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I have seen someone make the following statement.

For a polynomial over the real numbers with $f(x,x^2)=0 ~\forall x \in \mathbb{R}$ we can factor it $f(x,y)=(y-x^2)g(x,y)$ for some other polynomial $g$ in two variables over the real numbers. The reasoning that has been given for this is we can view $f$ as a polynomial just in $y$ and via the degree function we get:

$f(x,y)=(y-x^2)g_x(y)+r_x(y)$

Now $r_x$ must be constant for all $x$ and it follows: $0=f(x,x^2)=r_x(x^2)$. Now we get $f(x,y)=(y-x^2)g_x(y)$ as a result. We know that for all $x,$ $g_x$ is a polynomial in $y$ but the statement seems to imply that $g_x$ is a polynomial in $y$ and $x$ which is what seems fishy to me. Why would this be true and if not, can you even factor polynomials in this fashion?

(In the original statement I saw $g_x$ was just written as $g(x,y).$)

Bill Dubuque
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VeryGenericUsername
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1 Answers1

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One way to look at the situation is by writing $g(x,y) = f(x,y + x^2)$. Then $g(x,y)$ is a polynomial with $g(x,0) = 0$ for all $x$. Writing $g(x,y) = \sum_{i,j=0}^n a_{ij}x^iy^j$, the statement that $g(x,0) = 0$ is equivalent to saying that $\sum_{ i=0}^n a_{i0}x^i = 0$ for all $x$. This means that each $a_{i0}$ is zero, so that $$g(x,y) = \sum_{i=0}^n\sum_{j=1}^n a_{ij}x^iy^j$$ $$=y \sum_{i=0}^n\sum_{j=1}^n a_{ij}x^iy^{j-1}$$ Thus letting $h(x,y) = \sum_{i=0}^n\sum_{j=1}^n a_{ij}x^iy^{j-1}$ we have $$g(x,y) = y \,h(x,y)$$ Now replacing $y$ by $y - x^2$ one has $f(x,y) = g(x,y -x^2) = (y - x^2) h(x,y-x^2)$, which is of the desired form since $h(x,y)$ and therefore $h(x,y - x^2)$ is a polynomial.

Zarrax
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Jul 28 '24 at 17:24
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    It's not a duplicate answer. The 12 year old question you linked to has only one answer, and it's different from this one. I do believe there is value in seeing multiple approaches to popular questions. I would like to add that this constant closing of questions because similar ones have been asked before is a disincentive to participating in this site, and in my opinion will harm it in the long run. – Zarrax Jul 28 '24 at 18:32
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    The answer duplicates a proof of the factor (or remainder) theorem, i.e. $, y-a\mid f(y),$ if $,f(a)=0\ $ (here $,a = x^2).,$ Duplicating the proof in this special case (vs. invoking the theorem by name as in the linked dupe) greatly complicates comprehension of the essence of the matter. – Bill Dubuque Jul 28 '24 at 18:49
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    No, quite the opposite, as many students are more comfortable with a concrete example than an abstract generalization involving rings over a polynomial ring. Also, I am not invoking the remainder theorem; obviously my answer is not the same as the linked one. Please stop this. – Zarrax Jul 28 '24 at 18:54
  • The question is tagged abstract-algebra. There we strive to abstract out common algebraic structure (and methods) so they can be efficiently reused in diverse contexts. A beginner reading the above answer will likely have no clue that it is essentially a simple application of the ubiquitous factor or remainder theorem over the coef ring $\Bbb R[x],$ since the proof becomes more complicated when repeated inline in this context. As for "comfortable", even a high school student can understand how to write a polynomial $,g(x,y),$ as a polynomial in $y$ whose coef's are polynomials in $x\ $ – Bill Dubuque Jul 28 '24 at 19:06
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    I don't actually answer to you, and if I have reasons to think a new answer is warranted, I have the right to give one. You were out of line instructing me not to. Your reasoning is highly unconvincing to me, and I am not willing to discuss the matter any further. – Zarrax Jul 28 '24 at 19:18