Polynomial ring isomorphic to Quotient space

Question

I'm relatively new to rings, and I also don't have a very good understanding of quotient spaces, but I'd like to understand the following problem.

Show that for $A \in M_n(\mathbb R ),$ there is some $p \in \mathbb R [X]$ such that $\mathbb R [A]$ is isomorphic to $\mathbb R [X]/ \langle p \rangle$. Show that if $p,q \in \mathbb R [X]$ are quadratics with two distinct real roots then $\mathbb R [X]/ \langle p \rangle \cong \mathbb R [X]/ \langle q \rangle$. Justify which, if any, of $\mathbb R [A]$, $\mathbb R [B]$, $\mathbb R [C]$ and $\mathbb R [D]$ are isomorphic as rings. $$ A:= \left( \begin{matrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right), B:= \left( \begin{matrix} 4 & 1 \\ 0 & 3 \\ \end{matrix} \right), C:= \left( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} \right), D:= \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right). $$

I already struggle to understand the question... I know that $\mathbb R [X]$ are the polynomials with real coefficients, and we can also plug matrices in and evaluate the polynomials as we are used to. So, I guess $\mathbb R [A]$ is the space of all matrices we can get by adding powers of $A,$ etc. I definitely lack an intuition what $\mathbb R [X]/ \langle p \rangle$ is. Could somebody offer some insight? And then maybe a hint about how to solve that?

As for the second part, I guess if we have isomorphisms, they are in the same dimension. Since $A$ is diagonal and $C$ is not, I don't think $\mathbb R [A]$ and $\mathbb R [C]$ are isomorphic. But how would I go about solving that part as well?,

Answer 1

Hint: consider the epimorphism $\mathbf{ev}_A \colon\Bbb R [X] \to \Bbb R[A]$. What's its kernel?

You can check that almost by definition $\ker \mathbf{ev}_A$ is generated by the minimial polynomial $m_A$ of $A$. Hence $\Bbb R[X]/(m_A) \simeq \Bbb R[A]$. This will help you when computing the particular examples and see which are isomorphic. If case this says something to you, the idea is that by Cayley-Hamilton the extension $\Bbb R \to M_n(\Bbb R)$ is integral, hence we characterize each algebra $\Bbb R[A]$ with $A \in M_n(\Bbb R)$ as a quotient of $\Bbb R[X]$.

Hint: if $p = (X-\alpha)(X-\beta)$, note that $(X-\alpha)$ and $(X-\beta)$ are coprime.

Use the chinese remainder theorem to prove that $\Bbb R[X]/p \simeq \Bbb R[X]/(X-\alpha) \times \Bbb R[X]/(X-\beta)$. The latter are isomorphic to $\Bbb R$ via evaluation at $\alpha$ and $\beta$ respectively, so any such quotient is $\Bbb R^2$.

Answer 2

Unfortunately, I am not sure that the following explanation will mean very much to you, if you are not so familiar with ring theory; however, I will provide as much scaffolding as possible so that you can develop an understanding and eventually return to this post with full comprehension.

Let us begin with an $n \times n$ matrix $A$ with entries in $\mathbb R.$ Henceforth, let us use $I$ to denote the $n \times n$ identity matrix. Convince yourself that the collection of "polynomials" $$\mathbb R[A] = \{c_n A^n + \cdots + c_1 A + c_0 I \,|\, n \geq 0 \text{ is an integer and } c_0, c_1, \dots, c_n \in \mathbb R\}$$ in the matrix $A$ with real coefficients is a unital ring with respect to matrix addition and multiplication: the zero matrix is the additive identity; the identity matrix $I$ is the multiplicative identity. Consequently, the map $\varphi_A : \mathbb R[X] \to \mathbb R[A]$ defined by $$\varphi_A(c_n X^n + \cdots + c_1 X + c_0) = c_n A^n + \cdots + c_1 A + c_0 I$$ is a surjective unital ring homomorphism. By the First Isomorphism Theorem, therefore, we have that $\mathbb R[A] \cong \mathbb R[X] / \ker \varphi_A,$ where we denote by $$\ker \varphi_A = \{p(X) \in \mathbb R[X] \,|\, \varphi_A(p(X)) \text{ is the zero matrix} \}$$ the kernel of the ring homomorphism $\varphi_A.$ Until now, I hope that things have made sense.

But here is where you will have to cross-reference a few things. Considering that $\mathbb R$ is a field, it follows that $\mathbb R[X]$ is a Euclidean domain. Every Euclidean domain is a principal ideal domain (PID), so $\mathbb R[X]$ is a PID. By definition, therefore, for every ideal $J$ of $\mathbb R[X],$ there exists a polynomial $f(X)$ in $J$ such that $J = f(X) \mathbb R[X],$ i.e., for every polynomial $h(X)$ in $J,$ there exists a polynomial $g(X)$ such that $h(X) = f(X) g(X).$ By the First Isomorphism Theorem, we have that $\ker \varphi_A$ is an ideal of $\mathbb R[X],$ hence there exists a polynomial $p(X)$ in $\ker \varphi_A$ such that $\ker \varphi_A = p(X) \mathbb R[X] = \langle p(X) \rangle$ (in your notation). Consequently, the first claim is established. Given that $p(X)$ is not monic, its leading coefficient $C$ is a unit in $\mathbb R[X]$ (because $\mathbb R$ is a field), hence $\mu(X) = C^{-1} p(X)$ is a monic polynomial with $\langle \mu(X) \rangle = \langle p(X) \rangle.$ Often, in the literature, the monic polynomial $\mu(X)$ that generates $\ker \varphi_A$ is referred to as the minimal polynomial of $A$ because it is the unique monic polynomial of least degree such that $\mu(A)$ is the zero matrix.

Consequently, for any real matrices $M$ and $N$ with the same minimal polynomial $\mu(X)$ in $\mathbb R[X],$ we have that $\mathbb R[M] \cong \mathbb R[X] / \langle \mu(X) \rangle \cong \mathbb R[N].$ By the Chinese Remainder Theorem, if $\mu(X)$ is a product of $k$ distinct linear factors (i.e., $\mu(X)$ has distinct real roots), we have that $\mathbb R[X] / \langle \mu(X) \rangle \cong \mathbb R^k.$ We can therefore determine which rings $\mathbb R[A],$ $\mathbb R[B],$ $\mathbb R[C],$ and $\mathbb R[D]$ are isomorphic by computing the minimal polynomials of $A, B, C,$ and $D.$

Computing the minimal polynomial of a matrix can be difficult in certain situations; however, the matrix $A$ is diagonal, so its minimal polynomial is easily deduced. On the other hand, all three of $B, C,$ and $D$ are both upper-triangular, hence the determinants $\det(xI - B),$ $\det(xI - C),$ and $\det(xI - D)$ are easy to compute. We refer to $\chi_B(x) = \det(xI - B)$ as the characteristic polynomial of $B.$ Observe that if $B$ is an $n \times n$ matrix, then the degree of the characteristic polynomial of $B$ is $n.$ By the Cayley-Hamilton Theorem, the characteristic polynomial of a matrix annihilates the matrix (i.e., $\chi_B(B)$ is the zero matrix), hence by the above paragraph, we must have that $\mu_B(x) \,|\, \chi_B(x),$ where $\mu_B(x)$ is the minimal polynomial of $B.$ Consequently, if $B$ is an $n \times n$ matrix with $n$ small (e.g., $n \leq 3$), then one may compute the minimal polynomial of $B$ by finding the smallest product of the irreducible factors of $\chi_B(x)$ that annihilate $B.$