Invertible elements of a quotient ring of $ℝ[x]$.

Question

I was given the following problem as part of an algebra assignment:

Let: $$f(x)=15x^3+8x^2+5x+9 \in ℝ[x]$$ and $$I=\langle f(x)\rangle$$ the ideal generated by $f(x)$.

If $ℝ[x]/I$ is the quotient ring modulo I, then find all the invertible elements of $ℝ[x]/I$ and similarly all the invertible elements of the quotient ring $ℚ[x]/I$.

My approach so far: I imagine that we are not expected to come up with a certain set of specific elements, but rather describe them in their general form.

So, let $p(x)$ be an element of $ℝ[x]/I$. Then, $p(x)$ is of the form: $$ax^2+bx+c$$
Since $$deg(p(x))<deg(f(x))$$ and $a,b,c \in ℝ$, not all zero

Let $p(x)\in U(ℝ[x]/I)$, where $U(ℝ[x]/I)$ denotes the set of all invertible elements in $ℝ[x]/I$.

Then, let $$q(x)=a'x^2+b'x+c'$$ be its inverse. It follows that: $$(q(x)+I)(p(x)+I)=1_{ℝ[x]}+I=1+I$$
So, $$(q(x)p(x)-1)\in I$$ and if we use the expressions of $q,p$:

$$(a'x^2+b'x+c')(ax^2+bx+c)-1 \in I$$ $$aa'x^4+9ab'+a'bx^3+(ac'+bb')x^2+(bc'+cb')x+cc'-1\in I$$

If I understand correctly, this means that the expression above must be generated somehow by $f(x)$, and its degree must equal the degree of $f(x)$. But where does that lead? How can I continue-if this approach can indeed work-and find a general form for the inverses?

To be more specific, am I to find expressions of $a',b',c'$ in relation to $a,b,c$? Under which conditions will they apply?
Any help is appreciated.

Answer 1

If ℝ[x]/I is the division ring modulo $I$...

I guess you mean "quotient ring." "Division ring" means something else altogether: a division ring is a noncommutative field. This is important because this quotient is not a field (or division ring either.)

then find all the inverse elements of ℝ[x]/I

Do you mean find all invertible elements? It must be impossible to specify a general form of invertible elements, but you can say that the invertible polynomials are the ones coprime to $f(x)$.

similarly all the inverse elements of the [quotient] ring ℚ[x]/I.

Here you can work to show that $f(x)$ is irreducible over $\Bbb Q$, so the quotient is a field extension of $\Bbb Q$ of degree $3$. All nonzero elements are invertible.

Answer 2

In a way, the interesting case is where the base is $\Bbb R$ instead of $\Bbb Q$. You see that the derivative of your cubic polynomial is a quadratic with negative discriminant, so that there is no local maximum nor minimum. That is, your cubic polynomial represents a strictly monotone real-valued function, and so has only one real root, which I’ll call $\rho$. We can write $f/15=x^3+8x^2/15+x/3 +3/5=(x-\rho)(x^2+bx+c)$, where the quadratic factor has two (nonreal) complex roots $z$ and $\bar z$. Now, we expect $\Bbb R[x]/(f)$ to have the shape $\Bbb R\oplus\Bbb C$, with coordinatewise addition and multiplication, and indeed we get our isomorphism $\psi:\Bbb R[x]/(f)\to\Bbb R\oplus\Bbb C$ by defining $\psi(g)=\bigl(g(\rho),g(z)\bigr)$. Since $g$ has real coefficients, $g(\rho)\in\Bbb R$ and $g(z)\in\Bbb C$. The answer to your question of which $g$ are invertible is easy now: you need both $g(\rho)\ne0$ and $g(z)\ne0$.

For the less interesting case of $\Bbb Q[x]/(f)$, you know that every element is uniquely representable as a polynomial $a+bx+cx^2$, and this is invertible if and only if at least one of $a$, $b$, $c$ is nonzero.