$F$ be a Euclidean domain satisfying $\delta(a + b) ≤ \max(\delta(a),\delta(b))$ then $F$ is a field $K$ or $F \subset K[x]$ Let $F$ be a Euclidean domain with norm function $\delta$ satisfying the additional condition $\delta(a + b) ≤ \max(\delta(a),\delta(b))$. We have to prove that either $F$ is a field $K$ or $F \subset K[x]$ for some field $K$.
Now let's Define $K = \{a \in F :
\delta(a) ≤ 1\}$. I want to show $K$ is a subring of $F$. Now let $a,b \in K \implies \delta(a − b) ≤ \max(\delta(a),\delta(b)) ≤ 1 \implies a-b \in K$ and $\delta(ab)=\delta(a)\delta(b) ≤ 1 \implies ab \in K$ so $K$ is a subring of $F$. And $K$ also contain Identity as $\delta(e) = \delta(e^2) = \delta(e)\delta(e) \implies \delta(e) = 1$.  $K$ is subring of $F$ and contains identity.
Now we know a theorem that subring of field contains identity then it's a intregral domain.
From this I don't know how to show $K$ is field or $F \subset K[x]$
 A: It seems that $K=\{a\in F\mid \delta(a)=0\}$ is a field. The intuition for this is to think that $\delta$ acts like $\deg$, then the field $K$ is precisely the set of elements, $x$, in $F$ that have $\delta(x)=0$. Let's show this:

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*Closed under addition: Let $a,b\in K$, then $\delta(a+b)\leq \max\{\delta(a),\delta(b)\}=0$. Since $\delta: F\rightarrow\Bbb N$ we must have $\delta(a+b)=0$, so $a+b\in K$.


*Closed under multiplication: Let $a,b\in K$, then $\delta(ab)=\delta(a)\delta(b)=0$, so $ab\in K$.


*Existence of multiplicative identity & inverses: Let $a\in K$, and $1,q,r\in F$. We may write $$1=aq+r$$ where either $r=0$, or $\delta(r)<\delta(a)$. Since $\delta(a)=0$, we have $\delta(r)<0$ which is impossible. Therefore, $$1=aq$$ Now we need to show that $q\in K$, then it will follow that $1\in K$ also, since $K$ is closed under multiplication. Notice that $1=aq$ implies that $q=aq^2$, we thus have $\delta(q)=\delta(a)\delta(q^2)=0$, so $q\in K$.

To show that $F$ is a subring of $K[x]$, first define $$A=\{a\in F\mid\delta(a)\geq 0\}$$
Then $F=A$, since $F$ is an Euclidean domain. Now choose $a\in A$ with $\delta(a)=r>0$ as small as possible. Then define the homomorphism $$\phi:A\rightarrow K[x]$$ by $\phi(a)=x^r$, and fixes $K$. Now we need to check the following:

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*Additivity: First, for any $f,g\in A$ we may write $f=c_na^n+\ldots+c_0$ and $g=d_ma^m+\ldots+d_0$, where $c_n,\ldots,c_0, d_m,\ldots,d_0\in K$. To see this notice that we can write $f=c_na^n+\rho$, where $\delta(c_n)<\delta(a)$. This forces $c_n$ to be a unit, $c_n\in K$. From this we may continue the process, knowing that $\delta(f-c_na^n)\leq r(n-1)$. Then it is clear that $\phi(f+g)=\phi(f)+\phi(g)$.


*Multiplicativity: Also clear by the argument above.


*Injectivity: Suppose $f(x^r)=0$ then, $c_{n},c_{(n-1)},\ldots,c_0$ are all $0$. Thus $f(a)=0$.
Infact, $A\cong K[a]$.
