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A monic quadratic polynomial over a field is irreducible exactly when it has no root in that field. Testing the nine choices of its two coefficients in ${\mathbb{F}}_3$ givesFor clarity, a zero constant term gives a root at zero. Among constant term one, only the zero linear coefficient avoids roots at $1,2$. Among constant term two, precisely the linear coefficients one and two avoid roots. This exhausts all choices.

Put $f=x^4+x+2$. Its values at $0,1,2$ are respectively $2,1,2$, so it has no linear factor. If this quartic were reducible, absence of a linear factor would force a factorization into two irreducible quadratics. Division by the three candidates above gives remainders respectively $x,x+1,x+1$, all nonzero. Thus $f$ is an irreducible polynomial.

The quotient by an irreducible polynomial is a field consequently gives a field of $3^4=81$ elements. Its nonzero elements form a multiplicative group of a finite field of order $80$. In this field, $a^4=1+2a$. Successive squaring givesThe latter is not one: otherwise the nonzero polynomial $1+x+2x^3$ of degree below four would vanish at $a$, contradicting its minimal polynomial. By Lagrange's theorem, $(a^{16}{)}^5=a^{80}=1$. Since five is prime, its order is exactly five. Hence one answer isIts other nonidentity powers are the remaining primitive fifth roots. It is unnecessary to prove that $a$ itself generates the entire multiplicative group.

A monic quadratic polynomial over a field is irreducible exactly when it has no root in that field. Testing the nine choices of its two coefficients in ${\mathbb{F}}_3$ givesFor clarity, a zero constant term gives a root at zero. Among constant term one, only the zero linear coefficient avoids roots at $1,2$. Among constant term two, precisely the linear coefficients one and two avoid roots. This exhausts all choices.

Put $f=x^4+x+2$. Its values at $0,1,2$ are respectively $2,1,2$, so it has no linear factor. If this quartic were reducible, absence of a linear factor would force a factorization into two irreducible quadratics. Division by the three candidates above gives remainders respectively $x,x+1,x+1$, all nonzero. Thus $f$ is an irreducible polynomial.

The quotient by an irreducible polynomial is a field consequently gives a field of $3^4=81$ elements. Its nonzero elements form a multiplicative group of a finite field of order $80$. In this field, $a^4=1+2a$. Successive squaring givesThe latter is not one: otherwise the nonzero polynomial $1+x+2x^3$ of degree below four would vanish at $a$, contradicting its minimal polynomial. By Lagrange's theorem, $(a^{16}{)}^5=a^{80}=1$. Since five is prime, its order is exactly five. Hence one answer isIts other nonidentity powers are the remaining primitive fifth roots. It is unnecessary to prove that $a$ itself generates the entire multiplicative group.

The degree of a field extension $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower law states that for finite extensions $F\subseteq E\subseteq K$,If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so there is no proper intermediate field.

Let $K$ have 32 elements. Its prime field is ${\mathbb{F}}_2$, and its dimension over that field is five, since a dimension-$d$ vector space over ${\mathbb{F}}_2$ has $2^d$ elements. Every element outside ${\mathbb{F}}_2$ has minimal polynomial of degree five: its generated intermediate field has degree greater than one and dividing five.

Every element of $K$ satisfies $x^{32}=x$, by the multiplicative group of a field of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any irreducible polynomial $q$ over ${\mathbb{F}}_2$ gives a quotient ring ${\mathbb{F}}_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that polynomial and therefore has all five distinct roots in $K$.

The 30 elements of $K\setminus {\mathbb{F}}_2$ are thus partitioned into disjoint five-element root sets of the irreducible polynomials. Their number isNo classification of the six individual polynomials is needed; the tower degree and separability already count them all.

The splitting field in $\mathbb{C}$ is the smallest subfield containing $\mathbb{Q}$ and every root of the polynomial. The sixth roots of unity are powers of ${\zeta }_6=e^{i\pi /3}$, so the splitting field of $x^6-1$ is $\mathbb{Q}({\zeta }_6)=\mathbb{Q}(\sqrt{-3})$.

For $x^6+x^3+1=(x^9-1)/(x^3-1)$, the roots are exactly the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus its splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

Since $K=F(\lambda )$ and the extension is finite, the element $\lambda$ is algebraic. Its minimal polynomial is irreducible, and evaluation gives an isomorphism $F[X]/(m_{\lambda })\cong F[\lambda ]=F(\lambda )=K$. If its degree is $d$, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis. Comparing with $[K:F]=n$ gives $\boxed{\mathrm{deg}{m}_{\lambda }=n}$.

We first prove that every extension of two simple modules splits. ConsiderIf $n\ne m$, the central Casimir element has distinct scalars $n(n+2)$ and $m(m+2)$. On $V$ its minimal polynomial divides the product of the corresponding two distinct linear factors. Its eigenspaces are submodules, and the eigenspace for $m(m+2)$ maps isomorphically onto the quotient. It is the required complement.

