For an ideal $I$ of $K[x]$, $K[x]/I$ is finitely generated iff $I$ is nonnull

In a book on rational series, a blunt statement is made to the effect that:

For $K$ a field, $I$ an ideal of $K[x]$, $K[x]/I$ is finitely generated iff $I$ is nonnull.

The statement elaborates with the not-so-enlightening (to me) sentence

This is true since a nonnull ideal in $K[x]$ always has a finite codimension ⁽¹⁾, and the latter is equal to the degree of any generator of this ideal ⁽²⁾.

I gather that if (2) is true, then $K[x]$ may be finitely generated if $K$ itself is finitely generated, but this is as far as I can go.

As for (1), I have a feeling of why this is true, but no proof. Thus I need help in proving the whole statement :-) Thanks !

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The set of the classes of the powers of $x$ are clearly generators of $K[x]/I$. Recall that since $K$ is a field, $K[x]$ is euclidian so $I$ is generated by a polynomial, says $P$, of degree $n$.

Now consider any power $x^l$, with $l>n$. The euclidian division gives two polynomials $Q$ and $R$ such that $$x^l=QP +R$$ with $R$ of degree lesser than $n-1$. In particular the class of $x^l$ in $K[x]/I$ is generated by the classes of the $x^k$, with $0\leq k \leq n-1$, and thus the dimension of $K[x]/I$ is lesser than $n$. To finish, show that those classes are indeed a base, and check the case where $I$ is reduced to $0$.

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Hint $\$ Over any ring, the polynomial division algorithm works universally for monic polynomials. Thus modulo a monic $\rm\:f,\:$ every polynomial is congruent to one of degree smaller than $\rm\:deg\: f.\:$

Or, computationally, $\rm\:f = x^n\ +\ c_{n-1}\,x^{n-1}+\cdots + c_1\,x + c_0 = 0\:$ may be viewed as a rewrite rule

$$\rm\: x^n \equiv -c_{n-1} x^{n-1} -\cdots -c_1\,x -c_0\pmod f$$

By induction, this allows one to rewrite all powers of $\rm\:x\:$ to congruent polynomials of $\rm\:deg < n.\:$

This monic viewpoint will come to the fore when you later meet integral extensions.

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When one writes finitely generated, it is always good to specify the particular ring for which this is true.

E.g. $K[x]$ is always finitely generated as a module over itself (indeed, any ring with unit $R$ is finitely generated --- by $1$ --- as a module over itself).

The statement in your question is about finite generation over $K$, and one could replace finitely generated by finite dimensional (since a $K$-module is the same as a $K$-vector space, and being finitely generated is the same as being finite-dimensional).

So the statement could be made simpler and clearer by writing

For $K$ a field and $I$ an ideal in $K[x]$, the quotient $K[x]/I$ is finite-dimensional as a $K$-vector space if and only if $I$ is non-null.

The proof of the if direction is as indicated in the other two answers, and the conclusion they lead to is that if $I$ (which is necessarily principal --- the ring $K[x]$ is a PID) is generated by a degree $n$ polynomial, then $K[x]/I$ has dimension $n$ over $K$.

The proof of the only if direction is easy: if $I = 0$, then $K[x]/I = K[x]$, and it's easy to see that $K[x]$ is infinite dimensional over $K$.

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