I'm studying Lovett. "Abstract Algebra." Chapter 12. Multivariable Polynomial Rings. Section 3. The Nullstellensatz.

The book deduces The weak nullstellensatz from proposition 12.3.2. The author doesn't give proof for the weak nullstellensatz.

Proposition 12.3.2 $$\;$$ Let $$F$$ be a field and let $$E$$ be a finitely generated $$F$$-algebra. If $$E$$ is a field then it is a finite extension of $$F$$.

Theorem 12.3.3 (The Weak Nullstellensatz) $$\;$$ Let $$F$$ be a field and let $$R$$ be a finitely generated $$F$$-algebra. Let $$M$$ be a maximal ideal of $$R$$. Then the field $$R/M$$ is a finite extension of $$F$$. In particular, if $$F$$ is algebraically closed, then $$R/M \cong F$$.

I don't understand two parts (Actually there are more that I don't understand but I think I can solve the rest myself once I understand these two parts).

1) Why is $$R$$ a ring? To talk about the ideal in $$R$$, $$R$$ should be a ring. I think the author is assuming that $$(R, +, [,])$$ is a ring in the $$F$$-algebra structure $$(F, +, \times, R, + \cdot, [,])$$, where the first $$+$$ and $$\times$$ are addition and multiplication in $$F$$, the second $$+$$ is an addition in $$R$$, $$\cdot$$ is a scalar multiplication, and $$[,]$$ is an $$F$$-bilinear map in $$R$$. I know that if $$[,]$$ is associative then $$R$$ is a ring. But I can't prove that.

2) Why is $$R/M$$ an $$F$$-algebra? To deduce that $$R/M$$ is a fiinite extension of $$F$$ from proposition 12.3.2, $$R/M$$ should be a finitely generated $$F$$-algebra. I think the author is assuming that $$(F, +, \times, R/M, +, \cdot, [,])$$ is an $$F$$-algebra structure, where for the second $$+$$, $$+: R/M \times R/M \to R/M, (\overline{r_1}, \overline{r_2}) \mapsto \overline{r_1 + r_2},$$ $$\cdot: F \times R/M \to R/M, (f, \overline{r}) \mapsto \overline{fr},$$ and $$[,]:R/M \times R/M \to R/M, (\overline{r_1}, \overline{r_2}) \mapsto \overline{r_1 r_2}.$$ But if $$\cdot$$ is to be well-defined, shouldn't $$M$$ be a submodule of $$R$$? I don't see why $$M$$ is a submodule.

• How is an $F$-algebra defined? – Kenny Lau Dec 31 '18 at 10:07
• @Kenny Let $R$ be a commutative ring. An $R$-algebra is a pair $(M, [,])$ where $M$ is an $R$-module $M$ and where $[,]$ is a $R$-bilinear map $[,]: M \times M \to M$. – zxcv Dec 31 '18 at 10:39
• @Kenny $R$-bilinear map is defined as follows: Let $R$ be a commutative ring and let $M$, $N$, $P$ be three $R$-modules. A function $\varphi: M \times N \to P$ is called bilinear if (1) $\varphi(m_1+m_2,n)=\varphi(m_1,n)+\varphi(m_2,n)$ for all $m_1,m_2\in M$ and all $n\in N$; (2) $\varphi(rm,n)=r\varphi(m,n)$ for all $r\in R$, all $m\in M$, and all $n\in N$; (3) $\varphi(m,n_1+n_2)=\varphi(m,n_1)+\varphi(m,n_2)$ for all $m\in M$ and all $n_1,n_2\in N$; (4) $\varphi(m,rn)=r\varphi(m,n)$ for all $r\in R$, all $m\in M$, and all $n\in N$. – zxcv Dec 31 '18 at 10:39
• Is it how it is defined in Lovett? – Kenny Lau Dec 31 '18 at 10:43
• @Kenny Yes, I copied down the whole definition. – zxcv Dec 31 '18 at 10:45

In thm.12.3.3, $$R$$ should be assumed to be an associative $$F$$-algebra (with identity), hence it is a ring.
In the presence of identity element, every ideal $$M$$ of an $$F$$-algebra $$R$$ is also a submodule (subspace): $$\lambda\cdot m=[(\lambda\cdot 1),m]\in M$$ for any $$m\in M$$ and $$\lambda\in F$$.