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Chapter 5: Q11. (page 134)

Show that the ring in Exercise 8 is not a field.

Short Answer

Expert verified

There is no inverse for element [x].

Step by step solution

Determine function ring is not field

Consider the ring in that there is no inverse for the element [x]. Now assume $[a_{0} + a_{1} x]$ is an inverse. That means

$[x] \cdot [a_{0} + a_{1} x] = [a_{0} x] = [1]$

It is a contradiction as there is no $a_{0} \in ℚ$ , so that $x^{2}| a_{0} x - 1$ .

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Most popular questions from this chapter

Let p (x)be irreducible in F [X]. Without using Theorem 5.10, prove that if $[f (x)] [g (x)] = [0_{F}]$ in F [x]/(p (x)) then $[f (x)] = [0_{F}] o r [g (x)] = [0_{F}]$ . [Hint: Exercise 10 in section 5.1.]

a) Show that $\frac{ℤ_{2} [x]}{(x^{3} + x + 1)}$ is a field.

b) Show that the field $\frac{ℤ_{2} [x]}{(x^{3} + x + 1)}$ contains all three roots of $x^{3} + x + 1$ .

Question: In Exercises 5-8, each element of the given congruence-class ring can be written in the form $[a x + b]$ (Why?). Determine the rules for addition and multiplication

of congruence classes. (In other words, if the product role="math" localid="1649064856064" $[ax + b] [cx + d]$ is the class $[r x + s]$ , describe how to find r and s from a,b,c,d and similarly for addition.) $[x] / [x^{2} + 1]$

If $p (x)$ is reducible in $F [x]$ , prove that there exist $f (x) g (x) \in F [x]$ such that $f (x) \equiv 0_{F} (mod p (x))$ and $g (x) \equiv 0_{F} (mod p (x))$ but $f (x) g (x) \equiv 0_{F} (mod p (x))$ .

Find a fourth-degree polynomial in $ℤ_{2} [x]$ whose roots are the four elements of the field. $ℤ_{2} / (x^{2} + x + 1)$ , whose tables are given in Example 3. [Hint: The Factor Theorem may be helpful.]

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Show that the ring in Exercise 8 is not a field.

Short Answer

Step by step solution

Determine function ring is not field

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