0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $Q(u_1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.