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Let $F$ be a field. Prove that $\langle X,Y\rangle$ is a maximal ideal of $F[X,Y]$.

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Hi: I notice in the past several posts you're in the habit of just posting problem statements. To get better help, you should include whatever partial progress you've made, and it would be better if you phrased the post as a question and not as an imperative task. Posters who persist using this pattern often experience a backlash of downvotes, closures, and can become ignored. Don't let that happen: spend a little more time on your posts! –  rschwieb Dec 15 '13 at 14:15
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An small add to the above comment: you also didn't accept (and probably upvote) any answer so far. Although this is not mandatory, it is however the best way to show your gratitude to the answerers. –  user89712 Dec 15 '13 at 16:19

3 Answers 3

Hint: The following result is fundamental and you should always keep it in mind:

If $R$ is a commutative ring and $P$ is an ideal in $R$, then the quotient ring $R/P$ is a field if and only if $P$ is a maximal ideal.

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I can think of a few situations when having this in mind would be considered strange, or maybe even inappropriate. ;) –  tomasz Dec 15 '13 at 13:42
    
Is F[X,Y]/<X,Y> a field? I dont know. –  EuReka Dec 15 '13 at 13:43
    
You should compute that quotient. Hint: you are expecting to get a field... what object is a field in your hypothesis? –  Abramo Dec 15 '13 at 13:45
    
@EuReka : Are you familiar with quotient-ing a ring by some ideal? –  Praphulla Koushik Dec 15 '13 at 14:08
    
nnnnnnnnnnnnnnnnnnnnot getting. plz explain. –  EuReka Dec 15 '13 at 14:09

I am sorry to say this but it would be fruitful if you can identify definitions of terms you are using...

By $F[X]/(X)$ we mean $X$ is zero in $F[X]$, we would not have $X$ terms in $F[X]$..

What is $F[X]$??

Collection of all polynomials with coefficients in $F$

If $X=0$ then what would we left with?

$F[X]=\{a_0+a_1X+a_2X^2+\dots+a_nX^n : n\in \mathbb{N}; a_i\in F\}$

Now what would be $F[X,Y]$??

what would be result if we see $(X,Y)=0$ in $F[X,Y]$

As $X,Y$ are independent we would then have $X=0,Y=0$ when we say $(X,Y)=0$

So, Now what would we will be left with when we say $F[X,Y]/(X,Y)$??

I would prefer not to interrupt in some one else answer..

Hope this helps you some how...

Before commenting anything Please read what do we mean by quotient..

Good luck!

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Without using quotient operations:

Let $I$ be an ideal of $F[X,Y]$ containing $X$ and $Y$. Let $P$ be in $I$. Then $P = \sum_{n\geq 0, m\geq 0} a_{nm} X^n Y^m$. So $P = a_{00} + \sum_{m \geq 1} a_{0m} Y^m + \sum_{n \geq 1} a_{n0}X^n + \sum_{n \geq 1, m\geq 1} a_{nm} X^n Y^m$.

The second term is in the ideal generated by $Y$, the third in the ideal generated by $X$ and the fourth in both. Since $I$ contains these ideals, it shows that $a_{00} \in I$. If all $P$ in $I$ have $a_{00} = 0$, then $I$ is the ideal generated by $X$ and $Y$. If one $P$ has $a_{00} \neq 1$, then $a_{00}$ is an invertible and belongs to $I$, showing that $I = F[X,Y]$.

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