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Let $L/K$ be an algebraic extension, and let $a,b \in L$. Suppose that $f(X,Y) \in K[X,Y]$ is a polynomial such that $f(a,b)=0$.

Is it true that $$f(x,y)= P_{\text{min},a,K}(x) P(x,y) + P_{\text{min},b,K}(y) Q(x,y) $$ for some polynomials $P,Q \in K[X,Y]$ ?

I know that this is true if we consider instead $f(X) \in K[X]$, because $K[X]$ is an euclidean ring. My idea was to consider $K(X)[Y]$ which is euclidean, and then to come back to $K[X,Y]$, but this is probably not the best idea.

Thank you for your help!

Alphonse
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    To be clear (I hope), I think you are asking whether the ideal of $K[X,Y]$ that is the kernel of the evaluation $X:=a,Y:=b$ equals the sum of the ideals generated by the images under the embeddings $K[X]\to K[X,Y]$ and $K[Y]\to K[X,Y]$ of the kernel of the the evaluation $X:=a$ respectively the kernel of the the evaluation $Y:=b$. (It being fairly clear that it contains that sum.) – Marc van Leeuwen Sep 02 '16 at 09:42
  • HINT: $$f(a,b)=0 \Longleftrightarrow f(x,y) \in (x-a, y-b)K[x,y]$$ – Crostul Sep 02 '16 at 09:58
  • I think this seems to be a proper example: $K=\mathbb{Q}, L = \mathbb{R}, a=b=\sqrt{2}, f(X,Y)=X-Y$. For me it seems, this isn't in the generated ideal. @Crostul The problem is that $x-a \neq K[x,y]$ in general, hence don't give a solution for above. – ctst Sep 02 '16 at 10:02
  • Well, of course don't take $\mathbb{R}$ in the example above, but e.g. the algebraic closure of $\mathbb{Q}$ instead (forgot there $\mathbb{R}$ is not algebraic over $\mathbb{Q}$). – ctst Sep 02 '16 at 10:09
  • @ctst: $X-Y = X-\sqrt 2 - (Y-\sqrt 2)$ – Alphonse Sep 02 '16 at 10:11
  • @Alphonse $X-\sqrt{2} \notin \mathbb{Q}[X,Y]$! The minimal polynomial of $\sqrt{2}$ in our field is still $X^2-2$! – ctst Sep 02 '16 at 10:11
  • @ctst: But $X-\sqrt 2 = (X-\sqrt 2) \cdot 1 \in (X-\sqrt 2)\Bbb Q[X,Y]$ – Alphonse Sep 02 '16 at 10:12
  • @Crostul: how would you prove the part $\implies$ of your hint? – Alphonse Sep 02 '16 at 10:13
  • Let us continue this discussion in chat. – ctst Sep 02 '16 at 10:14
  • @Crostul: What about $f(x,y)=x+y^2-2,a=0,b=\sqrt 2$ ? We can't write $f(x,y)=xP(x,y)+(y-\sqrt 2)Q(x,y)$ with $P,Q \in \Bbb Q[X,Y]$. – Alphonse Sep 02 '16 at 10:21
  • Let $A=K[X],B=K[Y]$. Notice that if $f(a,b)=0$ then $$f(X,Y)=q_X(Y) (Y-b) + r_X(Y)$$ with $q_X, r_X \in A[Y] = K[X,Y], deg_Y(r) = 0$. Then $$r_X(Y) = q'_Y(X) (X-b) + r'_Y(X)$$ with $q', r' \in B[X] ; deg_X(r')=0$. You can then conclude that $r'$ is actually a constant, which is $0$ (from $f(a,b)=0$). – Watson Sep 29 '16 at 16:01
  • I found http://math.stackexchange.com/questions/1854588/ker-phi-a-1-a-n-for-a-ring-homomorphism-phi-rx-1-x-n-t :-) – Alphonse Jan 22 '17 at 16:08
  • Also http://math.stackexchange.com/questions/1352606 – Alphonse Feb 07 '17 at 13:42
  • As ctst mentioned above, the claim doesn't hold. But maybe it is true if we assume $K(a) \cap K(b) = K$ ? However, there would be a strange dissymmetry because we consider $$P_{min, a, K}(x), P_{min, b, K}(y)$$ and not $$P_{min, a, K}(y), P_{min, b, K}(x)$$ (there is no reason for choosing the $x$ for $a$ and $y$ for $b$...) – Alphonse Mar 03 '18 at 12:46
  • At least $K(a) \cap K(b) = K$ prevents from ctst's example. And at least it is true when $a,b \in K$ (see Crostul's hint). – Alphonse Mar 03 '18 at 12:47
  • Here is what I tried (which doesn't work - see ctst's comment): Consider $f \in A[Y]$ with $A = K[X]$. The ring $A[Y]$ is not a PID (since $A$ is not a field), but we can still perform an euclidean division by the monic polynomial $P_{min, a, K}(Y)$ (caution : $P_{min, a, K}(X) \in A$ is a constant non-monic polynomial in $A[Y]$ !) :

    $$f(X, Y) = q_1(X,Y)P_{min, a, K}(Y) + r_1(X,Y)$$

    Similarly, considering $r_1(X,Y)$ as an element of $B[X]$ where $B = K[Y]$, we have : $$r_1(X,Y) = q_2(X,Y)P_{min, b, K}(X) + r_2(X,Y)$$

    We have

    $$deg_Y(r_1) < deg_Y(f), deg_X(r_2) < deg_X(r_1)$$

     – Alphonse Mar 03 '18 at 12:47

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