Why is $\mathbb{Z}$ not a euclidean domain? What goes wrong with the degree function?

I know that $K[X]$ is a Euclidean domain but $\mathbb{Z}[X]$ is not.

I understand this, if I consider the ideal $\langle X,2 \rangle$ which is a principal ideal of $K[X]$ but not of $\mathbb{Z}[X]$. So $\mathbb{Z}[X]$ isn’t a principal ideal domain and therefore not an Euclidean domain.

But I don’t understand this, if I consider the definition of Euclidean domains. Basically, a Euclidean domain is a ring where I can do division with remainders. For polynomial rings, the Euclidean function should be the degree of the polynomials. What’s the crucial difference between $K[X]$ and $\mathbb{Z}[X]$ with respect to this?

I already did exercises involving polynomial division in $\mathbb{Z}[X]$, so clearly I must be missing something here.

Solutions Collecting From Web of "Why is $\mathbb{Z}$ not a euclidean domain? What goes wrong with the degree function?"

Try to divide something like $x+1$ by $2x+1$. If it were a Euclidean Domain, you should be able to write $x+1=q(x)(2x+1) + r(x)$ where $r(x)$ has to have degree 0. You can see why this is not possible to do by looking at the coefficient on the $x$ term, since $2$ is not invertible in $\mathbb{Z}$

In $K[x]$ the reason things go nicely is because every non-zero element in $K$ is invertible ($K$ being a field). So we can have
$$a(x)=b(x)q(x)+r(x) \qquad \text{ with } \qquad 0 \leq \text{deg}r(x) < \text{deg}b(x).$$

But one of the problems in executing this in $\mathbb{Z}[x]$ is that unless the polynomial is monic you may not have the degree inequality (which is one the requirements of Euclidean Norm function). For example. if you try to carry out division with $a(x)=x+1$ and $b(x)=2x$, then you cannot have the degree of the remainder less than $1$. Thereby it does not fulfill the required conditions of an Euclidean Domain.

Try perform a division with remainder between $X$ and $2$ in $\mathbb Z[X]$.

Take the two polynomial of $\mathbb{Z}[X]$, $P(X)=X^2+1$ and $Q(X)=2X$ and try to perform the division while keeping integer coefficients.

You can see the obstruction is coming from non invertible elements of the ring $\mathbb{Z}$

Suppose you have a Euclidean domain. I claim that any element $x$ with $\deg x = 0$ should be invertible. Indeed the claim of the division algorithm gives that I can write $1 = qx + r$ with $r = 0$ or $\deg r < \deg x$. The latter is impossible if the degree is $0$, so $r = 0$ and thus $qx = 1$, giving that it is a unit.

Thus the explicit obstruction is that there are integers with degree 0 that are not invertible in the polynomial ring.

In fact, let $R$ be an integral domain and consider the ring $R[x]$ with the degree function. Then if $R[x]$ were a Euclidean domain under the degree function, every $r \neq 0 \in R$ would be invertible in $R[x]$, but since $R$ is an integral domain it is easy to see this cannot happen without $r$ being invertible in $R$. So $R$ is a field.

Hint $\ $ An element $a\ne 0\,$ of $\,\rm\color{#c00}{minimal}\,$ Euclidean value is a unit, i.e. invertible (else $\,a\nmid b\,$ for some $\,b\,$ so $\,b\div a\,$ leaves remainder smaller than $\,a,\,$ contra minimality of $\,a).\,$ In particular, if the polynomial degree is a Euclidean function, then a nonzero element of $\,\rm\color{#c00}{degree\ 0}\,$ is a unit.

Remark $\ $ This is a special case of the general argument that ideals are principal in Euclidean domains, i.e. any element $\neq 0\,$ of $\,I\,$ of least Euclidean value must divide every other element, else the remainder would be a smaller element of $I.$ Above is the special case $\,I = (1).$

Similar ideas can be used to prove that certain quadratic number rings are not Euclidean, e.g. see the use of a “universal side divisor” in the proof of Lenstra linked in this answer. One can obtain a deeper understanding of Euclidean domains from the excellent exposition by Hendrik Lenstra in Mathematical Intelligencer 1979/1980 (Euclidean Number Fields 1,2,3)

In a PID, any irreducible element generates a maximal ideal. In $\mathbf Z[X]$, both $2$ and $X$, say are irreducible, but they’re no maximal, since:
$$\mathbf Z[X]/2\mathbf Z[X]\simeq(\mathbf Z/2\mathbf Z)[X]\quad\text{and}\mathbf Z[X]/X\mathbf Z[X]\simeq\mathbf Z,$$
which are not fields.