I wonder if $K$ is a field then $K[X]$ is a PID then the ideals of it have formula that is $<f(X)>$ when $f(X)$ is a polynomial with coefficent in $K$

Question

I see the solution of proving if K is a field then $K[X]$ is PID that is taking the minimal polynomial of ideal but I do not see they use the hypothesis K is field? Especially, I do not see the difference between K is ring and K is field. Thank you so much for you help.

Answer 1

The proof that $K[X]$ is a PID uses essentially that we can do polynomial division in $K[X]$.

In $\mathbb Z[X]$ however we cannot perform polynomial division, i.e. for $a=X$ and $b=2$ there are no $q,r$ with $\deg r < \deg a$ and $a=qb+r$.

Answer 2

the difference that is if $K[X]$ is a principal domain then $\forall a\in K^*$ the ideal $(a) + (X)$ is principal, so there i s $P\in K[X]$ such that $(a) + (X) = (P)$, this applique that $a$ and $X$ are in $(P)$, and so $a = PQ$ and $X = PR$ for some $Q,R\in K[X]$ , since $K[X]$ is one integral domain we have $dega = degP + degG $, and this shows that $P$ is a constant $b\in K$, so $X = b R $ and so as $ degX=1$ therefore necessarily $R = c X$ with $c$ a constant in $K$; the equality $ X = bcX $ shows that $bc = 1$ and so $b$ invertible and therefor $(a) + (X) = (1)=K[X]$; by Bézout's theorem there is $ M,N \in K[X]$ such that $aM + NX = 1$ this is true only if $N = 0$ (definition of polynomial) and $M$ is a constant $m\in K$ ; so then $am = 1$ and $a$ reversible in $K$ and so $K$ a field.