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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.

user26857
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Marc
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  • I think the problem is there are irreducibles of degree zero in there, the primes in $\Bbb Z$ – Gregory Grant May 26 '15 at 20:54
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    Try writing $x^2 + 1 = 2x * q(x) + r(x)$ with deg r < deg (2x) = 1 – jxnh May 26 '15 at 20:55
  • Although $\mathbb{Z}[x]$ is not Euclidean domain, still we are able to simulate polynomial division or gcd algorithm, because of Gauss's lemma which breaks the initial polynomial into content and primitive part, thus making division possible for the primitive part. See Modern Computer Algebra by Gathen. – Pranav Bisht Mar 01 '17 at 12:24
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    @PranavBisht: "Thus making division possible for the primitive part." That assertion is false. $2x+1$ and $3x+1$ are both primitive. How do you divide $2x+1$ by $3x+1$ with remainder? You seem to be conflating the fact that $\mathbb{Z}[x]$ is a UFD (and thus the gcd can be defined) with the Euclidean algorithm. You can't. For example, $\gcd(x,2)=1$, but there is no way to express $1$ as $xp(x)+2q(x)$. You can do division by $p(x)$ if the leading coefficient of $p(x)$ is a unit ($1$ or $-1$), which will imply $p(x)$ is primitive; but you cannot divide by *any* primitive polynomial. – Arturo Magidin Sep 22 '21 at 01:08
  • @ArturoMagidin Yes, I see. We cannot run Euclidean algorithm on $\mathbb{Z}[x]$ but only on $\mathbb{Q}[x]$. For computing gcd in $\mathbb{Z}[x]$, we use Gauss's Lemma and Euclidean division in $\mathbb{Q}[x]$. – Pranav Bisht Sep 22 '21 at 08:07
  • @PranavBisht: No, we don't. Gauss's Lemma yields that a polynomial in $\mathbb{Z}[x]$ is irreducible if and only it is irreducible over $\mathbb{Q}$, and from that you can deduce that $\mathbb{Z}[x]$ is a UFD. But you cannot do the Euclidean division in $\mathbb{Q}[x]$ and then try to lift it to $\mathbb{Z}$. *Try it with the example I gave*. They are both primitive, and if you do the division in $\mathbb{Q}[x]$, you get $2x+1 = \frac{2}{3}(3x+1) + \frac{1}{3}$. How do you go back to $\mathbb{Z}[x]$? – Arturo Magidin Sep 22 '21 at 14:18
  • @ArturoMagidin Yes, I got your point in the first comment that one cannot perform Euclidean division in $\mathbb{Z}[x]$ (even if one can do so for $\mathbb{Q}[x]$) and thank you for that. In my last comment, I only talked about *computing gcd* in $\mathbb{Z}[x]$ by first computing gcd in $\mathbb{Q}[x]$. That is, gcd of primitive parts in $\mathbb{Q}[x]$ and then multiplying by gcd of contents. In this example, I believe gcd is 1 for both $\mathbb{Q}[x]$ and $\mathbb{Z}[x]$. – Pranav Bisht Sep 23 '21 at 06:38

8 Answers8

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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 $\mathbb{Z}$

CPM
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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.

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

user26857
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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.

user26857
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jxnh
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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 $\mathbb{Z}$

marwalix
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Hint $ $ An element $a\ne 0\,$ of $\,\rm\color{#c00}{minimal}\,$ Euclidean value is a unit (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 proof that ideals are principal in Euclidean domains (generated by a minimal element), since any element $\neq 0\,$ of $\,I\,$ of least Euclidean value divides every other element, else the remainder is a smaller element of $I.$ Above is 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 $1,2,3$).

Bill Dubuque
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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.

Bernard
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$\mathbb{Z}[X]$ is not PID since $\langle 2,x\rangle$ is not principal. So if $\mathbb{Z}[X]$ were euclidian it would be PID.