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Stanley Yao Xiao
Stanley Yao Xiao
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Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - (\frac{b^2-4ac}{4a})y^2. $$$$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \left(\frac{b^2-4ac}{4a}\right)y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - (\frac{b^2-4ac}{4a})y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \left(\frac{b^2-4ac}{4a}\right)y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

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JSE
JSE
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Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \frac{b^2-4ac}{4a}. $$$$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - (\frac{b^2-4ac}{4a})y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \frac{b^2-4ac}{4a}. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - (\frac{b^2-4ac}{4a})y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

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Joe Silverman
Joe Silverman
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Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \frac{b^2-4ac}{4a}. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)