If a field K is algebraically closed, then every polymomiel with co- efficients in k has a root in K. © True. This is an equiral ent way of statling the defiaition of in algebiuically closed field. O True. If K is algebraically closed, theen every equation hos a solution in K, so eve ry polymomiel has a root in K. O False. This only applies to non-consbant poly nomicls. A nonzeu constant polynomiel such as 1 does not hove a root in K. O False. For example, the polynomisl T k-W +1 does not dEK have a root in K,

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.3: The Field Of Quotients Of An Integral Domain
Problem 10E: Since this section presents a method for constructing a field of quotients for an arbitrary integral...
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Which option is correct. THERE IS ONLY 1 CORRECT option. a,b,c or d!

If a field K is algebraically closed, then every polymomiel with co-
efficients in k has a root in K.
© True. This is an equiral ent way of stating the defiaition of
in algebrically closed field,
True. If K is algebraically closed, then every equation has a
solution in K, So every polynomiel haS a root in K.
False. This only applies to non-consbant poly nomiuls. A nonzero
constant polynomiel such as 1 does 2ot hove a root in K.
False, For example, the polymomiel TT k-w) +1 does not
have a root in K,
Transcribed Image Text:If a field K is algebraically closed, then every polymomiel with co- efficients in k has a root in K. © True. This is an equiral ent way of stating the defiaition of in algebrically closed field, True. If K is algebraically closed, then every equation has a solution in K, So every polynomiel haS a root in K. False. This only applies to non-consbant poly nomiuls. A nonzero constant polynomiel such as 1 does 2ot hove a root in K. False, For example, the polymomiel TT k-w) +1 does not have a root in K,
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