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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

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

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

- $\nu(a\cdot b)=\nu(a)+\nu(b)$ [i.e. $\nu$ is a homomorphism from the multiplication group of nonzero elements of $K$ to $\Bbb{Z}$]
- $\nu$ is surjective, and
- $\nu(x+y)\ge \min{[\nu(x),\nu(y)]}$, for all $x,y\in K^{\times}$ with $x+y\ne 0$

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

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