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An integral domain $R$ is a Euclidean domain if there is a degree function $\deg : R-\\{0\\} \to \mathbb{Z}_{\ge 0} $$$\deg : R-\{0\} \to \mathbb{Z}_{\ge 0} $$ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

An integral domain $R$ is a Euclidean domain if there is a degree function $\deg : R-\\{0\\} \to \mathbb{Z}_{\ge 0} $ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

An integral domain $R$ is a Euclidean domain if there is a degree function $$\deg : R-\{0\} \to \mathbb{Z}_{\ge 0} $$ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

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An integral domain $R$ is a Euclidean domain if there is a degree function $\deg : R-\\{0\\} \to \mathbb{Z}_{\ge 0} $ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

An integral domain $R$ is a Euclidean domain if there is a degree function $\deg : R-\\{0\\} \to \mathbb{Z}_{\ge 0} $ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

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A question about Euclidean domains

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?