Definition of an ideal in a L-language

Question

Let $\mathcal{L}_\text{ring}=\{0,1,+, \cdot, I\}$ where $0,1$ are constants, $+, \cdot$ are binary function symbols and $I$ is an unary relation symbol.

Give $\mathcal{L}$-formulas which express that the subset defined by $I$ is:

a) an ideal

b) a prime ideal

c) a maximal ideal

I would write down the following for a): $$I(0) \wedge \forall x \forall y(I(x) \wedge I(y) \rightarrow I(x+y)) \wedge \forall x \forall y(I(y) \rightarrow I(x\cdot y)\wedge I(y\cdot x))$$ Would this be right?

b) Can I just write? $$\forall x\forall y I(x\cdot y) \rightarrow I(x)\vee I(y)$$

c) The assistant-professor wrote: $$\forall x[\neg I(x) \rightarrow \forall y\exists i\exists u (y=i+(u\cdot x) \wedge I(i))]$$ Could someone explain this further, I don't really follow...

Answer 1

(a) Looks right to me. The solution to (c) looks like you're supposed to be considering commutative rings, in which case you'd only need one of $I(x\cdot y)$ and $I(y\cdot x)$.

(b) Looks right too, assuming you already know $I$ is an ideal -- except that the whole ring does not usually count as a prime ideal. So you should add as an additional condition that $\neg I(1)$.

(c) What we really want to do is to claim "there is no ideal that is a proper superset of $I$ but is not the entire ring". But we cannot do that directly here, because the language doesn't allow us to quantify over ideals. Instead the solution uses the trick of considering the smallest ideal that contains both $I$ and some element $x\notin I$. This ideal is $$ I+\langle x\rangle = \{ i + ux \mid i\in I, u\in R \} $$ and the formula claim that if $x$ is not in $I$ this is the entire ring. It should be clear that this is true if $I$ is actually maximal (just show that the above set is always an ideal). On the other hand, if $I$ were not maximal, then there would be an ideal $J$ with $I\subsetneq J\subsetneq R$. In that case we can choose $x\in J\setminus I$, and then it we can see that $i+ux$ must be in $J$ for every $i\in I, u\in R$. Thus, $I+\langle x\rangle$ is not all of $R$.

Instead of quantifying over all $y$ it would be sufficient to test for $1$ being in $I+\langle x\rangle$, that is $$ \forall x[\neg I(x)\to \exists i\exists u(1=i+ux \land I(i))] $$

Answer 2

Your answer for (a) is fine. Your answer for (b) is almost right: you also need to specify that the ideal isn’t the whole ring, so you want

$$\forall x\,\forall y\big(I(x\cdot y)\to I(x)\lor I(y)\big)\land \exists x\big(\neg I(x)\big)\;.$$

Now let’s take a look at (c); the question is why

$$\forall x\Big(\neg I(x)\to\forall y\,\exists i\,\exists u\big(y=i+(u\cdot x)\land I(i)\big)\Big)\tag{1}$$

expresses the fact that the subset defined by $I$ is a maximal ideal.

Let $R$ be a commutative ring with $1$and let $I$ be an ideal in $R$. Suppose that $I$ is maximal. Then by definition if $J$ is an ideal of $R$ such that $I\subseteq J\subseteq R$, then either $J=I$ or $J=R$. Suppose that $x\in R\setminus I$. Then $I+Rx$ is an ideal of $R$, and $I\subseteq I+Rx$, since $0\in Rx$. But $$x=0+1\cdot x\in I+Rx$$ as well, so $I+Rx\ne I$, and therefore $I+Rx=R$. Thus, for each $x\in R\setminus I$ there are an $i\in I$ and a $u\in R$ such that $x=i+(u\cdot x)$. Thus, $(1)$ must hold if the set defined by $I$ is to be a maximal ideal.

If $(1)$ does hold, it’s not hard to check that the set defined by $I$ is either a maximal ideal or all of the ring. Call that set $I$ and the ring $R$; $(1)$ says that $I+Rx=R$ for all $x\in R\setminus I$. If $I$ is not maximal, let $J$ be an ideal such that $I\subsetneqq J\subsetneqq R$, and let $x\in J\setminus I$. Then $I+Rx\subseteq J$, so $I+Rx\ne R$. The only problem is that $(1)$ doesn’t exclude the possibility that $I=R$, so we need to replace it with

$$\forall x\Big(\neg I(x)\to\forall y\,\exists i\,\exists u\big(y=i+(u\cdot x)\land I(i)\big)\Big)\land\exists x\big(\neg I(x)\big)\;.$$