Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Algebra / 3

Use ordinary matrix addition and multiplication on $R=K[A]$. Sums and products of polynomials in $A$ remain such polynomials, and they commute because powers of the same matrix commute. The identity is $A^0=I$. Evaluation $\varphi (\sum c_i{X}^i)=\sum c_i{A}^i$ is a surjective unital algebra homomorphism over $K$. It is unique among maps fixing the scalar matrices and sending $X$ to $A$.

The scalar condition is needed for the word “unique”: as an unrestricted ring-homomorphism assertion over an arbitrary field, prescribing only the image of $X$ does not suffice. For example, over $\mathbb{C}$ with a real scalar matrix $A$, conjugating every coefficient gives another unital ring homomorphism with the same image of $X$. The intended polynomial-evaluation map is the $K$-algebra map just defined.

The kernel of a ring homomorphism is an ideal of the principal ideal domain $K[X]$. It is nonzero because finitely many powers of $A$ must become linearly dependent in the finite-dimensional matrix space. Its unique monic generator is precisely the minimal polynomial $m$: division by $m$ shows that a polynomial killing $A$ has zero remainder, by minimality. Hence $\ker \varphi =(m)$ and $R\cong K[X]/(m)$.

If $\gcd (f,m)=1$, Bézout's identity gives $uf+vm=1$; evaluation gives $u(A)f(A)=I$. Conversely an inverse in $R$ has form $g(A)$, so $fg-1\in (m)$, which forces $\gcd (f,m)=1$. Thus invertibility in this ring is exactly coprimality with the minimal polynomial.

The displayed matrix is block diagonal with a $1\times 1$ block $1$ and a $2\times 2$ block $B$ having characteristic polynomial $X^2-4X-5=(X-5)(X+1)$. The latter block is not scalar, so its minimal polynomial has degree two and is exactly that polynomial. Taking the least common multiple of the block minimal polynomials gives $m=(X-1)(X-5)(X+1)$ over $\mathbb{R}$.

Ideals of $K[X]/(m)$ correspond to ideals of $K[X]$ containing $(m)$, namely the principal ideals generated by the monic divisors of $m$. Every one of the three distinct linear factors may be included or omitted, soThis includes both the zero ideal and the whole ring. Equivalently the Chinese remainder theorem identifies this ring with ${\mathbb{R}}^3$.

Reduce the matrix modulo two. It is $A=1\oplus \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, so $A\ne I$ but $(A-I{)}^2=0$. Thus the minimal polynomial is $\boxed{m=(X+1{)}^2}$. Its monic divisors are $1,X+1,(X+1{)}^2$, giving exactly $\boxed{3\text{ ideals}}$ in the quotient.

This ring has the four elements $0,I,A,I+A$. Coprimality with $X+1$ is equivalent to the representing polynomial taking value one at $1$. Thus precisely $I$ and $A$ are units: $\boxed{2\text{ invertible matrices}}$. Indeed $A^2=I$, whereas $(I+A{)}^2=0$.