# Maximal Ideals in Polynomial Ring Over a Field

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.

• Look at irreducible polynomials. – Randall Dec 29 '17 at 20:24
• @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. – user193319 Dec 29 '17 at 20:28

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?

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.

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 $F[x]$.

• Given the level OP says they are at, it is probably pretty nontrivial to prove that the kernel of this homomorphism is $(x-a)$. – Eric Wofsey Dec 29 '17 at 20:31
• The proof is a straighforward consequence of the division algorithm in $F[x]$. – carmichael561 Dec 29 '17 at 20:32
• 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. – Eric Wofsey Dec 29 '17 at 20:36

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.

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.