Now suppose $n=m=l$. Use generalized weight spaces for $H$; their possible weights are $l,l-2,\dots ,-l$, because the characteristic polynomial on an extension is the product of those on its two factors. Lift the quotient's highest vector to $w$ in the generalized weight space of representation weight $l$. Then $ew=0$ and $f^{l+1}w=0$, since these would lie in generalized weight spaces of weights $l+2$ and $-l-2$, which are absent. Also $(H-l)w=a{v}_0$ for a highest vector $v_0$ in the submodule $A_l$. The commutator identity givesThus $a=0$. The vector $w$ is an actual highest-weight vector. Its lowering chain maps to the full basis of the quotient, so it generates an $A_l$ complementary to the original submodule. This proves splitting also when the two simple types coincide, including $l=0$.

The same splitting holds for a semisimple kernel and simple quotient by induction on the number of kernel summands. Write the kernel as $U_1\oplus U_2$ with $U_1$ simple. Modulo $U_2$, split the extension by the simple quotient; the inverse image of that chosen quotient copy is an extension with kernel $U_2$. Split it inductively, obtaining a submodule in $V$ meeting the entire original kernel trivially.

Finally induct on $\dim V$. A maximal proper submodule has smaller dimension and hence is semisimple by induction; the quotient is simple by maximality. The just-proved splitting completes the induction. Hencewith only finitely many nonzero multiplicities. This proof does not assume the complete-reducibility theorem it is establishing.

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.

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

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 $I_2$.

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.

For a finite extension of number fields $L/K$ and a nonzero prime ideal $\mathfrak{p}$ of $K$, the primes of $L$ above $\mathfrak{p}$ correspond to the orbits ofAn embedding determines a prime by pulling back the local absolute value on a field. Embeddings with the same prime extend to embeddings of the same completion and are conjugate. If $L=K(\alpha )$, these orbits are the irreducible factors of its minimal polynomial over $K_{\mathfrak{p}}$.

If $L/K$ is a finite Galois extension of local fields, ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, and $f$ is the minimal polynomial of $\alpha$, then

Let an invertible matrix $T$ have monic invariant factors of a linear operator $f_j$ of degrees $d_j$. Their constant coefficients are nonzero, since their product is the characteristic polynomial and $\det T\ne 0$. On a cyclic block, $f(T)=0$ implies $T^{-d}f(T)=0$, so $f^{\vee }(T^{-1})=0$, where $f^{\vee }(t)=t^{d}f(1/t)/f(0)$ is monic of degree $d$. The vector $v$ remains cyclic for $T^{-1}$: multiplying $v,T^{-1}v,\dots ,T^{-(d-1)}v$ by the invertible map $T^{d-1}$ gives the reversed original cyclic basis. Thus this polynomial is the minimal polynomial of the inverse block.

Normalized reciprocation preserves divisibility because $(fg{)}^{\vee }=f^{\vee }{g}^{\vee }$ when both constant coefficients are nonzero. Consequently the rational canonical form of $T^{-1}$ is the direct sum of the companion matrices $C(f_j^{\vee })$. The normalization by $f_j(0)$ is essential to make each reciprocal polynomial monic.

Let $U=aT+bI$ on a nonzero finite-dimensional vector space over a field, with $a\ne 0$. An invariant block for $T$ is also invariant for $U$. If $v,Tv,\dots ,T^{d-1}v$ is its cyclic vector basis, then $v,Uv,\dots ,U^{d-1}v$ is another basis: the binomial theorem gives a triangular change-of-basis matrix with diagonal $1,a,\dots ,a^{d-1}$. The monic polynomial $g(t)=a^{d}f((t-b)/a)$ annihilates this block, and its degree is $d$, so it is the block's minimal polynomial. In the new cyclic basis the block is $C(g)$.

The substitution preserves divisibility of the monic invariant factors of a linear operator. Apply it to every block to obtain the rational canonical form of $U$. If $a=0$, then $U=bI$ is a scalar matrix instead: all invariant factors are $t-b$, one for each dimension. This exceptional case cannot be obtained by dividing by $a$.

Let $A$ act on a nonzero finite-dimensional vector space over a field $F$, with minimal polynomial an irreducible polynomial $f$ of polynomial degree $d$. The nonconstant invariant factors of a linear operator all divide $f$, so each equals $f$. Their degrees sum to the dimension of a vector space, giving $d\mid n$ and rational canonical form consisting of $n/d$ copies of the companion matrix $C_f$. Equivalently, make the vector space a module over the field $F[X]/(f)$ by letting $X$ act as $A$; a basis over this field gives the same block decomposition.

The minimal polynomial of a real matrix, initially computed over the complex numbers, has real coefficients. Its coefficientwise complex conjugate is also a monic polynomial of the same degree annihilating the matrix. Uniqueness of the minimal polynomial forces the two polynomials to agree.

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.