sides are zero, so any $\xi \in (a,b)$ works. Such an interior point exists because $n\ge 1$ and the interpolation nodes are distinct. The selected $\xi$ generally depends on $x$; the theorem makes an existence assertion, not a regularity assertion about that dependence.

The monomial $x^k$ lies in ${\Pi }_n$ and already fits the data. Uniqueness of polynomial interpolation therefore givesApply this result with a nonnegative integer $m\le n$. Its Lagrange interpolation representation is the polynomial identityThe numerator in each summand is a monic polynomial of degree $n$, so the coefficient of $x^n$ on the right is exactly ${\sum }_i{x}_i^m/{\prod }_{j\ne i}(x_i-x_j)$. The coefficient on the left is zero for $m<n$ and one for $m=n$. Comparing coefficients proves the Lagrange leading-coefficient identityThe intended exponent range is $m\ge 0$: arbitrary negative exponents would not give this polynomial identity. For $m=0$, $x_i^0$ denotes the value of the constant polynomial $1$, including at a zero interpolation node.

Clearing the denominators in the rational interpolation conditions givesThese are $n+1$ linear equations for $(m+1)+\ell =n+1$ unknowns. The $i$th row of the coefficient matrix, in the order $a_0,\dots ,a_m,b_0,\dots ,b_{\ell -1}$, isThis rational interpolation linearization is equivalent to the original fraction conditions only if $r(x_i)\ne 0$ at every interpolation node. If $r(x_i)=0$, the cleared equation merely forces $q(x_i)=0$, leaving the fraction undefined.

For $f_i=x_i^{m+1}$, the coefficient row becomesThis is the Vandermonde matrix, whose determinant is ${\prod }_{0\le i<j\le n}(x_j-x_i)\ne 0$. Equivalently, a vector in its null space would define a degree-at-most-$n$ polynomial vanishing at all $n+1$ distinct interpolation nodes, which is zero by the preceding uniqueness proof. Thus the cleared linear system has exactly one solution.

There is also a constructive description of that solution. Define the monic polynomialThe polynomial $x^{m+1}r(x)-q(x)$ is monic of degree $n+1$ and vanishes at every interpolation node. It must therefore equal $W(x)$. By polynomial division, write uniquelyThe quotient $R$ is monic of degree $\ell$, andis the unique coefficient solution. This does not assert that $R$ is nonzero at every interpolation node.

For a counterexample with $m=0$, $\ell =1$, choose $x_0=0,x_1=1$ and the specified data $f_i=x_i$, so $f_0=0,f_1=1$. The cleared equations reduce togiving the unique solution $q(x)=0$, $r(x)=x-1$. But at $x_1=1$ the quotient is $0/0$. Indeed, any genuine rational interpolant of this prescribed form would need $r(0)\ne 0$; then $q(0)/r(0)=0$ forces the constant numerator to be zero. At a nonzero denominator its value at $1$ would also be zero, never one. Cancelling the common factor in the cleared candidate would merely give the identically zero rational function, which still fails the data. Hence

An element is algebraic over $F$ when some nonzero polynomial in $F[x]$ vanishes at it. Its minimal polynomial is the monic polynomial of least positive degree with this property; scaling a least-degree nonzero vanishing polynomial makes it monic.

If $m_1,m_2$ are two such polynomials, they have the same least degree. Their difference has smaller degree because both are monic, and it also vanishes at $a$. Minimality forces the difference to be zero. Hence the monic minimal polynomial is unique.

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are

The three answers, in the given order, are no, yes, yes.

For $\mathbb{Z}[x]$, consider the ideal $I=(2,x)$. Its members have even constant coefficient, so $1\notin I$ and the ideal is proper. If $I=(g)$, then $g$ divides the constant $2$, forcing $g$ to be constant by additivity of polynomial degree in this integral domain. Since $g$ also divides $x$, it must divide the coefficient $1$, giving $g=\pm 1$. That would make $I$ the whole ring, a contradiction. Thus $\mathbb{Z}[x]$ is not a principal ideal domain, and part (a) shows it cannot be a Euclidean domain. This is the nonprincipal ideal generated by a prime and an indeterminate obstruction.

Evaluation at $i$ gives the ring isomorphismIndeed, division by the monic polynomial $x^2+1$ leaves a unique remainder $a+bx$, and its value $a+bi$ is zero only when both integer coefficients vanish. Thus the quotient maps bijectively, with operations preserved, onto the Gaussian integers. These form a Euclidean domain for the Gaussian integer norm $N(a+bi)=a^2+b^2$, a standard Euclidean example allowed in this part. Geometrically, rounding the real and imaginary coordinates gives a quotient error with squared modulus at most $1/2<1$.

