0
$\begingroup$

Okay. I showed that $(x)$ is a maximal ideal in the polynomial ring $F[x]$, where $F$ is some field. Now I have been asked to find another maximal ideal in $F[x]$. I tried showing that $(x+1)$ is a maximal ideal, but I had to luck. I could use a hint. I don't know very much about polynomials at this point (e.g., the degree of a polynomial hasn't even been defined yet; also, I don't know that $F[x]$ is a PID). I do know, e.g., that if $F[x]/I$ is a division ring, where $I$ is an ideal, then $I$ is a maximal ideal. I tried this for $I=(x+1)$, but I had no luck.

$\endgroup$
2
  • $\begingroup$ Look at irreducible polynomials. $\endgroup$ – Randall Dec 29 '17 at 20:24
  • $\begingroup$ @Randall Isn't $x+1$ an irreducible polynomial? By the way, I know no nice theorems about irreducible polynomials. I only have the definition of an irreducible element: $c$ is irreducible iff $c \neq 0$ is not a unit and $c=ab$ implies either $a$ or $b$ is a unit. Beyond that, I don't know much about irreducible elements. $\endgroup$ – user193319 Dec 29 '17 at 20:28
1
$\begingroup$

Hint: Instead of describing another maximal ideal explicitly using generators, just try to describe it as a kernel of some surjective homomorphism to a field. The ideal $(x)$ is the kernel of surjective homomorphism $\varphi:F[x]\to F$ given by $\varphi(f)=f(0)$. Can you think of any other homomorphisms $F[x]\to F$ defined similarly, which you could take the kernel of?

An answer is hidden below.

For any element $a\in F$, there is a homomorphism $ev_a:F[x]\to F$ given by $ev_a(f)=f(a)$. This is surjective, since for any $b\in F$, $ev_a$ sends the constant polynomial $b$ to $b$. So for any $a$, the kernel of $ev_a$ is another maximal ideal in $F[x]$. These maximal ideals are all different from each other, because $x-a\in\ker(ev_a)$ but $x-b\not\in\ker(ev_a)$ for all other $b\in F$. In fact, $\ker(ev_a)=(x-a)$ so you can get your proposed maximal ideal $(x+1)$ as $\ker(ev_{-1})$, but this takes some work to prove.

$\endgroup$
0
$\begingroup$

The ideal $(x-a)$ is maximal for any $a\in F$. Indeed, the homomorphism $F[x]\to F$ defined by $x\mapsto a$ is surjective and has kernel $(x-a)$. Since $F$ is a field, this implies that $(x-a)$ is maximal. If $F$ is algebraically closed, then these are the only maximal ideals in $F[x]$.

$\endgroup$
3
  • $\begingroup$ Given the level OP says they are at, it is probably pretty nontrivial to prove that the kernel of this homomorphism is $(x-a)$. $\endgroup$ – Eric Wofsey Dec 29 '17 at 20:31
  • 1
    $\begingroup$ The proof is a straighforward consequence of the division algorithm in $F[x]$. $\endgroup$ – carmichael561 Dec 29 '17 at 20:32
  • $\begingroup$ They claim to not have officially learned about the degree of polynomials yet; I doubt the division algorithm is something they can use without proving it from scratch. $\endgroup$ – Eric Wofsey Dec 29 '17 at 20:36
0
$\begingroup$

Imagine there is a larger proper ideal $I $ that contains $(x+1)$. It would then contain some polynomial $p\not\in (x+1)$. Divide $p $ by $x+1$: $p=(x+1)q+c$ where $c\in F $.

  • If $c=0$, then $p=(x+1)q\in (x+1)$ - a contradiction.

  • If $c\ne 0$, note $c=p-(x+1)q\in I $, which would imply $c^{-1}c=1\in I $ and then every polynomial would be in $I $ and $I $ wouldn't be proper - a contradiction again.

$\endgroup$
0
$\begingroup$

we can prove easily that $F[x]\cong F[x+1]$, so $F[x]/(x) \cong F[x+1]/(x+1)\cong F$, and we get $(x+1)$ is a maximal ideal.

$\endgroup$

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.