1
$\begingroup$

A theorem states that given an irreducible polynomial $f(x) \in F[x]$, the field $K = F[x]/(f(x))$ contains a root $\alpha$ of f(x).

The proof says that if we let $\alpha = \overline{x} \in F[x]/(f(x))$ then $f(\alpha) = \overline{f(x)} = 0$ so $\alpha$ is indeed a root of $f(x)$ in K.

The proof is what I don't understand - how is it valid? How does setting $\alpha = \overline{x}$ work? It is clear that $\overline{f(x)} = 0$, but how does this show that there exists a root (which is a constant) of f?

$\endgroup$
  • 1
    $\begingroup$ $\bar{x}$ means the coset of $x$ in the quotient ring $F[x]/(f)$ $\endgroup$ – Nick Mar 25 '17 at 22:14
2
$\begingroup$

It's the equivalence class of the monomial $x$ in the quotient ring $K=F[x]/(f)$ where basically the element $f$ will be equal to $0$.

So, $f$ had no roots in the field $F$, but we want a (necessarily bigger) field where it has at least one root. And we do it by a construction: first we adjoin a formal element to $F$ - that's how $F[x]$ enters the picture, and then we force this new element to be a root of $f$, by taking the quotient.

In formulas, suppose $f=a_0+a_1x+a_2x^2+\dots$, and taking $\alpha:=\bar x$, we will have $$f(\alpha) = a_0+a_1\bar x+a_2\bar x^2+\dots=\overline{a_0+a_1x+a_2x^2+\dots}=\overline{f}=0$$ where in the middle we used that taking the quotient respects the ring operations.

Can you catch where we use that $f$ is irreducible?

$\endgroup$
1
$\begingroup$

When you define the polynomial ring $K[x]$, you impose no relations on $x$: all the powers of $x$ are distinct and none of them can be expressed as a linear combination of other powers of $x$. However, there is no reason why we cannot define a different ring which is the same as $K[x]$ except for the additional condition that $x$ satisfies $f(x) = 0$: all the ring axioms are readily verified. This ring contains a root of $f$ by definition. Moving from $K[x]$ to this ring is exactly what you accomplish when you quotient by $(f(x))$.

It is very important to realize that adjoining new relations to a ring (that respect the ring structure) and quotienting by an ideal are the exact same thing.

$\endgroup$

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.