Finally,via $f(x,y)\mapsto f(0,y)\mathrm{mod}2$. Every polynomial over ${\mathbb{F}}_2$ occurs. The kernel consists exactly of polynomials whose $x$-independent part has even coefficients: write $f(x,y)=xg(x,y)+h(y)$ to see that this is precisely $(2,x)$. Hence the induced quotient map is a bijective ring homomorphism. Its target is a Euclidean domain with Euclidean function $\mathrm{deg}$: a polynomial ring over a field is Euclidean by the division algorithm in part (a).

Regard $\mathbb{R}[X,Y]$ as the polynomial ring $(\mathbb{R}[X])[Y]$. The coefficient ring $\mathbb{R}[X]$ is a Euclidean domain, hence a principal ideal domain. The element $X$ is a prime element: the quotient ring $\mathbb{R}[X]/(X)\cong \mathbb{R}$ is an integral domain.

The monic polynomial $Y^2+X$ has its other coefficients divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the indeterminate, proves that $Y^2+X$ is an irreducible polynomial.

Now use the automorphism of a polynomial ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

Since $A^2+I=0$, its minimal polynomial $m_A$ divides $X^2+1$. The polynomial $X^2+1$ is an irreducible polynomial over the rational numbers: a reducible quadratic would have a rational root, whereas $r^2+1>0$ for every $r\in \mathbb{Q}$.

Because $n>0$, the constant polynomial one evaluates to the nonzero identity matrix $I$, so $m_A\ne 1$. Both $m_A$ and $X^2+1$ are monic polynomials; hence the only remaining divisor is

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

By polynomial division, write $f(t)=q(t){\chi }_A(t)+r(t)$ with $\mathrm{deg}r<n$. Because ${\chi }_A$ is a monic polynomial, this division works even over a commutative ring. Evaluating at $A$ makes the multiple of ${\chi }_A(A)$ vanish, leaving $f(A)=r(A)$. This explains the reduction in two-coefficient reduction of matrix powers without multiplying out every intermediate matrix.

Over a field, the minimal polynomial $m_A$ divides ${\chi }_A$ and gives the strongest such universal reduction: use the polynomial remainder modulo $m_A$ instead. For the scalar matrix $2I_2$, the characteristic polynomial is $(t-2{)}^2$, while the minimal polynomial is only $t-2$. Every matrix power is then a scalar multiple of $I_2$.

Dividing a polynomial $f$ by $X-a$ leaves the constant polynomial remainder $f(a)$. Polynomial division makes the polynomial remainder constant because its polynomial degree must be below one; polynomial evaluation at $a$ identifies that constant. In particular, it is zero exactly when $a$ is a root of a polynomial of $f$, which is the factor theorem.^[ref]

In the worked example, division by $X-1$ leaves $6$ and division by $X+2$ leaves $12$. Division by their product leaves $-2X+8$, a polynomial remainder that stores both values at once. The argument also holds over a commutative ring for $a$ in that ring, because $X-a$ is a monic polynomial.

For a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$, factor its minimal polynomial as $m_T={\prod }_i{p}_i^{a_i}$ with distinct monic polynomials $p_i$ that are irreducible polynomials. ThenEach $V_i$ is an invariant subspace, since $T$ commutes with every polynomial in $T$. These are canonical vector subspaces, determined by $T$, without requiring any eigenvalues in $F$.

For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The coprime polynomials $m_i,M_i$ admit a Bézout's identity for polynomials $b_i{M}_i+c_i{m}_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The polynomial ${\sum }_i{b}_i{M}_i-1$ is divisible by every $m_i$, hence by their product, so ${\sum }_i{E}_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the direct sum and identify $E_i$ as its projection operators. Finally, if the minimal polynomial of any restriction had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.

A monic polynomial of degree $m$ over ${\mathbb{F}}_q$ that is an irreducible polynomial is primitive in the finite-field sense when one of its roots generates the multiplicative group of a finite field ${\mathbb{F}}_{q^m}^{\times }$. All its roots are Frobenius conjugates and then have the same order $q^m-1$. This notion differs from a primitive polynomial over an integral domain, which concerns the common divisor of its coefficients. Primitive finite-field polynomials provide explicit field representations and primitive cyclic Reed-Solomon codes or BCH codes.

Rational interpolation seeks a rational function $q/r$ with prescribed bounds on the polynomial degrees of its numerator and denominator and prescribed values at distinct interpolation nodes. A rational interpolant must have a well-defined value at each interpolation node. With a specified numerator-denominator representation, this requires $r(x_i)\ne 0$ for every $i$. A normalization such as a monic polynomial denominator removes the common scaling freedom but does not guarantee existence or uniqueness. Multiplying by a zero denominator can create spurious solutions to the rational interpolation linearization.