What is the Meaning of $Q[u_1] \cong Q[x]/\langle f(x)\rangle$?

Question

Suppose that $f(x) = ax^2 + bx + c \in Q[x]$ is irreducible. Then $\sqrt{(b^2 − 4ac)} \neq 0$, so that $f$ has two distinct roots -

$$\begin{array}{*{20}c} {u_1 = \frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}}}{, u_2 = \frac{{ - b - \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$$

I found in an article (similar examples can be found in other texts) -

$$Q[u_1] \cong Q[x]/\langle f(x)\rangle.$$

Now I understand -

$\langle f(x)\rangle$ is an ideal, $Q[x]/\langle f(x)\rangle$ is a quotient ring, elements of $Q[x]/\langle f(x)\rangle$ are cosets, i.e. $Q[x]/\langle f(x)\rangle$ is the set of polynomials in $Q[x]$ of degree less than $\text{deg}(f(x))$, i.e. degree $2$.

On the other hand, $Q[u_1]$ is created by adjoining $u_1$, which is an element of degree 2 (the closed expression of $u_1$ has square root and square), to $Q_1$.

Then how is it possible that $Q[u_1] \cong Q[x]/\langle f(x)\rangle?$ What does $Q[u_1] \cong Q[x]/\langle f(x)\rangle$ mean? What are the elements of $ Q[x]/\langle f(x)\rangle$ here? Plz explain elaborately with demonstration. Thanks.

Answer 1

The substitution homomorphism $\phi(X)\mapsto \phi(u_1)$ from $\mathbb{Q}[X]$ to $\mathbb{C}$ has kernel $\langle f(X)\rangle$ and image $\mathbb{Q}[u_1]$; so the result follows from the Isomorphism Theorem.

Answer 2

Well, put $u_1= x + \langle f(x)\rangle$ (the residue class of $x$ mod $f(x)$).

Then $Q[u_1]$ becomes $Q[x]/\langle f(x)\rangle$.

It is clear that $u_1$ is a zero of $f(x)$ in the quotient ring $Q[u_1]$.

As a note, if $f(x)$ is irreducible the ring adjoint $Q[u_1]$ equals the field extension $Q(u_1)$.