Have to prove that $(x-a, y-b)$ is a maximal ideal of $K[x,y]$ . I looked for a proof but it involved algebraic geometry. I don't know algebraic geometry. I have seen one proof in math stacks exchange ,in which the quotient ring is shown to be Field. I am posting this question for more elementary proof (easy proof ) ,if there is one.
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1What is K[X,Y] ? – Qwerty Jun 07 '16 at 16:36
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2You could compute the quotient, and see that it is a field. – Watson Jun 07 '16 at 16:41
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1Hint: Consider the map that evaluates a polynomial at $(a,b)$. – Tobias Kildetoft Jun 07 '16 at 16:54
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1For the converse (that maximal ideals are of this form) one usually uses algebraic geometry, e.g., Hilbert's Nullstellensatz - see here. However, your question is elementary. – Dietrich Burde Jun 07 '16 at 17:38
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I find it hard to believe that you saw a proof using algebraic geometry. This is very simple (that follows from that fact that it's algebra but I see how to do it).
It's enough to show that $$(x-a,y-b)=\{p\in K[x,y]:p(a,b)=0\}.$$
For the less obvious of the two inclusions, assume that $p(a,b)=0$. Now $x^n=((x-a)+a)^n=\dots$ and similarly for $y^n$, so there exists a polynomial $P$ with $$p(x,y)=P(x-a,y-b).$$So $P(0,0)=p(a,b)=0$. So $P$ has vanishing constant term. Every other term is divisible by $x$ or by $y$, so there exist $Q$ and $R$ with $$P(x,y)=xQ(x)+yR(y).$$So $$p(x,y)=(x-a)Q(x-a)+(y-b)R(y-b),$$hence $p\in(x-a,y-b)$.
David C. Ullrich
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you are saying that xb-ya belongs to the ideal because we can write x as x-a+a,and similar for y. Am I correctly getting your logic? – low iq Jun 08 '16 at 02:51
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C.Ullrich does (x-a,y-b) mean polynomials whose terms are product of (x-a) and (x-b)? please give me the general form of an element in K[X,Y]. – low iq Jun 08 '16 at 07:38
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@lowiq Oh dear. If you don't know the definition of $K[x-a,y-b]$ then asking how to prove things about it is premature. The definition is in the book... – David C. Ullrich Jun 08 '16 at 12:54
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as far as I understand, the polynomial ring in two variables can be obtained as:K[x,y]=k[x][y] .I asked for example of a polynomial that belongs to the ideal (x-a, y-b) so that I can understand solution clearly. – low iq Jun 08 '16 at 13:37
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@lowiq Right. The $K[x-a,y-b]$ in my last comment was a typo for $(x-a,y-b)$, sorry. – David C. Ullrich Jun 08 '16 at 13:48
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there is one proof of this question using some isomorphism theorem on this wetsite.i know we can prove it by showing that the quotient by ideal is a field.can you tell me what more result is needed to prove it this way? – low iq Jun 09 '16 at 09:59
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This one is very elementary, using only the definition of maximal: After applying some automorphism, it suffices to deal with $I=(x,y)$. Take some $f \notin I$. Then $f$ has non-zero constant coefficient, i.e. $f=xg+yh+c$ for some non-zero constant $c$. We deduce:
$$c = f - xg -yh \in (I,f),$$
i.e. $(I,f)=(1)$. Thus we have shown $I$ is maximal, since we have shown that any larger ideal is the whole ring.
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The ideal generated by $I$ and $f$, you might also find this notated as $I+(f)$. – MooS Jun 07 '16 at 17:58
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C belongs to that ideal. And so it is generated by 1.but how does it prove its maximal? Because 1 generates whole ring? – low iq Jun 08 '16 at 02:50
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since it is field and unit element belongs to ideal ,it generates the whole ring. – low iq Jun 09 '16 at 04:10