# Polynomial rings over a field and maximal/prime ideals

Let $F$ be a field , I want to prove that every proper nontrivial prime ideal of $F[x]$ is maximal.

My definitions of prime/maximal ideals are as follows:

$N$ is a prime ideal of $R$ iff $ab \in N \implies a \in N$ or $b \in N$.

Two definitions for maximal:

$I$ is a maximal ideal of $R$ if there is no proper ideal $N$ of $R$ properly containing $I$.

or

$I$ is a maximal ideal of a commutative ring $R$ iff $R/I$ is a field.

There is another definition my professor did in class involving cosets and units but I can't recall it.

As far as I am aware there are non-commutative fields, which makes this problem a little tricky.

• It is I think clear that if $f$ is not irreducible, the ideal $I$ generated by $f$ is not maximal. To show that if $f$ is irreducible, then $I$ is maximal, suppose to the contrary there is a proper ideal $J$ strictly above $I$. Let $g$ be a generator of $J$. Then $f$ is a no-trivial multiple of $g$, meaning $f$ is not irreducible. – André Nicolas Mar 13 '14 at 1:21
• why is it true that $f$ is not irreducible if $f$ is a non-trivial multiple of g? – terrible at math Mar 13 '14 at 1:28
• @AndréNicolas okay, i've let $a \in I$ and made the deduction that $a=gf$ for some $g \in F[x]$ and I am now trying to reason about the degree of $a$ and $g$, since I have a definition that f is irreducible iff it can't be expressed as a product of lower degree polynomials. $deg(a) \geq deg(f)$ clearly but I don't know what I can say about $deg(g)$ or $deg(a/g)$. – terrible at math Mar 13 '14 at 2:18
• You have changed notation. I am not happy about that, and will keep the previous notations. Recall that $f$ is a generator of $I$. If $J$ is a proper extension of $I$ not equal to $F[x]$, let $g$ be a generator. Note that $g$ is not a constant polynomial, so it has degree $\ge 1$. Because $J$ is a proper extension of $J$, we have $f=gh$ where $h$ is not a constant polynomial, so has degree $\ge 1$. Since the degree of $f$ is the sum of the degrees of $g$ and $h$, it follows that the degree of $g$ is less than the degree of $f$, and the same is true for the degree of $h$. The end. – André Nicolas Mar 13 '14 at 2:28
• @AndréNicolas I got it! HEre's my proof. Let $f = gh$ be a factorization of $f$ in $F[x]$. $f$ generates $N$ so clearly $gh \in N$, thus either $g \in N$ or $h$ in N. This gives us that either $g$ or $h$ has $f$ as a factor. But this means its impossible for the degree of $g$ and $h$ to be less than the degree of $f$, so $f$ cannot be expressed as a product of lower degree polynomials and is this irreducible, proving that $N = <f>$ is a maximal ideal. – terrible at math Mar 13 '14 at 2:29

Proof:

Let $I$ be a nontrivial prime ideal for $F[x]$. Since $F$ is a field, that means $F$ is a Euclidean Domain which also implies $F$ is a PID. So $I$ is a principal ideal which is generated by $f$ for some $f \in F[x].$

$I$ is maximal if and only if $f$ is irreducible.

Think you can go from there?

• never actually heard the term Euclidian domain - what is it? (wikipedia is not being particularly helpful) also, don't you mean $F$ is a field not $F[x]$ ? (i'm pretty sure we aren't gauranteed that the polynomial ring is a field.) In any case, to be an irreducible polynomial in $F[x]$ means you have no roots in F, and F is a field so I have inverses.. perhaps I can take any polynomial in I and show it has to be irreducible. – terrible at math Mar 12 '14 at 23:52
• dpmms.cam.ac.uk/~par31/notes/ed.pdf This should give you an idea what a Euclidean domain is. – user133458 Mar 12 '14 at 23:55
• ah, that's a neat property. so then I know $I$ is just all $(g)(f)$ for $(g \in F[x])$. But I know that I is prime, so if $gf \in I$ then $g \in I$ or $f \in I$ . $gf \in I \implies (gf-I) = 0$.. I think i need to redo this argument without this $gf$ business :P let me try again in a new comment – terrible at math Mar 13 '14 at 0:00
• $I$ is generated by $f$ so $I = gf$ for $g \in F[x]$. Let $a,b \in F[x]$ and consider $(a+I)(b+I) = 0, ab + I = 0$ This is true if and only if $ab \in I$. But $I$ is prime so if $ab \in I$ then either $a \in I$ or $b \in I$. Okay, I think I am stuck: how can I translate this into something about the irreducibility of $f$? – terrible at math Mar 13 '14 at 0:04
• you there? i'm still confused :( – terrible at math Mar 13 '14 at 0:57