# If $F$ is an algebraically closed field, then every maximal ideal of $F[x_1,…,x_n]$ is of the form $(x_1-c_1,…,x_n-c_n)$

I have problem to understand all the proof of the theorem :

If $F$ is an algebraically closed field, then every maximal ideal of $F[x_1,...,x_n]$ is of the form $(x_1-c_1,...,x_n-c_n)$.

Proof : Let $\mathfrak m\subset F[x_1,...,x_n]=:R$ be a maximal ideal of $R$ and let $$\varphi: R\longrightarrow R/\mathfrak m=:k$$ the projection. Since $\mathfrak m$ is maximal, $k$ is a field. Since $\mathfrak m$ is a proper ideal, $1\notin \mathfrak m$, and thus $F\cap \mathfrak m=(0)$.

Q1) $\mathfrak m$ is a proper ideal of $R$, not of $F$, so how can we have $F\cap \mathfrak m=(0)$ since $\mathfrak m$ is even not in $F$ ?

In particular, $\varphi|_F$ is an injection. Let $\bar x_i$ the residue classes of $x_i$ in $k$. Then, using $\varphi|_F$, we have that $k$ is the extension of $k$ by elements $\bar x_1,...,\bar x_n$.

Q2) I don't understand this fact. I agree that $\varphi|_F$ is a field homomorphism, but why (and what does it mean) that $k$ is the extension of $F$ by the elements $\bar x_1,...,\bar x_n$ ?

Since $F$ is algebraically closed, if $\bar x_i$ is algebraic over $F$, then it's contained in $F$. By reordering the variable, we may assume the the first $j$ are not algebraic, that is $k=F[\bar x_1,...,\bar x_j]$ where $\bar x_i$ is tanscendental of $f$ for $i=1,...,j$.

If $j=0$, then $k=F$. Then, let $c_i$ the residue classes of $x_i$ in $k\cong F$. Then, $\mathfrak m=\ker \varphi\supset (x_1-c_1,...,x_n-c_n)$, and since $\mathfrak m$ is maximal, we get $m= (x_1-c_1,...,x_n-c_n)$.

Q3) I don't understand why $m=\ker \varphi\supset (x_1-c_1,...,x_n-c_n)$.

For the rest, it's fine for the moment.

• There seems to be a typo (or mistake) in (2). I think it should say "...ww have that $\;k\;$ is an extension of $\;F\;$ by elements..." Check this, please. – DonAntonio Dec 24 '16 at 15:59
• For Q1, if $\mathfrak{m}$ contained a nonzero element of $c \in F$, it would contain $1$ because we can multiply by $c^{-1}$. If $\mathfrak{m}$ contains $1$, it is all of $R$. – Matthew Leingang Dec 24 '16 at 16:03
• In 1) some kind of abuse of notation is going on. $F$ is included in $F[x_1,\dots,x_n]$ in the obvious way. – Maffred Dec 24 '16 at 16:26
• For 2) informally speaking if we have $x^2 +3y^2$, applying "class" we get $[x^2 +3y^2]\equiv[x]^2 + 3[y]^2$. That's it. – Maffred Dec 24 '16 at 16:29
• For 3) $\phi(x_i-c_i) = \phi(x_i) - \phi(c_i)= c_i - c_i=0$ (I'm abusing of $k \simeq F$) thus $(x_i - c_i) \in \ker \phi= \frak m$. – Maffred Dec 24 '16 at 16:46

we know that any ideal of the form $(x_1-c_1,...,x_n-c_n)$ is maximal, and $F \cong k$ .Let $b_i=x_i(mod m) \in k$ and $a_i=\varphi^{-1}(b_i)$ then $x_i-a_i \in ker\varphi=m$ so $(x_1-c_1,...,x_n-c_n)\subseteq ker\varphi$. and from kronecker theorem: we have that $k$ is extension of $F$. if $F\cap m \ne (0)$ then $m=F$ contradiction.