# [SOLVED]Discrete valuation Ring which is a subring of a field K Problem

#### Cbarker1

##### Active member
Dear Everyone,

I am stuck in the middle of a proof. Here is the background information from Dummit and Foote Abstract Algebra 2nd ed.:

Let $K$ be a field. A discrete valuation on $K$ on a function $\nu$: $K^{\times} \to \Bbb{Z}$ satisfying
1. $\nu(a\cdot b)=\nu(a)+\nu(b)$ [i.e. $\nu$ is a homomorphism from the multiplication group of nonzero elements of $K$ to $\Bbb{Z}$]
2. $\nu$ is surjective, and
3. $\nu(x+y)\ge \min{[\nu(x),\nu(y)]}$, for all $x,y\in K^{\times}$ with $x+y\ne 0$
The set $R=\left\{x\in K^{\times}| \nu(x)\ge 0\right\} \cup \left\{0\right\}$.

The definition from the book for a subring of ring $P$ (in general) states is a subgroup of $P$ that is closed under multiplication.

Prove $R$ is a subring of $K$ which contains the identity.

Here is my work:

Proof: Since $1\in K$, then $1\in R$ due to the fact that $\nu(1)\ge 0$. Let $a,b\in R$. Then $\nu(a\cdot b^{-1})=\nu(a)+\nu(b^{-1})=\nu(a)+{\nu(b)}^{-1}\ge 0$ (here is where I am stuck).

Thanks,
Cbarker1

Last edited:

#### Euge

##### MHB Global Moderator
Staff member
Hi Cbarker1 ,

It is important to note that a subring of a ring is an additive subgroup. Here, we must show that (1) $R$ contains $0$, (2) $R$ is closed under subtraction, and (3) $R$ is closed under multiplication. Claim (1) is trivial by definition of $R$. For claim (2), use the fact that $\nu(a - b) \ge \min\{\nu(a), \nu(-b)\}$ if $a \neq b$ and $\nu(-b) = \nu(b)$. For claim (3), consider that $\nu(ab) = \nu(a) + \nu(b) \ge 0$ provided $\nu(a) \ge 0$ and $\nu(b) \ge 0$.

#### Cbarker1

##### Active member
Hi Cbarker1 ,

It is important to note that a subring of a ring is an additive subgroup. Here, we must show that (1) $R$ contains $0$, (2) $R$ is closed under subtraction, and (3) $R$ is closed under multiplication. Claim (1) is trivial by definition of $R$. For claim (2), use the fact that $\nu(a - b) \ge \min\{\nu(a), \nu(-b)\}$ if $a \neq b$ and $\nu(-b) = \nu(b)$. For claim (3), consider that $\nu(ab) = \nu(a) + \nu(b) \ge 0$ provided $\nu(a) \ge 0$ and $\nu(b) \ge 0$.

Claim (2) (WTS:$a-b \in R$): Let $a,b \in R$. Then $\nu(a-b) \ge \min\{\nu(a),\nu(-b)\}\ge \min\{\nu(a),\nu(b)\}\ge 0$ provided that $a\ne b$. Thus, $a-b\in R$.
is this the right direction of the proof?

thanks
Cbarker1

Last edited:

#### Euge

##### MHB Global Moderator
Staff member
You're on the right track, but to make the argument precise, suppose, without loss of generality, that both $a$ and $b$ are nonzero (since $x \in R \implies -x \in R$) and $a \neq b$ (or else the statement $a - b \in R$ is immediate). Then since $a, b\in R$, $\nu(a) \ge 0$ and $\nu(b) \ge 0$, so that $\nu(a - b) \ge \min\{\nu(a), \nu(-b)\} = \min\{\nu(a),\nu(b)\} \ge 0$. Hence $a - b\in R$.