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Question

Let F[X] be the polynomial ring in one variable over a field F. Then which of the following statements are true?

F[X] Polynomial Ring over a Field

Let $F$ be a field. The symbol $F[X]$ represents the set of all polynomials in a single variable $X$ with coefficients taken from the field $F$. For example, if $F$ is the field of real numbers $\mathbb{R}$, then $\mathbb{R}[X]$ includes polynomials like $2X^3 - 5X + 1$, $X^2 + \sqrt{2}$, etc.

$F[X]$ with the usual polynomial addition and multiplication forms a ring. Since $F$ is a field, it is an integral domain, and this property carries over to $F[X]$, meaning $F[X]$ is also an integral domain.

Euclidean Domain Property of F[X]

A Euclidean domain is an integral domain $R$ equipped with a Euclidean function $v: R \setminus \{0\} \to \mathbb{N} \cup \{0\}$ such that for any two elements $a, b$ in $R$ with $b \neq 0$, there exist elements $q, r$ in $R$ such that $a = bq + r$, where either $r=0$ or $v(r) < v(b)$. This is essentially the division algorithm.

For the polynomial ring $F[X]$, we can use the degree of a polynomial as the Euclidean function. The degree of a non-zero polynomial $f(X)$, denoted by $\text{deg}(f(X))$, is the highest power of $X$ with a non-zero coefficient.

The polynomial division algorithm states that for any two polynomials $f(X)$ and $g(X)$ in $F[X]$ with $g(X) \neq 0$, there exist unique polynomials $q(X)$ (quotient) and $r(X)$ (remainder) in $F[X]$ such that:

$\qquad f(X) = q(X)g(X) + r(X)$

where either $r(X) = 0$ or $\text{deg}(r(X)) < \text{deg}(g(X))$.

This property exactly matches the definition of a Euclidean domain, with the degree function $\text{deg}(f(X))$ serving as the Euclidean function $v$. Thus, $F[X]$ is a Euclidean domain.

PID and UFD Properties of F[X]

There is a well-established hierarchy among different types of integral domains:

Euclidean Domain $\implies$ Principal Ideal Domain (PID) $\implies$ Unique Factorization Domain (UFD)

  • A Principal Ideal Domain (PID) is an integral domain where every ideal is principal, meaning every ideal can be generated by a single element.
  • A Unique Factorization Domain (UFD) is an integral domain where every non-zero, non-unit element can be written as a product of prime (or irreducible) elements, uniquely up to associates and order.

Since we have established that $F[X]$ is a Euclidean domain, it automatically follows from the hierarchy that $F[X]$ is also a Principal Ideal Domain (PID) and a Unique Factorization Domain (UFD).

Analyzing the Statements

Let's evaluate each statement based on our findings:

  • Statement 1: F[X] is a UFD. This is true because $F[X]$ is a Euclidean domain, and every Euclidean domain is a UFD.
  • Statement 2: F[X] is a PID. This is true because $F[X]$ is a Euclidean domain, and every Euclidean domain is a PID.
  • Statement 3: F[X] is a Euclidean domain. This is true as shown by the existence of the division algorithm using the degree function.
  • Statement 4: F[X] is a PID but is not an Euclidean domain. This is false. As we showed, $F[X]$ is indeed a Euclidean domain. A domain can be a PID but not a Euclidean domain (e.g., $\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$), but $F[X]$ is not such an example.

Therefore, the statements that are true are: F[X] is a UFD, F[X] is a PID, and F[X] is a Euclidean domain.

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Important Questions from Rings & Ideals

  1. If the ring R is a commutative ring with unity, then the polynomial ring R[X] is-

  2. Let R = (Z2 × Z2, +,.) forms a ring of module 2 such that (a, b) + (c, d) = (a + c, d + d) and (a, b) (c. d) = (a.c, b.d) for (a, b), (c, d) ∈ Z2 × Z2 then-

  3. The set of all units in a ring R with unity forms ______.

  4. Let C[0, 1] be the ring of all real valued continuous function on [0, 1].

    Let A = {f ∈ C[0, 1] ∶ \(f\left( \frac{1}{4}\right)=f\left( \frac{3}{4}\right)\) = 0}. Then which of the following statements are true? 

  5. Which of the following statements is NOT true?

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