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If $A$ is a $2 \times 2$ real matrix, then what is $\mathbb{R}[A]$ isomorphic to as a ring?

I have proven that $\mathbb{R}[A] \simeq \frac{\mathbb{R}[x]}{\langle p(x)\rangle}$ where $p(x)$ is the minimal polynomial of $A$.

Then if $p(x)=x+a$, we have $\mathbb{R}[A] \simeq \mathbb{R}$.

If $p(x)=x^2$ such that $A$ is a nilpotent, then $\mathbb{R}[A] \simeq \frac{\mathbb{R}[x]}{\langle x^2\rangle}$

What if $p(x)=x^2+ax+b$?

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    It almost feels like you're too focused on the problem you're trying to solve, and you've forgotten that you can solve problems of the form "describe the quotient of a polynomial ring over a field by a polynomial". – 2017-01-24
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    @AdamHughes Good point I totally ommited. Thanks, deleting. – 2017-01-24
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    @DonAntonio cheers! – 2017-01-24

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Let $p(x) = q_1(x)^{e_1}\ldots q_r(x)^{e_r}$ be a product of powers of distinct, irreducible factors of $p(x)$. Then by the Chinese remainder theorem,

$$\Bbb R[x]/(p(x))\cong \bigoplus_{i=1}^r \Bbb R[x]/(q_i(x)^{e_i})$$

In your specific case, if $x^2+ax+b=q_1(x)q_2(x)$ is a product of distinct irreducibles

$$\Bbb R[A]\cong \Bbb R\oplus \Bbb R$$

as a ring. That is to say the multiplication is that of the direct sum, i.e. $(a,b)\cdot (c,d)=(ac,bd)$.

If $p(x)$ is irreducible, then the roots are complex and

$$\Bbb R[A]\cong \Bbb C.$$

Finally if $p(x) = q(x)^2=(x-a)^2$ then

$$\Bbb R[A]\cong \Bbb R\oplus\Bbb R$$

as a set with the ring multiplication given by $(a,b)\cdot(c,d)= (ac, bc+ad)$.

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    I know that Chinese Remainder Theorem works when $$ and $$ are coprime (i.e. $+=R[x]$. If $=$ and $=$, how to get $1\in R[x]$ from their sum? – 2017-01-24
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    @Kenneth.K consider $1={1\over 3}\bigg((x-3)-(x-6)\bigg)\in \langle x-3\rangle+\langle x-6\rangle$. Remember ideal sums mean sums of everything in the ideals, in particular scalar multiples of the generators. – 2017-01-24