In abstract algebra, an ideal is a subset of a ring that is closed under addition, subtraction, and multiplication by elements of the ring. An ideal can be seen as a generalization of a subring, and it plays an important role in the study of algebraic structures. In particular, prime ideals are a special class of ideals that have important applications in number theory, algebraic geometry, and commutative algebra. In this article, we will discuss how to show that a given ideal is prime in a ring.

Definition of Prime Ideal

Let R be a commutative ring with identity, and let I be an ideal of R. We say that I is a prime ideal of R if the following conditions hold:

1. I is not the whole ring R.
2. If a and b are elements of R such that ab is in I, then either a is in I or b is in I.

Intuitively, a prime ideal is a subset of a ring that is “closed” under multiplication and that “detects” factors. In other words, if a product of two elements is in the ideal, then at least one of the factors must be in the ideal.

Examples of Prime Ideals

Let’s consider some examples of prime ideals in various rings:

1. In the ring Z of integers, the ideal (p) generated by a prime number p is a prime ideal. This follows from the fact that Z is a principal ideal domain (PID), which means that every ideal is generated by a single element. If ab is in (p), then p divides ab, and since p is prime, p divides either a or b, which means that a or b is in (p).
2. In the ring F[x] of polynomials over a field F, the ideal (x) generated by the variable x is a prime ideal. This follows from the fact that F[x] is a unique factorization domain (UFD), which means that every polynomial can be factored into irreducible factors in a unique way. If fg is in (x), then the constant term of fg is zero, which means that either f or g has zero constant term. Without loss of generality, we can assume that f has zero constant term. Then f = xh for some polynomial h, and we have fg = xhg. Since F[x] is a UFD, we can factor hg into irreducible factors, and since x is irreducible, it must divide one of the factors, which means that either f or g is in (x).

How to Show that an Ideal is Prime

To show that an ideal I is prime in a ring R, we need to verify the two conditions in the definition of a prime ideal:

1. I is not the whole ring R.
2. If a and b are elements of R such that ab is in I, then either a is in I or b is in I.

The first condition is usually easy to verify, since it is usually clear whether an ideal is proper or not. The second condition is more subtle, and it often requires some algebraic manipulation and cleverness.

Here is a general strategy for showing that an ideal I is prime in a ring R:

1. Assume that ab is in I, where a and b are elements of R.
2. Suppose that neither a nor b is in I.
3. Try to derive a contradiction from this assumption.
4. Use the fact that I is an ideal to manipulate the elements a, b, ab, and I.
5. Use any additional information about R and I to simplify the algebraic expressions.

To show that an ideal is prime in a ring, we need to prove that it satisfies the definition of a prime ideal. The definition of a prime ideal is as follows:

An ideal $P$ in a ring $R$ is prime if for any two elements $a,b\in R$, if their product $ab$ is in $P$, then at least one of $a$ or $b$ is in $P$.

In other words, a prime ideal is an ideal such that if the product of two elements is in the ideal, then at least one of the elements must be in the ideal.

To show that an ideal $P$ is prime in a ring $R$, we need to prove the following two conditions:

$P$ is not the whole ring $R$.
If $ab \in P$ for some $a, b \in R$, then either $a\in P$ or $b\in P$.
We will illustrate how to prove that an ideal is prime in a ring with an example.

Example: Let $R = \mathbb{Z}[x]$ be the ring of polynomials with integer coefficients, and let $P$ be the ideal generated by the polynomial $p(x) = x^2 – 3$. Show that $P$ is a prime ideal.

First, we need to check that $P$ is not the whole ring $R$. This is easy to see, since the constant polynomial $1$ is not in $P$, as $1$ cannot be expressed as a multiple of $x^2 – 3$.

Next, we need to show that if $ab \in P$ for some $a, b \in R$, then either $a \in P$ or $b \in P$. Suppose that $ab \in P$. Then there exist polynomials $f, g \in \mathbb{Z}[x]$ such that $ab = f(x)(x^2-3) + g(x)$. We want to show that either $a \in P$ or $b \in P$.

Suppose that $a$ is not in $P$. Then $a$ can be written as $a = q(x)(x^2 – 3) + r(x)$, where $q(x), r(x) \in \mathbb{Z}[x]$ and either $r(x) = 0$ or $\deg(r(x)) < \deg(x^2 – 3)$. Similarly, we can write $b = s(x)(x^2 – 3) + t(x)$, where $s(x), t(x) \in \mathbb{Z}[x]$ and either $t(x) = 0$ or $\deg(t(x)) < \deg(x^2 – 3)$.

Multiplying $a$ and $b$ together, we get
\begin{align*}
ab &= (q(x)(x^2 – 3) + r(x))(s(x)(x^2 – 3) + t(x))\
&= q(x)s(x)(x^2 – 3)^2 + (q(x)t(x) + r(x)s(x))(x^2 – 3) + r(x)t(x).
\end{align*}

Since $ab \in P$, we have $ab = h(x)(x^2 – 3)$ for some $h(x) \in \mathbb{Z}[x]$. Comparing the coefficients of $x^3$ on both sides of the equation, we see that $q(x)s(x) = 0$. 

Another important property of prime ideals is that they are always maximal ideals in a PID. To prove this, suppose that $P$ is a prime ideal in a PID $R$, and let $M$ be an ideal such that $P\subseteq M\subseteq R$. We want to show that either $M=P$ or $M=R$. Since $R$ is a PID, we can write $M=(m)$ for some $m\in R$. Since $P\subseteq (m)$ and $P$ is prime, we must have either $P=(m)$ or $P\subseteq (m)$.

If $P=(m)$, then $M=(m)=P$, and we are done. Otherwise, we have $P\subseteq (m)$. Since $P$ is prime and $R$ is a PID, $P$ is a maximal ideal in $R$. Therefore, either $(m)=P$ or $(m)=R$. If $(m)=P$, then $M=(m)=P$ and we are done. If $(m)=R$, then $M=(m)=R$, and we are done.

This shows that any prime ideal in a PID is maximal. However, the converse is not true in general: not every maximal ideal in a PID is prime. For example, in the PID $\mathbb{Z}$, the ideal $(p)$ generated by a prime number $p$ is maximal but not prime, since $\mathbb{Z}/(p)$ is a field and hence an integral domain, but $(p)$ is not a prime ideal of $\mathbb{Z}$.

In conclusion, determining whether a given ideal is prime in a ring involves showing that the quotient ring by the ideal is an integral domain. This can be done using various techniques such as direct computation or checking properties of elements in the quotient ring. Prime ideals also have the property of being maximal in a PID, but not all maximal ideals in a PID are prime.